Phyllotaxis
the spirals are counted, not admired
The collection

Every essay

One idea per essay, ordered so that the earlier ones set up the later ones — but nothing here depends on being read in sequence. Page 1 of 2.

Branching and transport

A network built to move fluid for the least work obeys a cube law at every junction. That is checkable on a real tree, though not from the angles: the same cost fixes those too, and the rival rule turns out to predict every one of them.

A branching tree in which every junction obeys the cube law. r₀³ = r₁³ + r₂³ at all 63 junctions, to 2e-16. The widths in the drawing are the radii the law gives, not widths chosen to look right.

The cube law

A branching network built to move fluid for the least work obeys one relation at every junction — the cube of the parent radius equals the sum of the cubes of the daughters. It is a minimisation result, it is checkable on a real tree, and it is the rare biological rule with a derivation.

6 figures The exponent fitted from the junctions, rather than assumed. Sweeping k and asking where r₀ᵏ = Σ rᵢᵏ holds best gives 3.000 — Murray's 3, recovered rather than imposed.

Fitting the exponent

Assuming the exponent is three and reporting the error says how far the data is from that assumption. Fitting the exponent and reporting what it comes out as says what the network is doing — and an estimator has to be shown returning something other than three, or it is not a fit.

6 figures An L-system after 4 rewrites of two rules. X → F[+X]F[-X]+X and F → FF, walked by a turtle turning 22.5°. 130 segments, and not one of them knows anything about light, water or auxin.

L-systems describe, they do not explain

Two rewriting rules and a turtle produce something indistinguishable from a plant, and there is no plant in it — no light, no water, no auxin, no mechanics. That is worth demonstrating precisely because the output is so convincing.

6 figures The uninformative sample does not say so — it says something else. Above: how many junctions of a given asymmetry it takes to distinguish an exponent of 3 from an exponent of 2, at 2% measurement error on each radius. An even fork needs one; a junction whose small daughter is a twentieth of the large one needs 2195. Below: 50 junctions from each end of the range, on a synthetic tree built at exactly 3. The even forks return 3.01; the twigs return 1.69, with an interval no wider — because 47% of them measure as a parent thinner than its own larger daughter, and dropping those keeps only the half where the noise ran the right way.

Which junctions say anything

Da Vinci's rule and Murray's law differ by 12% at an even fork and by a tenth of a per cent at a twig. So one even fork settles which is right, and two thousand twigs do not — a factor of two thousand across a tree, decided entirely by the shape of the junction and not by how carefully it is measured.

8 figures The uninformative sample does not say so — it says something else. Above: how many junctions of a given asymmetry it takes to distinguish an exponent of 3 from an exponent of 2, at 2% measurement error on each radius. An even fork needs one; a junction whose small daughter is a twentieth of the large one needs 2195. Below: 50 junctions from each end of the range, on a synthetic tree built at exactly 3. The even forks return 3.01; the twigs return 1.69, with an interval no wider — because 47% of them measure as a parent thinner than its own larger daughter, and dropping those keeps only the half where the noise ran the right way.

A sample that is confidently wrong

Fifty lopsided junctions from a tree built at an exponent of exactly 3 return 1.7, with an interval that excludes 3 and excludes 2 as well. The sample carrying almost no information does not give a wide answer — it gives a narrow wrong one, and the cause is a selection nobody applies on purpose.

8 figures A tree built at an exponent of 3, measured band by band. Six bands of daughter ratio, 300 junctions in each, all from trees built at an exponent of exactly 3 with the same 2% measurement error. The median implied exponent falls from 3.00 at an even fork to 1.58 at a twig, and the share of junctions that have any exponent at all falls from 100% to 53% over the same range. The rule across the top is a single fit over 200 junctions spanning the whole range: 2.85. A mixed sample is safe because least squares already weights by leverage — 42% of it sits in the most symmetric band and 0.06% in the twigs. The sample that is not safe is the one a person can reach.

The band decides the answer

A fit over a whole tree's junctions returns the exponent the tree was built at, even though most of its junctions are from bands that on their own return 1.6. Least squares is already weighting by leverage. The dangerous sample is not the mixed one — it is the one a person can reach.

8 figures A tree built at 3, measured to 2%, reads 2.957 on the informative band. The exponent recovered from 100 junctions of a tree built at exactly 3, against the relative error placed independently on the parent and on both daughters — half a per cent is a machined section under a microscope, one to two per cent is callipers on a clean branch, five is a branch that is not round, ten is a radius read off a photograph. Each line is a band of daughter ratio and the whiskers are the central 90% of 300 replicate samples. Every band is displaced downward at every error and never upward: at 2% the informative band read 2.957 and the informative band read 2.957. Below, the same rows with the displacement and the spread drawn as separate bars, because only one of them falls when more junctions are measured.

The exponent an error moves

Every real measurement of a branch radius carries error and no synthetic tree does, so the question is what a symmetric error does to a fitted exponent. It does two things — a bias and a spread — and the bias runs downward at every error level and in every band, by an amount derivable from the daughter ratios alone.

7 figures One junction's bias moves 1.36-fold across the range its leverage moves 308,352-fold, at an exponent of 3. The bias a single junction of daughter ratio γ contributes to a least-squares fit, as a multiple of the squared measurement error, drawn against the leverage that junction carries — both computed from the expansion about a true exponent of 3 rather than fitted to anything. Across the whole range from an even fork to a twentieth the bias moves by a factor of 1.36 and the leverage by a factor of 308,352, and the uninformative junction contributes the larger share: 6.00σ² at γ = 0.05 against 4.50σ² at an even fork.

The fragile junctions are the informative ones

That is the obvious worry once the radii are uncertain, and it is false. Across the whole range of asymmetry a junction's contribution to the bias moves by a factor of 1.36 while its leverage moves by a factor of 308,352, so the junction that says nothing damages the answer as badly as the one that says everything — and a sample is spoiled by counting rather than by weight.

7 figures Between 3% and 5% of radius error, no sample size answers — 50 junctions among them. One row per error level. The pale bar is the sample sizes whose interval is narrow enough to state a claim from — half-width under ±0.25 and excluding 2 — and it starts where precision arrives. The second bar is the sample sizes whose interval still contains the 3 the tree was built at, and it ends where the displacement overtakes the width. Where the two overlap there is a usable window; at 5%, 7%, 10% they do not overlap at all, so below 50 junctions the answer is too wide to state and above 30 it no longer contains the truth.

The window that closes

The spread of a fitted branching exponent falls as the reciprocal root of the sample and its displacement does not fall at all, so there is a count past which every further junction buys confidence and no accuracy. Between three and five per cent of radius error the count arrives before the answer does, and no sample size both states a claim and contains the truth.

7 figures Murray's law and Da Vinci's become one measurement at 12% of radius error on the informative band. Two trees measured the same way: 50 junctions of daughter ratio 0.6–1, 300 replicate samples at each of 15 error levels, one tree built at exactly 3 and one at exactly 2. Each shaded band is the central 90% of the recovered exponents. Both run downward, but the tree at 3 runs down faster — -113σ² against -29σ² — so the two close on each other. At 11% they are still apart; at 12% the bands overlap and one study's answer could have come from either tree; at 20.5% the means cross, and above it a tree built at 3 measures lower than a tree built at 2.

Where three and two become one

Two trees, one built to obey Murray's law and one to obey Da Vinci's, are measured through the same fifty junctions with the same instrument. At twelve per cent of error on each radius the two answers overlap, and above twenty and a half the tree built at three measures lower than the tree built at two.

8 figures The fork the cost chooses at a daughter ratio of 1: 37.47 and 37.47 degrees. Three ends held fixed, three weights fixed by the radii, and the branch point put where the total cost is least. At the optimal radius the pumping term is exactly half the upkeep term, so a segment's weight is its own cross-section and the three weights here are 1.5874, 1.0000, 1.0000. Minimising directly over the position — a 41×41 grid re-centred and shrunk 220 times, told nothing about any formula — puts the daughters at 37.4673° and 37.4673° from the parent's own forward direction, against the closed form's 37.4673° and 37.4673°.

The angle the cost chooses

Murray's exponent falls out of minimising a cost over the radius of a tube. The same cost minimised over the position of the branch point instead fixes both fork angles, and 281 networks minimised on a grid told nothing about any formula agree with the closed form to under two ten-thousandths of a degree.

7 figures The three weights at an exponent of 2 and 3: one closes a triangle and one is a straight line. A branch point minimising a weighted sum of three lengths has an interior solution only when the three weights close a triangle, and the weight on a segment here is its own cross-section. At an exponent of 3 the three areas clear that condition by 0.4126, and the triangle they close is what the two fork angles are read off. At an exponent of 2 the parent's area is exactly the daughters' areas summed — that is what area conservation says — so the slack is -4.44e-16, the triangle collapses onto a line, and the fork closes to 0.0000 degrees. Leonardo's rule does not predict a different angle here; it predicts no angle.

A rule that predicts everything

Leonardo's rule says a fork conserves cross-section, and cross-section is exactly the weight the branch point is minimised against. So the three weights land on the boundary of the triangle inequality, the cosine comes out at one to the last bit, and the rule predicts no angle at all — and the free constant its own derivation leaves behind then walks the prediction across every angle a fork could have.

7 figures What insisting on a fork angle costs: 43.3 degrees for one per cent of the network. The three ends are held where the optimum wants them, the branch point is moved everywhere inside them, and the cheapest network at each total angle is kept. The optimum sits at 74.93°. Everything within one per cent of the least cost runs 55.9° to 99.2° — a span of 43.3°, which read back as exponents covers 2.44 to 5.34 — and within a tenth of a per cent it still runs 13.6°. The prediction is steep in the exponent and the cost is nearly flat in the angle; they are the same curve read along its two axes.

An optimum too flat to reach

One per cent of a branching network's cost buys forty-three degrees of fork angle, covering exponents from 2.44 to 5.34, while the angle the theory predicts moves only fourteen and a half degrees across every daughter ratio there is. The prediction is steep and the cost is flat, and those are the same curve read along its two axes.

7 figures The two trees this site draws, at 30° and 32° to a side, against the cost's 37.47° and 37.47°. Two trees of 63 segments each, 5 generations deep and 31 junctions apiece, with every junction's radii taken from r₀³ = r₁³ + r₂³ exactly and every junction's angle taken from a constant. Read as an exponent through cos(θ/2) = 2^(2/p − 1), the drawn angles say 2.5237 and 2.6239, in pictures whose widths are built at exactly 3. The cost that fixed those widths wants 37.47° and 37.47° at this daughter ratio, 74.93° in total, and the misses cost 0.573% and 0.292% of the network — which is why a fixed angle can sit in a figure about a minimisation and never look wrong.

The trees drawn at no angle

Two branching figures in these essays set every junction's radii from the cube law exactly and every junction's angle from a constant nobody derived. Read as exponents the drawn angles say 2.52 and 2.62, in pictures whose widths say exactly three — and at a lopsided fork the drawing puts a daughter thirty-four degrees from where the same cost puts it.

7 figures What a swelling does to the exponent read from the informative band, by where the swelling is. 50 junctions of daughter ratio 0.6–1, built at exactly 3 and read with no random error, but with one or more radii measured fat. A parent read fat lowers the reading: 1% gives 2.864, 3% gives 2.631, 10% gives 2.075, and at 11.3% the tree reads Da Vinci's 2. Daughters read fat raise it: 1% gives 3.150, 3% gives 3.499; from 8% some junctions have a daughter measured wider than their parent, which no exponent balances, and the line stops. All three radii read fat by one factor return 3.000 at every swelling — the unswollen reading exactly.

A swelling at the fork

A branch thickens where it forks, so a parent measured just below a junction and daughters measured just above it carry three different amounts of the same swelling. A swelling that fattens all three alike moves a fitted exponent by exactly nothing. A parent read one per cent fat moves it by as much as 3.6 per cent of random error on every radius, in a sign known in advance, and a tenth of a radius turns a tree built at Murray's three into one that reads Da Vinci's two with no noise at all. Added to the noise, it does not bring the two rules together any sooner: the two errors do not add.

8 figures Adding error to carry a fitted exponent back to none, at 12%. At 12% of error on every radius each replicate is refitted with more error added at four levels, and the dots are the means over 300 replicates, the tree at 3 above and the tree at 2 below. Uncorrected they read 2.035 and 1.676. Carried back to no error through the mean points, the tree at 3 reads 2.695 by a quadratic curve, 2.380 by a linear curve, 2.914 by a rational curve; the tree at 2 reads 1.961 by a quadratic one, 1.877 by a linear one, 1.982 by a rational one. The curve carried back is a choice the method does not make.

A correction that keeps the overlap

The duel between a tree built at Murray's exponent and one built at Da Vinci's ended by saying the displacement is the geometry, and that no better estimator removes it. Correcting every replicate by simulation-extrapolation removes 92 per cent of the tree at three's displacement at five per cent of error and 68 per cent at twelve, and the error at which the two means cross leaves the measured range altogether. It pays in spread — the corrected readings are twice as wide at twelve per cent — so the error at which the two trees' intervals overlap does not move. Of the duel's two numbers, the inversion was the estimator's and the overlap is the question's.

9 figures Measured radii against the tips each branch carries, at 12% of error on every radius. One replicate of a 50-junction tree, 101 segments, every radius measured with 12% of relative error and drawn against the number of tips that segment carries. The count has no error in it, so the slope is not displaced: the tree built at 3 gives a slope of 0.3404, an exponent of 2.937, and the tree built at 2 a slope of 0.5071, an exponent of 1.972. On the same measured radii the junction-by-junction fit reads 2.218 and 1.736.

A count carries no error

Fitting r₀ᵏ = Σrᵢᵏ junction by junction puts a measured radius on both sides of every equation, and at twelve per cent of error a tree built at Murray's three and one built at Da Vinci's two stop being told apart, however the fit is corrected. Fit the same measured radii against the number of tips each branch carries instead — a count, which nobody measures with error — and the two trees read 2.996 and 1.997 at twelve per cent and 3.015 and 2.001 at thirty, never overlapping. The twelve per cent belonged to the junction fit, not to the tree. The count fails in its own way, and the way is stated.

8 figures One tree sized for flow and for stress, with each branch 2^(−1/2) the length of its parent. The same symmetric tree, 8 generations deep, each generation's branches 2^(−1/2) the length of the one before and turned 30° at every fork, sized two ways and drawn to one trunk width. On the left each branch's radius cubed is proportional to the tips it feeds — Murray's flow rule — and every junction conserves r³. On the right each branch is sized so that the same load on every tip bends it to the same stress at its base, radius cubed proportional to the sum of its lever arms to its tips; its trunk junction conserves r to the power 1.967, its outermost junctions 1.349, against a deep-tree limit of 2.000. The two trees thin at different rates from the same trunk.

A cube law with a lever arm

Murray's exponent of three comes from moving fluid for the least work, and Da Vinci's two has had no derivation here, only the name of the mechanical answer. Size every branch so that the same wind on every tip bends it to the same stress, and a junction conserves r to the power 3/(1 + log₂(1/λ)), where λ is how much shorter each branch is than its parent. A crown that fills a plane gives exactly two; halving lengths gives one and a half; no shortening gives three. Murray's flow rule gives three at every λ, so the lengths of a tree's branches say which mechanism sized it.

8 figures The fork angle Da Vinci's rule predicts, against the size of the fork, in one tree. A tree has one value of the constant Da Vinci's rule leaves free, so a fork's share of it falls as the square of the fork's size and the angle the rule predicts changes with size. For even forks it is 119.1° at a relative size of 0.1, 101.7° at a relative size of 0.5, 74.9° at a relative size of 1, 44.6° at a relative size of 2, 9.6° at a relative size of 10, where Murray's rule gives 74.93° at every size; for daughter ratio 0.5 it is 118.9° at a relative size of 0.1, 101.3° at a relative size of 0.5, 77.6° at a relative size of 1, 48.9° at a relative size of 2, 10.9° at a relative size of 10, where Murray's rule gives 77.58° at every size. From a total of 100° to 20° at an even fork is a factor of 8.93 in radius.

One constant for every fork

Da Vinci's rule leaves a free constant in the cost that sets a fork's angle, and running it over its range walks the predicted angle from nothing to 120 degrees, through Murray's 74.93. So no single fork can refute the rule. But the constant is one number for a whole tree, and a fork's share of it falls as the square of the fork's size — the constant is a radius axis. A tree spanning a factor of ten in radius must show forks from 29.4 degrees at its biggest to 111.6 at its smallest, a spread wider than one fork's flatness can hide, while Murray's angle is the same at every size.

8 figures The exponent read from a tree that has lost tips, tips lost one at a time. The 50-junction tree built at 3 and at 2, 12% of error on every radius, 300 replicates of loss and measurement at each level. Shaded: the central 90% of the exponent read against the tips still there. Dotted: the mean read against every tip ever grown. Solid: the junction fit's mean over the junctions that survive. The count's two intervals are still apart with 70% of the tips gone, reading 2.366 and 1.576, and overlap by 80%. The junction fit's intervals overlap by 20%. At the heaviest loss drawn, 90%, the scars still read 2.998 and 1.997.

A count that has lost tips

Radii read against the tips each branch carries keep Murray's three apart from Da Vinci's two where junction fits cannot, because a count has no measurement error in it. A count of the tips a tree has is not a count of the tips it grew. Losing them lowers both trees' readings by one factor that belongs to the losses and not to the rule, so the count stops being right long before it stops telling the trees apart: on fifty junctions at twelve per cent of error, to seventy per cent of the tips lost one at a time, and only to about a quarter lost in whole limbs. Counting scars repairs single losses exactly. Nothing countable repairs a shed limb.

7 figures The fork angle the transport cost predicts on a tree sized by stress, against the size of the fork. A tree sized so that equal loads on its tips bend every branch to one stress conserves rᵖ with p set by how much shorter each branch is than its parent, λ. On such a tree the cost that fixes a fork's angle carries one constant, entering a fork of size s as s^(2p − 6), so the predicted angle changes with size unless p is three. With the constant set so that a fork of relative size one opens at Murray's 74.9°: At λ = 0.707 (p = 2.000) an even fork opens at 119.1°, 111.6°, 74.9°, 29.4°, 9.6° at relative sizes of 0.1, 0.32, 1, 3.2 and 10; at λ = 0.74 (p = 2.091) an even fork opens at 114.5°, 106.4°, 74.9°, 41.1°, 30.1° at relative sizes of 0.1, 0.32, 1, 3.2 and 10; at λ = 0.794 (p = 2.250) an even fork opens at 106.3°, 97.6°, 74.9°, 53.9°, 46.4° at relative sizes of 0.1, 0.32, 1, 3.2 and 10; at λ = 0.87 (p = 2.498) an even fork opens at 92.7°, 85.5°, 74.9°, 66.0°, 61.5° at relative sizes of 0.1, 0.32, 1, 3.2 and 10; at λ = 0.95 (p = 2.793) an even fork opens at 78.4°, 76.6°, 74.9°, 73.4°, 72.2° at relative sizes of 0.1, 0.32, 1, 3.2 and 10. Murray's rule opens every fork at 74.9°.

Forks on a tree sized by stress

Da Vinci's rule predicts that a tree's forks open wider as they get smaller, because the constant in its angle cost is one number for a tree and enters each fork scaled by its size. A tree sized for equal bending stress conserves an exponent set by how much shorter each branch is than its parent, and on such a tree the same constant enters each fork as its radius to the power 2p − 6. The trend survives at every length ratio short of one and shrinks with it: 105 degrees a decade of radius for a crown filling a plane, 83 at a length ratio of 0.74, 53 for a crown filling a volume, 13 at 0.9. Only a crown shortening about as fast as a volume-filling one fans wider across a tenfold range than one fork's flatness can hide.

7 figures One planar crown sized for a load on its tips and for the weight of its own wood. The same symmetric crown, 9 generations deep, each branch 2^(−1/2) the length of its parent and turned 30° at every fork, sized so that every branch is bent to one stress, drawn to one trunk width. On the left the load is on the tips and the trunk junction conserves r to the power 1.980; on the right the load is the weight of the wood, found by iterating the radii until they stop moving, and the trunk junction conserves r to the power 0.969. A crown sized for its own weight thins much faster from the trunk, because a branch's weight grows with the square of its radius.

A crown that carries its own wood

Sizing every branch so that equal loads on the tips bend it to one stress gives a crown filling a plane Da Vinci's exponent of two. Move the load onto the wood and the sizing becomes a fixed point, because a branch's load now depends on the radii being solved for. Under the wind on its wood a planar crown still conserves two, but only as a limit its trunk is two tenths short of at fifteen generations. Under its own weight it conserves one — radius rather than area, the stress-similarity law that radius goes as length squared — and a crown carrying leaves and wood reads the leaves' two near its twigs and the wood's one at its trunk, with the handover set by how much of the trunk's load the wood carries.

7 figures The number of growing points after each season, for buds that wait no season, one, two, three or four. From one mature apex, each season every mature apex makes a new bud, and a bud branches only after it has waited its delay. With no delay the counts run 1, 2, 4, 8, 16, 32, 64, 128, 256, 512 and settle into growing by 2.0000 a season; with one season the counts run 1, 2, 3, 5, 8, 13, 21, 34, 55, 89 and settle into growing by 1.6180 a season; with two seasons the counts run 1, 2, 3, 4, 6, 9, 13, 19, 28, 41 and settle into growing by 1.4656 a season; with three seasons the counts run 1, 2, 3, 4, 5, 7, 10, 14, 19, 26 and settle into growing by 1.3803 a season; with four seasons the counts run 1, 2, 3, 4, 5, 6, 8, 11, 15, 20 and settle into growing by 1.3247 a season. On a logarithmic axis each settles into a straight line whose slope is its growth rate, the positive root of x^(d+1) = x^d + 1.

A count set by a delay

An L-system describes a plant and forbids nothing, because none of its parameters is anything a plant has. One branching grammar is the exception: a mature apex makes a new bud every season, and a bud waits d seasons before it branches. Its counts grow at the root of x^(d+1) = x^d + 1, a delay of one season gives Fibonacci's numbers and nothing else does, and the fourth count already separates a one-season wait from every longer one. So a Fibonacci count in a branching plant is a measurement of how long its buds wait. It is also a fragile one: if one bud in ten waits two seasons instead, eleven counts in a row come out Fibonacci's three times in a thousand.

7 figures The rate a branching count grows at, against the chance a growing point dies. Every point dies with probability q each season and the survivors rewrite as before, so the expected counts obey x^(d+1) = (1 − q)·x^d + (1 − q)^(d+1). Substituting x = (1 − q)y returns the deathless equation exactly, which makes every line here straight: the rate is the deathless root multiplied by the survival. No delay runs from 2.0000 to one at q = 0.5000; one season runs from 1.6180 to one at q = 0.3820; two seasons runs from 1.4656 to one at q = 0.3177; three seasons runs from 1.3803 to one at q = 0.2755; four seasons runs from 1.3247 to one at q = 0.2451. Below the marked line a lineage shrinks.

A count that loses its growing points

The branching grammar behind the Fibonacci claim has no deaths in it, and a stem that loses shoots is the common case. Giving every growing point a chance q of dying each season leaves the counts a linear recurrence and does exactly one thing to it: the growth rate becomes the deathless root multiplied by 1 − q, at every delay and every death chance, to the last bit a double holds. So each waiting time has a death chance above which its lineage shrinks — a half with no wait, 0.3820 at one season, 0.2451 at four — and a longer wait tolerates less. What does not survive is the count itself: a plant losing one growing point in ten a season shows eight Fibonacci counts in a row one time in ten thousand, against one time in eight for a bud that occasionally waits an extra season.

8 figures How many scars a lineage carries for every growing point still alive. Deaths arrive at q times the standing count and the count grows by x a season, so the scars settle at q/(x − 1) — the curves. The dots are the ratio the expected counts actually reach after eighty seasons, and they agree to within 1.0 per cent. A bud waiting no delay at q = 0.1 carries 0.1250; a bud waiting one season at q = 0.1 carries 0.2192; a bud waiting two seasons at q = 0.1 carries 0.3135; a bud waiting three seasons at q = 0.1 carries 0.4128. Each curve runs to infinity at its own threshold, where the living stop outgrowing the dead.

What a scar is worth

Counting the scars a dead shoot leaves does not put a branching count back on the sequence it would have had. A scar records a growing point and a dead growing point takes every branch it would have made, so living points plus scars reach 39.2 per cent of the deathless count after twenty seasons at one death in twenty, and 1.9 per cent at one in five — falling without limit rather than closing. What the scars restore is the other number. Scars per living point settle at q/(x − 1) exactly, so a rate with a scar share beside it recovers the death chance and then the waiting time, where a rate alone is reached by a one-season wait losing a tenth, a two-season wait losing 0.64 per cent and no wait at all losing 27.2 per cent.

8 figures Sizing for equal bending and sizing for equal stress cross at one length ratio. Equal stress holds r³ against the sum of a load's arms and gives 3/(1 + ℓ); equal deflection holds r⁴ against the sum of the arms squared and gives 4/(1 + 2ℓ), with ℓ = log₂(1/λ). Setting them equal gives 3(1 + 2ℓ) = 4(1 + ℓ), whose only root is ℓ = 1/2 — the crown that fills a plane, λ = 0.707107 — and there both are exactly two. Below that ratio the stiffness rule reads the lower exponent of the two and above it the higher, so the two criteria size the same crown at one length ratio in the whole family and it is the one Da Vinci's rule names.

A crown sized for how far it bends

Stress is one criterion for sizing a branch and stiffness is another. Holding every branch to the same deflection as a share of its own length sizes r to the fourth against the sum of each load's arm squared, where equal stress sized r cubed against the arm, and the junctions of a deep crown then conserve 4/(1 + 2·log2(1/λ)). A single cantilever under its own weight comes out at radius as length to the three halves — McMahon's elastic similarity, fitted here rather than assumed — against the square that equal stress asks for. And the two criteria agree at exactly one length ratio out of the whole family: λ = 2 to the minus a half, the crown that fills a plane, where both give exactly two.

8 figures Three ways of sizing a crown, and the exponent each one conserves. Murray's flow rule sizes r³ against the tips a branch feeds and conserves three at every length ratio, reading nothing of the lengths at all. Equal bending stress conserves 3/(1 + ℓ) and equal deflection 4/(1 + 2ℓ), where ℓ = log₂(1/λ). So an exponent measured on a tree names a rule only with a length ratio beside it, and even then not everywhere: the stress and stiffness curves meet at λ = 0.7071, the stiffness curve passes three at λ = 0.8909, and the stress curve reaches three only as the branches stop shortening.

Three rules, one exponent

A measured branching exponent is quoted as evidence for a sizing rule, and it cannot be. Murray's flow rule conserves three at every length ratio and reads nothing of the lengths at all; equal stress conserves 3/(1 + l) and equal deflection 4/(1 + 2l), where l is log2(1/lambda). So an exponent names a rule only with a length ratio beside it, and even then not everywhere: of ninety-six length ratios between 0.3 and 0.99, thirteen have two rules within five hundredths of each other at a precision of 0.05, in three bands with three different reasons — stress against stiffness where they cross at the planar crown, stiffness against flow where the stiffness curve passes three at 0.8909, and stress against flow only as the branches stop shortening.

8 figures Each junction's exponent against its daughters' lengths, measured exactly and with a two per cent error in every radius. One thirteen-generation crown at a mean length ratio of 2^(−1/2), lengths spread by a half-width of 0.3 in log, its junctions five to ten generations above the tips — every third one drawn. Each junction appears twice, its exponent under equal stress and under equal bending, against the sum of its two daughters' log length ratios; the lines are the least-squares fits. Measured exactly, the stress exponents follow a slope of 0.298 and explain 62% of their scatter, the bending exponents a slope of 0.858 and 93%. With every radius read 2% wrong, the clouds swell — the stress scatter explained falls to 20% — and the slopes read 0.288 and 0.866. The error lands in the residual, and the slope is where the rule is.

The lengths that name the rule

A real crown has no single length ratio, and giving every fork a spread of daughter lengths does not blur what a sizing rule conserves: Murray's flow rule still conserves three at every junction, and the two mechanical rules keep their mean exponent, moved only as the square of the spread. What the spread adds is a second number. Each junction's exponent follows its daughters' summed log length with a slope of 0.30 under equal stress and 0.86 under equal bending at the planar crown, where the exponents are both two — and a two per cent error in every radius moves neither slope, while it swamps the scatter that looked like the obvious instrument.

8 figures Which tilts buckling sizes and which bending sizes, deep in the crown and near its tips. Crowns at the planar length ratio, forks turned 20°, sized by the larger of buckling and bending, at five balance angles. For each, the upper bar runs from vertical to the widest tilt at which buckling sizes a branch seven or more generations above the tips, and on from the narrowest tilt bending sizes; the lower bar is the same reading within six generations of the tips. Balanced at 30°, buckling sizes the deep crown to 20° and bending from 40°, and near the tips buckling reaches 60°; balanced at 40°, buckling sizes the deep crown to 40° and bending from 60°, and near the tips buckling reaches 60°; balanced at 50°, buckling sizes the deep crown to 40° and bending from 60°, and near the tips buckling reaches 60°; balanced at 60°, buckling sizes the deep crown to 60° and bending from 80°, and near the tips buckling reaches 60°; balanced at 70°, buckling sizes the deep crown to 60° and bending from 80°, and near the tips buckling reaches 60°. In the deep crown the two never overlap at any generation, so the criterion is chosen by direction; near the tips the arms have not converged, the bending term is smaller, and the upright core is wider.

A crown that would rather not buckle

A column held below the load at which it buckles and a cantilever held to a fixed deflection need the same radius at every length, because both hold the bending stiffness against a load times a length squared — so the three halves of elastic similarity is also the buckling law, and a crown whose loads all run along its branches conserves the same exponent under either. Gravity does not run along branches. It divides by the cosine of each branch's tilt, so a buckling junction's exponent is set by the direction its parent points, nothing past level is sized at all, and a crown sized by the larger of the two criteria splits by direction into an upright core and a spreading shell whose boundary junctions conserve more than either rule gives.

8 figures Fork angles on crowns sized by the larger of flow and stress, handing over at six different generations. Thirteen-generation crowns at a length ratio of 0.707, every branch sized by the larger of flow and stress, twigs at the radius the transport cost prefers, with the two radii equal at generations 3 to 8. Handing over at 3, the forks from the trunk open at 38, 51, 67, 75°; handing over at 4, the forks from the trunk open at 27, 38, 51, 67, 75°; handing over at 5, the forks from the trunk open at 19, 27, 37, 51, 67, 75°; handing over at 6, the forks from the trunk open at 13, 18, 26, 36, 50, 66, 75°; handing over at 7, the forks from the trunk open at 8, 12, 17, 24, 35, 49, 66, 75°; handing over at 8, the forks from the trunk open at 4, 6, 10, 15, 22, 32, 46, 65, 75°. Every fork from the handover outward opens at Murray's 74.9°. No fork on any of the six opens wider than that, and none turns back: the angle rises through the stress-sized generations and stops.

A trend that stops at Murray's angle

A crown sized by whichever of flow and bending stress asks for the thicker branch is sized by stress at its trunk end and by flow at its twigs, and the twigs fix the constant that was free in the fork-angle prediction: a twig at the radius the transport cost prefers spends exactly half its upkeep on pumping. On such a crown the fork angle does not change sign at the handover. It rises through every stress-sized generation, meets Murray's 74.93° at the handover and stays there, and no fork anywhere opens wider. The trend turns back only when the twigs are thinner than the cost wants — past a pumping share of (λ^(−2/3) − 1)/(1 − λ^(4/3)), 0.70 at the planar crown and closing on one half as branches stop shortening.

8 figures Every pair of death chances that reproduces each count, for one plant. A plant with a wait of two seasons, apices dying at 0.05 and buds at 0.15: a rate of 1.3351, 0.3098 scars per living point and 0.2225 of its scars left by apices. Each line is every pair of chances at that wait reproducing one count; the dashed lines either side are the same count one per cent high and low. The rate's and the scar share's lines cross at the plant's own pair, at an angle of 11.6°, so a one per cent error lets the pair slide along them — apex chances from 0.004 to 0.097 and bud chances from 0.100 to 0.198. The line for scars sorted by kind crosses the rate's at 51.5°, and read with it the same error leaves 0.044 to 0.056 and 0.134 to 0.166.

Two ways to die, three things to count

Giving a branching plant's waiting buds a death chance of their own leaves its counts a linear recurrence, but breaks the collapse onto the survival: the rate becomes the apex survival times the root of y^(d+1) = y^d + r^d, where r is the bud survival over the apex survival. The one-chance reading then names the wrong waiting time on 171 of 477 plants with waits of two to four seasons, shorter when the buds are the fragile ones and longer when the apices are. The two chances are separable from a rate and a scar share, exactly — but the two counts' loci cross at eight to sixteen degrees, so a one per cent error lets the chances wander by a factor of two. A third count is owed, and it is the scars sorted by kind.

8 figures

Stems and cones

A stem is a cylinder, and on a cylinder the spiral counts are the same the whole way up — which on a disc they never are. Two numbers fix the lattice, both come back out of the points exactly, and lowering one of them climbs the Fibonacci ladder.

A stem unrolled: 28 nodes at 137.51° with a rise of 0.090 circumferences. The counter is shown these coordinates and the circumference, and finds 2 parastichies one way and 3 the other. The faint strips left and right are the same stem: a family leaving one edge re-enters at the other.

A stem is a cylinder

The sunflower is the photograph, and it is the hard case. Nearly all real phyllotaxis happens on a stem, where the geometry is a lattice on a cylinder with two parameters — and where the spiral counts, which on a disc change with radius, are the same the whole way up.

8 figures The same counter, on a stem and on a disc. The stem returns 2 and 3 in all three bands. The disc returns 21/34, 34/55, 55/89 — three answers to one question, which is why a published count needs to say where it was taken.

Counting up the stem

The same counting machinery, pointed at a stem instead of a seed head, returns one answer three times where the head returned three answers. That contrast is a measurement rather than a preference, and it is the one the whole cylindrical argument rests on.

8 figures Six stems built, forgotten and recovered. Each row is a lattice built from a divergence and a rise, counted by machinery shown only the coordinates, and reconstructed from the counts and the two hop lengths. The worst error in the recovered angle is 3.0e-13°.

Two numbers out of the points

A seed head's divergence angle can be recovered from its spiral counts only to within an interval, because a range of angles gives the same counts. On a stem the counts come with lengths attached, two measurements pin two unknowns, and the lattice comes back to the last digit it was built with.

8 figures Every transition as the rise falls. The pair climbs 1/2 → 2/3 → 3/5 → 5/8 → 8/13 → 13/21. Consecutive transitions are 0.382, 0.382, 0.383, 0.382 of the previous rise — 1/φ² is 0.3820.

The Fibonacci ladder

Lower the rise on a cylinder and the parastichy pair climbs — 1 and 2, then 2 and 3, then 3 and 5 — each rung the sum of the two before it. Nothing in the arithmetic mentions Fibonacci, the transitions sit at computable rises, and consecutive ones stand in the ratio 1/φ².

8 figures Where the disc's counts change, predicted from a cylinder. The dashed lines are the transition radii the cylinder's ladder gives through h = c²/4πr², with nothing fitted. The dots are what the blind counter returns from the disc: 15 of 16 bands agree, and the ones that do not sit on a transition.

A disc is a cylinder

Vogel's seed head makes the rise fall as one over radius squared, so a disc is not one lattice but a family of them. Feed that into the cylinder's ladder and it predicts where a sunflower's spiral counts change — with nothing fitted, and against a counter that never sees either model.

8 figures The plane of stems: divergence across, rise up. Each shade is one parastichy pair. The marked points are the lattices where three families are equally short — the forks — and the Fibonacci ones run up the middle towards 137.51°.

The forks are exact

Where a stem's pattern has to choose between two futures, three spiral families are equally short and the lattice is exactly equilateral. A numerical solver found those points; the numbers it returned turned out to be rational, and chasing that gave a closed form — including the fact that every fork sits at a rational divergence, and the golden angle at none of them.

8 figures A cone unrolled: 105 nodes at 137.51°, sector 126°. The counter is shown these coordinates and the sector angle and nothing else. Near the apex it finds 3 and 5; near the base, 8 and 13. The rise falls as one over the distance from the apex, so the pair has to climb.

A cone has a rise that falls

A stem holds one parastichy pair for ever and a seed head changes its pair with radius. A cone does both — it is a cylinder whose rise falls as one over the distance from the apex, and the same blind counter that finds one answer up a stem finds four up a cone.

8 figures The rise falls as one over the distance from the apex. A straight line of slope −1 on log axes. The dashed horizontals are the transition rises of the cylinder's ladder, computed with no cone anywhere in them; where they cross, the count changes. Consecutive crossings are 2.62, 2.62, 2.62, 2.62 apart — φ² is 2.618.

Transitions a factor of φ² apart

The ladder's rungs are a factor of 1/φ² apart in rise. A disc's rise falls as one over radius squared and a cone's as one over distance, so the same rungs land a factor of φ apart on a seed head and a factor of φ² apart on a cone — measured, on both, by a counter that has never heard of either.

8 figures The exponent sets the spacing, and only the exponent. The curve is φ^(2/p), drawn from the ladder's ratio of 1/φ² and nothing else. The dots are measured: each surface built, a blind counter walked up its axis, the places its answer changed recorded. A cone that elongates and a paraboloid that fills sit on the same point at p = 1 — so the spacing identifies neither the shape nor the way material arrives, only their product.

The shape and the law

A cone's transitions are a factor of φ² apart and a disc's a factor of φ, and the temptation is to read the ratio as the shape. It is not. Five surfaces built and counted show that the ratio measures one exponent, and that the exponent is the shape multiplied by the way material arrives.

8 figures What is visible in the outer part of a 4000-element organ. Both surfaces have the same ladder in element number — the rise is 1/(2πi·flare) on a cone and 1/(4πi) on a disc, and c and the internode step both cancel. What differs is where the elements are. Counting outside 50 per cent of the extent, a cone shows 1 change and a disc 2, because half a cone's length holds half its elements and half a disc's radius holds three quarters of them.

Why a cone can be counted once

A pineapple is described as 8 and 13 and the description holds. A sunflower is described as 34 and 55 and the description is a statement about one annulus. Both organs have the same ladder in element number — what differs is where an organ puts its elements.

8 figures Three organs, and the exponent each has at each place. Left: the meridian of a cone, an ogive of arc radius 12, and a spherical cap of radius 40, each scaled to its own length. Right: d log r / d log s along it. The cone sits at 1 the whole way; the ogive starts near 2 at its blunt tip and falls to 0.23 by the end of the 100 per cent shown; the cap starts at 1 and falls slowly.

An organ has no single exponent

The earlier work measured that a surface whose circumference grows as a power of arc length puts its transitions a fixed factor apart, and checked it on five surfaces. Every one of them had a single exponent, and no organ does — a fir cone is an ogive, whose exponent runs from 2 at the tip to nearly 0 at the shoulder.

4 figures One exponent fitted to an organ that has 4 of them. Each dot is one step between consecutive rings, reporting 2 ln φ / ln(s′/s) — the exponent that step would have if the organ had one. They run from 1.980 to 1.697. The line is what a single fit returns, 1.891, which is their harmonic mean of 1.880 and sits below their plain average of 1.887.

What one exponent reports

Fit a single shape exponent to an organ that has four of them and it returns a real quantity — the harmonic mean of what its individual steps report. Harmonic means sit below arithmetic ones, so the fit understates, systematically, in a known direction, and invisibly.

4 figures At 1.0 per cent on each ring, 5 rings show the drift. The gaps between consecutive rings are 1.626, 1.631, 1.657, 1.763. Two rings give one gap and no way to disagree with itself; three give two gaps and a fit with nothing left over. The question is how many gaps it takes for their spread to exceed what the measuring error can explain, and the answer depends on the error as much as on the organ.

How much of a cone to measure

Two rings cannot show a varying exponent — not with difficulty, but in principle, because one gap determines one exponent with nothing left to disagree. Four or five can, if each is found to within a per cent. At three per cent this specimen cannot be told from a power law however many of its rings are recorded.

6 figures The memory of a divergence sequence, at 0.75° of scatter. With no noise at all the lag-one correlation is 0.54: the rule corrects itself, so a lattice arrives with a memory in it. Matched at the same recorded scatter, placement noise leaves -0.10, jostle noise leaves 0.66, field noise leaves 0.47. The band is ±0.13, which is what an uncorrelated sequence of this length gives.

The sequence has a memory

Every measurement this collection has made of a stem's divergence angles throws the order away. A spread is invariant to shuffling. Put the angles back in order and there is a large correlation between one and the next — 0.54 with no noise at all — which is the rule correcting itself, and which nothing had looked at.

8 figures What the sequence sees that the scatter cannot. Each point is an ensemble at one amplitude, placed at the scatter it produces. A stem at three quarters of a degree of scatter has a lag-one correlation near zero if its noise arrived after the primordium was placed, and near 0.7 if it arrived before — and no measurement of the scatter can tell those apart. The separation closes above about a degree, because what the other two kinds preserve is the correlation of a lattice.

What one angle says about the next

A tenth of a degree of placement noise moves a stem's divergence scatter from 0.50° to 0.62°, which nobody would report. It takes the correlation between consecutive angles from 0.54 to below zero. The other two kinds of noise, at scatters where no measurement can separate them, leave it at 0.6.

8 figures The order of the angles carries the count. Three stems, each held at a fixed rise so the pattern sits on one rung of the ladder. At a rise of 0.032 the positions count 3 and 5 spirals and the angles peak at 3; At a rise of 0.013 the positions count 5 and 8 spirals and the angles peak at 5; At a rise of 0.005 the positions count 8 and 13 spirals and the angles peak at 8. Each panel marks the peak and its multiples; the pale strip is what an uncorrelated sequence of this length gives.

The order carries the count

Take the divergence angles off a stem, throw away every coordinate, and autocorrelate what is left. The result is periodic at the smaller parastichy number — peaks at it and at every multiple of it. A list of angles, with no picture and no position in it, carries the spiral count.

8 figures The memory belongs to the rise, not to the lattice. The lag-one correlation of a noiseless rising stem, against how fast it climbs the ladder. Below about sixty nodes per rung it is negative; above it, 0.54, 0.74, 0.58, 0.58, 0.55 — flat across a fivefold change in rate. The horizontal line is the same rule with the rise held FIXED, where the correlation is -0.68. So the +0.74 the earlier work called the sequence's own memory is the pattern chasing an equilibrium that is moving under it.

The memory was the rise

The earlier work measured a lag-one correlation of 0.54 in a noiseless divergence sequence and called it the sequence's own memory. Hold the rise fixed and there is no sequence at all — every angle identical — and under a disturbance the correlation is negative. The 0.54 belongs to the pattern chasing an equilibrium that is moving under it.

8 figures Two combs, at a rise of 0.005. The autocorrelation of 760 divergence angles from one stem held at a rise of 0.005. The filled teeth are the lags at multiples of 8; the open teeth are the second comb, at the same spacing offset by 5. Reading the spacing off the first and the offset off the second gives the pair 8 and 13, which is what the position counter reports for the same stem — from angles alone, with no coordinate anywhere in the calculation.

The second comb

The autocorrelation of a divergence sequence has peaks at the smaller parastichy number and at every multiple of it. It also has a second set of peaks, at the same spacing, offset by the difference of the pair — so a list of angles with no coordinate in it returns both numbers rather than one.

8 figures Two combs, at a rise of 0.013. The autocorrelation of 760 divergence angles from one stem held at a rise of 0.013. The filled teeth are the lags at multiples of 5; the open teeth are the second comb, at the same spacing offset by 3. Reading the spacing off the first and the offset off the second gives the pair 5 and 8, which is what the position counter reports for the same stem — from angles alone, with no coordinate anywhere in the calculation.

A harmonic is a step taken twice

The spectrum contains the larger parastichy number, their sum, and echoes of the smaller one, and no ranking of peak heights separates them. What separates them is arithmetic: a harmonic is a multiple of the spacing and a family is not, and the two kinds sit in different residue classes.

8 figures Both statistics, on the same stems, at a rise of 0.005. Five seeded stems at each disturbance, held at a fixed rise. Bars are how many returned the pair the position counter finds; open portions are refusals. The pair comes out from 0.1 to 0.25, and across that whole range the lag-one correlation of the same sequences is -0.33, -0.58, -0.59 — decisive, negative and flat. There is no trade between the two: one stem supplies both. Below the window the sequence has locked onto the sampling grid and is a cycle rather than a sample; above it there is no lattice left, at 116° of scatter.

Two readings from one stem

Three note left with the work in a row have recorded that the two statistics of a divergence sequence want opposite plants — one quiet, one disturbed. Measured on the same stems they do not. The conflict was in the interpretation of a sign, and the window in which both are readable is wide.

8 figures Which arrangements carry a comb, and what each one reports. The largest comb mean in five arrangements at a rise of 0.005, all read by the same instrument at the same length, with the sampling band of 0.073 marked. Only the first is a placement rule; the other four are kinematic lattices with no rule in them, differing from one another only in how their azimuth errors are structured. Independent errors and errors with a memory leave nothing to read. A repeating error puts up a comb and names a partner that is not the lattice's. Errors inherited from the contact neighbours reproduce both the comb and the pair.

The comb was never the rule

A control is only as strong as the alternative it builds, and the earlier work built one that varied the rule while holding the disturbance fixed at independence. Five rounds of the angle-sequence thread, with what each claimed and what still stands — and why the next evidence has to come from an intervention rather than from a longer stem.

8 figures Every rung has a window it can be read in. For each rung of the ladder, the range of lag windows in which the pair can be read: at least three times the smaller parastichy number, so the main comb has three teeth inside the window, and less than three times the larger, so the larger number cannot itself be a candidate spacing. The band is [15, 24) at 5/8, [24, 39) at 8/13, [39, 63) at 13/21. It is non-empty at every rung, because it is empty only when the larger number is no larger than the smaller. The line at 30 is the window the earlier work used everywhere: it sits inside the 8/13 band and outside the 13/21 one, which is what was mistaken for a ceiling.

The rung was not the instrument

The earlier work said the pair readout has a ceiling one rung above where it works, that this is arithmetic rather than statistics, and that no amount of stem fixes it. The arithmetic is right and gives a band of lag windows that is never empty; what was actually stopping the reading was an eight-node seed and a grid of 384 azimuths.

8 figures The ratio is a U across every rung, and its floor is the number that was reported. The ratio of the second comb to the main comb, on five stems at each of 15 rises spanning two rungs, against the ladder's own coordinate for where each rise sits inside its rung. Both rungs give the same shape: a floor of 0.71 and 0.79 about two thirds of the way up, climbing towards the transition at either end. The dashed line is a transported disturbance with no rule in it at 1.29, which does not vary with the rise at all — a kinematic lattice's angle sequence has no rise in it. Where the rule's curve crosses that line the two accounts are indistinguishable.

The ratio was the floor of a curve

One number was left standing between a placement rule and a transported disturbance, measured at one rise, with the explanation that the geometry there happens to favour the larger parastichy number. Swept across two rungs the number is a U — a floor of about 0.79 two thirds of the way up a rung, climbing past 2.8 as a transition approaches — and the geometry is flat exactly where the curve is steepest.

8 figures At 384 azimuths the ratio is 0.62; converged it is 0.82. The comb ratio and the recorded divergence scatter at a rise of 0.005, against how finely the rule samples the circle when it takes its minimum. At 384 azimuths — the grid every flat run in these essays uses, and the grid that earlier work's 0.65 was measured on — the step is 0.94°, which is larger than the 0.25° disturbance the stems carry. The quantisation is white noise, it dilutes both combs, and it does not dilute them equally. The ratio settles at 0.82 from 1152 azimuths up, and the scatter loses 0.19° that belonged to the grid rather than to the stem.

The grid was in the number

The rule places each organ at the least of a profile sampled at a fixed number of azimuths, and every flat run in these essays samples 384 of them — a step of 0.94°, against a disturbance of a quarter of a degree. The quantisation is the larger of the two, it is white, and it moves the discriminator from 0.79 to 0.62. The convergence study this collection had asked for and never done, in the place it turned out to matter.

8 figures The damage is the sharing; the forgery is the history. Three disturbances of the same size, measured four ways. The two left columns are stems grown by the placement rule and jostled at 0.25° per organ: a disturbance shared between the contact neighbours scatters the lattice by 0.71° against white noise's 0.57°, and one inherited from them — the same sharing, passed on again at every organ — by 0.97°. The two right columns are kinematic lattices with no rule in them at all, where the whole question is what a disturbance can manufacture. The inherited one returns the pair on 8 seeds of 8 with a main comb of 0.205 against a band of 0.073; the shared one, at the same coupling and the same scatter, returns it on 1 and makes a comb of 0.099, which is the band. So sharing an error with the organs you touch does the damage, and only passing it on and on forges the evidence.

What the sharing costs a lattice

A disturbance inherited from the contact neighbours destroys a stem's lattice at half the displacement independent noise needs, and it moves the comb ratio a fifth of the way to a forgery's. Take the inheritance out and keep the sharing, and the damage stays and most of the ratio shift goes — so the two effects have different causes.

8 figures The block is one of the two numbers, and not always the smaller. Every lattice a single organ was removed from, one row each: golden, rise 0.020 carrying 3/5; golden, rise 0.013 carrying 5/8; golden, rise 0.008 carrying 5/8; golden, rise 0.005 carrying 8/13; Lucas, rise 0.020 carrying 4/7; Lucas, rise 0.013 carrying 4/7. The two open ticks on each line are that lattice's own parastichy numbers; the filled dots are the blocks the stems that never recovered settled into, one per offset that failed. The claim this table was built to test is that the block is the smaller of the two, which held at the two rises it was first measured at. It does not hold here: 3/5 gives 5, 5/8 gives 5 and 8, 4/7 gives 4 and 7. What survives is weaker and still worth something — every filled dot but 1 sits on one of that row's own ticks, so the orbit carries a count of the lattice it was cut from, and which of the two it carries is decided by the offset rather than by the pattern.

Two accounts of one number

A stem that never recovers from an ablation settles into a repeating block whose length was the smaller of its two spiral counts, at both arrangements it had been measured at. Two different explanations predicted exactly that and could not be told apart. Swept across four rises on the ordinary branch the premise itself fails: at one arrangement the block is the larger number, at another both appear, and the rule that seemed to be there was two measurements.

8 figures One rule, one rise, two branches that stay where they were put. The top 70 organs of two stems grown by the same placement rule at the same rise of 0.013, differing only in the stretch of ideal lattice each was started from. The left one was seeded at the golden angle and settles at 136.781° with the pair 5/8; the right one was seeded on the Lucas lattice and settles at 99.785° with 4/7. Neither drifts towards the other: 0.73° and 0.28° from where each was seeded, over four hundred organs. That is what makes an intervention on the right-hand stem a measurement about a different lattice rather than about a different rule — and 4 and 7 are not Fibonacci numbers, which is the property the experiment needs.

A stem on the other branch

Every stem an organ had been cut from carried Fibonacci counts, which is why two rival explanations of the block a wrecked stem settles into had never disagreed. A stem seeded on the Lucas lattice carries four and seven at the same rise, under the same rule. Cut, it settles on seven — the larger number, and not a Fibonacci one.

8 figures A bijugate stem answers in pairs, once the half turn is taken out. Removing one organ from a stem grown by a rule that places two at a time, at a rise of 0.0065, where the pattern counts 6 and 10 and has rotational symmetry of order 2. The open circles are the displacement of the next organ as it comes out of the arithmetic, which reaches 180° at offsets where the stem has demonstrably not been disturbed — the two organs of a whorl are interchangeable, so calling the other one "next" is a relabelling and not a movement. The filled circles are the same numbers read modulo 180°, which is the only way a 2-jugate divergence is defined. Read that way the response is a run of equal pairs — 68.0°, 68.0°, 42.9°, 42.9° — ending at 10, the larger parastichy number, with everything past it under 0.5°. Two organs of one whorl give the same answer to the last digit, which is the pattern's symmetry showing up in an experiment.

What a cut costs a whorl

A bijugate pattern is an ordinary lattice seen twice over, so the account that says a wrecked stem's repeating block is the repeat unit of the lattice underneath has a specific prediction here: three and five. It gets six and ten. And the thing a single missing organ does destroy on a whorled stem is the one property its counts cannot see.

5 figures The same rule, the same rise, two lattices, two fronts. How many organs back a single removal is still felt, on a stem seeded onto the golden lattice and on one seeded onto the Lucas lattice, at seven rises. Everything but the seed is identical at each rise — the rule, the spacing, the heights, the azimuth grid — and the two fronts differ at every one of them. Which branch has the wider front changes hands four times going down the range, so no function of the rise gives the column. The golden branch carries 5/8 at four of these rises, across a factor of two in the rise, and its front is eight at all four.

Seven rises and two seeds

One organ removed from a stem is felt out to the larger of its two spiral counts. Every test of that has confounded the count with the rise, because on one branch the two move together. Grow a second branch beside the first at the same rise and they come apart — and doing it at seven rises turns a matched pair into a design whose last column changes hands four times.

8 figures The newest member of the front is the weakest. For every cell of the design whose rung boundary is inside the range, how far the next organ moves when the organ exactly as many places back as the larger parastichy number is removed — the offset that arrived when the stem entered this rung — against how far below that boundary the stem sits. Each line is one lattice on one branch. The horizontal line is the threshold that decides whether an offset counts as felt, and the three cells below it are the three whose front reads one offset short. Nothing is a different kind of thing: the boundary is a step everywhere, and near the top of a rung its last stair is shallow.

The front that reads one short

Eleven cells of a fourteen-cell design put the boundary exactly at the larger spiral count. Three put it one offset earlier, and the tempting move is to lower the threshold until all fourteen agree. Measured instead of tuned, the three turn out to be the three cells nearest below their own rung's boundary — and the last offset of a front is weak because it has only just arrived.

8 figures The offset past the front that is felt anyway. The four cells of the design whose response has a hole in it: a run of felt offsets, a stretch of quiet, and then one isolated offset well outside the front at which a removal moves the next organ by tens of degrees. The open circle on each row is the count coming in at the next rung of that branch's ladder, and the filled point is the isolated offset. It sits one inside the incoming count on every row, including on the Lucas branch, where the incoming counts are 7 and 11 rather than the Fibonacci numbers the rule was found on.

The hole on the other branch

Near a transition, the run of offsets a stem notices stops being a run: there is quiet past the front and then one isolated offset, felt as hard as anything inside it. Where that offset sits was pinned down on Fibonacci lattices, where the numbers to check it against are 5, 8 and 13. On the Lucas branch they are 4, 7 and 11 — and the rule holds there too.

8 figures Everywhere a cut of one to five organs can send a 5/8 stem. Every settled divergence reached by any arrangement of up to five organs removed from a stem at the 5/8 rung, on one axis. There are six of them and no more. three are slips of the lattice the stem was cut from: each keeps the lag-5 family intact and sits a whole number of turns of it from the next, which is the ladder marked below the axis with rungs 72.0 degrees apart. The other three keep no lag at all and sit near a fraction of a turn, marked above: 175.0 degrees near 1 of 2 turns, 190.0 degrees near 1 of 2 turns, 235.0 degrees near 2 of 3 turns. A stem that is cut either slides along the ladder it was on or leaves it for a lattice with files in it.

A file has to close

The three destinations counted with a shared factor sit near a half turn, a half turn and two thirds. Measuring how near is the trap: by distance from the fraction, the golden angle is closer to two fifths than two of them are to anything, and would be reported as having five files it does not have.

8 figures Which rises are a lattice, from 0.04 to 0.13. How much the divergence wanders over the last sixty organs, at each rise up the coarse ladder. 13 of the 19 settle, scattering between 0.0000 and 0.3356 degrees. six do not: from 0.09 to 0.115 the divergence sticks on exactly 135.0000 degrees, which is three eighths of a turn, and wobbles about it by 0.79 to 1.60 degrees. A counter shown either kind returns the same pair, so the counts cannot tell them apart. The line is the threshold a rise has to pass before a cut is made on it, and it sits in the gap rather than among the measurements.

A stem coarse enough to cut

Below the 3/5 rung is a 2/3 rung, and it runs from a rise of 0.050 to 0.120. It is not a lattice across all of it: from 0.090 to 0.115 the divergence stops settling and sticks on exactly three eighths of a turn, wobbling by a degree and a half — while a counter goes on reporting 2/3 as though nothing had happened.

8 figures What a two-organ cut does at each rise of the 2/3 rung. At every rise the coarse rung is a lattice on, all 36 arrangements of two organs removed, with the ones that never repair counted and split by where they end up. 22 of 324 cuts across the rung reverse the stem's handedness onto the mirror of the divergence they were cut from. 40 fall instead into a cycle whose mean is half a turn, which the lattice they came from has no number for. 4 rises give only the first, 4 give only the second, and at a rise of 0.075 both happen in the same table, which is what says the fate belongs to the cut and not to the rise.

The shallower front turns over

If reversing a stem means rearranging its whole front, then a stem with a shallow front should reverse more often. Measured across three rungs and four hundred and seventy-three cuts: 6.8 per cent at a front of three, 4.7 at five, and none at all at eight — where the nearest approach is two tenths of a degree away and stays there.

8 figures Which offsets give short hops, at a rise of 0.013. The two lowest points are at 5 and 8, and those are the parastichy numbers. Offset 1 is high because a hop of one node is at least the rise, which is what makes a stem easier to count than a disc.

Where a handover sits

Inside every rung there is a rise at which the two contact steps change places, so that the shorter hop belongs to the other family below it. Six of the eight rungs on this ladder have one, each has exactly one, and every one of them sits in the coarse half.

8 figures Two lines across the 5/8 rung, crossing once. The divergence the rule settles on, against the divergence at which the two contact steps would be exactly the same length. The second is arithmetic on the lattice and no stem is grown for it. Across this rung the balanced line moves 2.281 degrees and the rule's own line moves 1.262, so the shallower line crosses the steeper one, and it does so exactly once at a rise of 0.0154 — 16 per cent of the way down from the coarse end. That crossing is the handover: above it one family has the shorter step and below it the other does. So a rung has one handover, its position is fixed by the arithmetic rather than by any experiment, and a sweep of the rise carries a stem across it at a place nobody chose.

Two lines that cross once

The divergence at which a lattice's two contact steps are exactly equal is a curve across each rung, computable from the geometry with nothing grown. The rule's own settled divergence is a second, shallower curve, and where they cross is where the step ordering changes hands.

7 figures A divergence that does not move across the 5/8 band. Measured at every rise of a band on the golden branch, where a counter returns 5 and 8 spirals throughout. The settled divergence moves by 0.0469 degrees across the whole band, which is a fraction of the azimuth grid step and a fortieth of the slide across the rung it sits in. The ratio of the two contact steps does move: it falls to 1.0013 and the ordering changes hands at a rise of 0.0156, so above that rise the shorter step belongs to the 5 family and below it to the 8 family. Two of the three quantities that vary along a rung are therefore held here and the third is not, which is what makes the ends of this band a matched pair.

A band that holds the angle still

Around every handover the settled divergence has a shallow floor, so a run of rises either side of it share a divergence to a twentieth of a degree while their two contact steps change places. That is a matched pair with one quantity varying, and it is the design the ablation thread had no way to state.

8 figures How many organs a stem needs before it is on a lattice. One row per rise, one mark per starting angle, placed at the organ from which every later divergence stays within a degree and a half of the run's own final value. Where a stem settles at all it does so between 0 and 290 organs in, against the 400 every ablation run here grows before it cuts anything. Not one row needs the length it is given. What changes down the table is the count on the right: how many of the nine starting angles reach a lattice at all, which falls from 7 at the coarse rises to 1 at the finest.

How long a stem takes to settle

Every result here is grown on a stem that has settled onto a lattice, and settling has always been tested for and never timed. Timed, it takes between nothing and two hundred and ninety organs — against the four hundred every ablation run grows before it cuts anything, and the nine hundred the noise runs carry.

8 figures The same table, grown 2.7 times as long. Every rise and every starting angle, grown to 1200 organs and then to 3200. The two middle columns are how many starting angles reached a lattice at each length, and they are the same column: of the 72 pairs of runs, 72 are identical organ for organ and 0 settle at the longer length after failing at the shorter one. Tripling the budget buys nothing anywhere. What the fine rises are short of is not run length: the share of starting angles that reach a lattice at all falls from 7 of 9 to 1, so the arrangements a stem could fall into have mostly stopped existing.

A wall and not a budget

Below a rise of about 0.005 this collection's stems stop settling onto a lattice, and the limit has been written up four times without anybody asking which kind of limit it is. Grown three times as long, the table is identical row for row: not one stem that failed to settle succeeds. The floor is a wall.

8 figures The settled divergence down the golden branch. Every rise from 0.07 down to 0.00482, plotted against the divergence the rule settles on, with each rung drawn in its own stroke and the branch's limit angle marked. The curve does not slide: it turns three times in four rungs, climbing across one and falling across the next, so a value it takes on one rung it takes again on another. That is what makes a matched pair possible — two rises, different counted pairs, one angle — and it is the whole reason the design exists on this branch. The widest excursions from the limit angle, coarse rung first, are 3.195°, 3.352°, 0.961°, 0.422°.

The angle the ladder returns to

Down the golden branch the settled divergence climbs across one rung and falls across the next, turning three times in four rungs. That is why the same angle is reached at two different rises — and why the Lucas branch, which turns once, almost never offers the same thing.

5 figures Which rungs of the golden branch share a divergence. One row per pair of rungs. A pair whose divergence ranges overlap has a rise on each rung where the rule settles on the same angle; a pair whose ranges do not overlap has none, whatever the search. On this branch four of six pairs match, and three of those match to 0.0000° — the same value of a quantity read on a grid of 1,536 azimuths. The rises differ by factors of 1.48 to 5.09, so the design holds one angle while changing everything the rise controls.

Two rungs, one angle

Five pairs of rises settle on the same divergence while a counter returns different pairs at them, and four of the five agree to 0.0000° — the same value of a quantity read on a grid of 1,536 azimuths. The rises differ by factors of 1.48 to 5.09.

5 figures A divergence that does not move across the 5/8 band. Measured at every rise of a band on the golden branch, where a counter returns 5 and 8 spirals throughout. The settled divergence moves by 0.0469 degrees across the whole band, which is a fraction of the azimuth grid step and a fortieth of the slide across the rung it sits in. The ratio of the two contact steps does move: it falls to 1.0013 and the ordering changes hands at a rise of 0.0156, so above that rise the shorter step belongs to the 5 family and below it to the 8 family. Two of the three quantities that vary along a rung are therefore held here and the third is not, which is what makes the ends of this band a matched pair.

The angle is not the actor

Cut an organ out of two stems that settled on the same divergence and return different counted pairs, and the family left standing is different at every one of the four pairs where both stems wreck. The angle is held to a hundredth of a degree underneath.

8 figures Shared counted numbers against shared survivors. One row per matched pair, over both branches. The third column is the counted numbers the two rungs have in common and the fourth is the families both stems leave standing; on every row the two are the same set. The row whose rungs share no counted number is the one whose stems share no survivor, which is what makes this a claim about an intersection rather than a restatement that a survivor is usually a contact family. four rows, and the empty case is one of them.

Where the survivors meet

At a matched pair the two stems keep exactly the counted numbers their two pairs have in common — the 5 where 3/5 meets 5/8, the 8 where 5/8 meets 8/13, the 7 where 4/7 meets 7/11, and nothing at all where 3/5 meets 8/13. Four rows, including the empty one.

8 figures Which rungs of the golden branch share a divergence. One row per pair of rungs. A pair whose divergence ranges overlap has a rise on each rung where the rule settles on the same angle; a pair whose ranges do not overlap has none, whatever the search. On this branch four of six pairs match, and three of those match to 0.0000° — the same value of a quantity read on a grid of 1,536 azimuths. The rises differ by factors of 1.48 to 5.09, so the design holds one angle while changing everything the rise controls.

The last of three quantities

The pair, the divergence and the step ordering move together when the rise is swept, and for a long time no result could be attributed to any of them. Two designs later, two are ruled out as sufficient and the third has never been held still — because holding it is what a rung already does.

8 figures Where the two contact steps change places, on every rung. One row per rung of the two branches, drawn from its coarse end to its fine one on a logarithmic axis. The mark on each row is the rise at which the two contact steps change places — the rung's handover — and the shaded stretch is the band around it over which the settled divergence holds still. six of the eight rungs have a handover and every one of those sits between 6 and 40 per cent of the way down its rung, never past the middle. The two rungs without one are the coarsest on each branch, whose crossing is above the range this ladder reaches.

Four crossings nobody visited

Six rungs of the ladder carry a handover and two of them had a band built on them. The other four are here: 70, 126, 16 and 124 rises wide, found by sweeping at a ratio rather than at a fixed step in the rise, which is why the fine ones had been stepped over.

7 figures What each band holds, and what it moves. One row per band. The counted pair is held by construction and the settled divergence is held to a twentieth of a degree; the quantity a band exists to move is which of the two contact steps is the shorter. five of the six do move it — the ordering changes hands exactly once inside, and at both ends the two steps differ by enough for an ordering to mean anything. The remaining one changes hands three times and its two steps are never more than 0.0 per cent apart, so it holds all three quantities and is a control rather than an experiment.

A band that moves nothing

One of the six bands holds the counted pair, holds the divergence, and does not move the ordering: its two contact steps stay within four parts in a thousand of each other across the whole of it, so the ordering changes hands three times and neither end has one worth the name.

7 figures Two ways of predicting how wide a band is. A band ends where the settled divergence has moved 0.05° from its value at the handover, so the width should follow from how fast the divergence changes there. Reading that rate as the rung's average slope predicts widths that are wrong by factors of 0.20 to 5.92 — wrong in both directions, so no constant rescues it. Reading it as a curvature about a stationary point gives 0.41 to 1.08, with five of the six inside a third. The difference between the two is the difference between a curve and its average, and a band is exactly where the two are least alike.

How wide a band should be

A band ends where the divergence has slid a twentieth of a degree, so its width should follow from how fast the divergence slides. Predicted from the rung's slope that is wrong by factors of 0.20 to 5.92; predicted from a stationary point it is 0.41 to 1.08, and the outlier is the rung that has no stationary point.

8 figures The families left standing, on both sides of every handover. One row per band, with every offset cut at rises spread across it and always at both ends and at the handover itself. The last column is every family left standing anywhere on that band. On three of the four bands that wreck at all it is a single family, unchanged across a rise at which the two contact steps swap places — so the step ordering is not what decides which family survives, and the result now rests on four counted pairs rather than on two. The shortest-hop reading scores 44 of 114 across the whole set, which is what a reading looks like when the quantity it is stated over is not in the mechanism.

The ordering on six bands

A hundred and fourteen wrecked cuts across four bands, and at every offset of every one of them the family left standing is the same immediately above the handover and immediately below it. Where the answer does change — on the widest band, at three offsets — it changes somewhere else.

4 figures Every rise of the 8/13 band, cut at every offset. One column per rise of the band, coarse on the left and fine on the right, and one row per offset that wrecks anywhere on it. A filled cell is the family the cut stem keeps; a pale cell is an offset that recovers at that rise and has no survivor to report. The band holds 126 rises and 1890 cut stems. three of the six offsets change their answer somewhere inside, three never do, and the vertical rule is the handover — the rise where the two contact steps change places. Not one of the 19 changes is at it.

Every rise of a band

A band is cut at nine rises because the quantity it was built to test is a constant, and a constant is checked at the ends and at the crossing. On the widest band that quantity turned out not to be constant, which makes nine the wrong number. This is all hundred and twenty-six.

8 figures A period fitted to the speckle, at every period it could have. Each mark is one candidate period, drawn at the share of rises it gets right when it is given its best phase and its best family in each residue class — the most generous reading of periodic there is. The flat rule is what saying nothing gets: name the commonest family and stop. The best period scores 76 per cent against 76 for no period at all, a gain of 0 points over 123 rises, so the alternation the coarse design reported is not a period being sampled badly.

The alternation is not a period

Nine sampled rises gave 8, 4, 8, 4 at one offset of one band, and a period was the obvious thing to look for. At full resolution it is thirteen islands one to three rises wide, with gaps of 1, 2, 3, 6, 7, 8, 9, 16, 31, 44 and 48 — and a fitted period buys exactly nothing.

8 figures Offset 7 across the 8/13 band, rise by rise. The family this one offset keeps at each of the band's 126 rises, coarse on the left. It wrecks at 126 of them and keeps the 4-family and the 8-family at different rises. The ticks below mark one islands — runs of 2 rises where the coarse family comes back inside the fine one. The handover is the taller rule and the change of answer is nowhere near it.

Three offsets, three crossings

The claim the band design rests on is that the survivor does not change where the two contact steps change places. It holds at full resolution: nineteen changes and not one at the handover. Where they are is three different rises, eight, nineteen and twenty-nine below it.

8 figures Offset 4 across the 8/13 band, rise by rise. The family this one offset keeps at each of the band's 126 rises, coarse on the left. It wrecks at 98 of them and keeps the 8-family throughout. The ticks below mark no island: the answer changes once and stays changed. The handover is the taller rule and the change of answer is nowhere near it.

The offsets that never change

Three of the six offsets that wreck anywhere on the band keep the same family at every rise they wreck at — 98, 22 and 81 rises of the 126. And which offsets wreck at all is a function of the rise, which no reading of a band had drawn.

8 figures Which offsets wreck across the Lucas 7/11 band. One row per offset and one column per rise, with a mark where a single removal at that offset wrecks the stem. The set is not the same at every rise: on this band one offset wrecks at only 27 of its 124 rises, in several separate stretches, while others wreck at all of them. So a census taken at one rise of a band and a census taken at another are censuses of different sizes, and every claim of the form "at every offset that wrecks" is quantified over a set the rise decides.

A band with nothing inside it

Five offsets wreck on the Lucas 7/11 band and every one of them keeps the same family at every rise it wrecks at. There are no islands, no transition region and no period to look for, which is what makes the picture from the other band a picture of that band.

8 figures The hops the correction is fitted over, with the new one at the near end. Each cluster of rows placed by the lag it kept and the angle of the hop that lag keeps. The four the correction was fitted over run from 19.5 to 39.1 degrees; the new one sits at 12.78 degrees, a third smaller than any of them. A fifth point beyond the near end of a fitted range is a test of the fit, where a fifth point between two old ones would mostly have been a restatement.

A fifth cluster

A correction to the exchange's size was fitted over four hop clusters and its own file said so. A search turned up a fifth, at a hop smaller than any of the four, and the rule is right on it — which is what a prediction being confirmed looks like when the confirmation is worth having.

4 figures What is the same at 0.00998 and at 0.00997. Six quantities read on the two rises the transition sits between. Five of them are identical: the two walls the slot has, the divergence the intact stem settles to, the block the cut opens, which of the three removals wreck the stem, and the lags the doubled cut leaves rigid. The sixth is how far the first organ placed after the doubled cut moves, and it goes from 163.59 to -9.14 degrees. The lattice is the same on both sides; where one organ goes is not.

A transition and not a slope

The question was whether a fourth cell's cost declines smoothly to nothing or falls in one step. It falls in one step, and the answer decides whether a word in the collection names something or is a threshold on a continuum.

8 figures Twenty rises at the fine end of both branches, cut at every offset. One row per rise searched, coarse at the top of each block. The bar names the counted pair the stem shows and the numbers on the right are the lags its wrecking cuts leave standing. A pale row is a rise whose settled divergence has left the branch it was started from by more than 20 degrees, which is what happens below the ladder's finest rung — the pairs there are 2 and 4, 8 and 16, 11 and 22, which are not two consecutive terms of any additive sequence. One rise on the Lucas branch keeps a lag of 11 while still on it.

One rise below the census

Ten lattices were cut at every offset and their surviving lags came back as four numbers. One rise further down a rung the census already sweeps, three cuts keep a lag of eleven — which is a fifth number, on a lattice nothing about was unusual except that nobody had cut it.

4 figures Where the doubled cut's run finishes, at every rise of the sweep. The divergence the wrecked run ends at, rise by rise. It takes only a handful of values and jumps between them at rises one part in a thousand apart, 4 times inside the 9 rises of the finest sweep alone — where the walls, the block, the intact stem's divergence and the rigid lags are all held. So the end of a wrecked run is not a stable quantity on this rung, and no statement about where the stem finishes is available on either side of the transition. The first organ's displacement is the reproducible half of the same measurement.

The end of a wrecked run

The obvious follow-up to a transition in where one organ goes is whether the stem also finishes somewhere different. On this rung the question has no answer: a wrecked run's final divergence takes four values and changes between rises a thousandth apart, three times inside a nine-rise sweep.

8 figures Twenty rises at the fine end of both branches, cut at every offset. One row per rise searched, coarse at the top of each block. The bar names the counted pair the stem shows and the numbers on the right are the lags its wrecking cuts leave standing. A pale row is a rise whose settled divergence has left the branch it was started from by more than 20 degrees, which is what happens below the ladder's finest rung — the pairs there are 2 and 4, 8 and 16, 11 and 22, which are not two consecutive terms of any additive sequence. One rise on the Lucas branch keeps a lag of 11 while still on it.

The lag that never survives

A correction to the exchange's size rests on four hop clusters, and a fifth would be the first real test of it. The prediction was written for a golden lattice at a lag of eleven. No golden lattice on this ladder reaches one, and the reason is a fact about the rule rather than about the search.

8 figures The two widest bands on the ladder, each cut at every rise. One block per band, one row per offset that wrecks anywhere on it, one column per rise, coarse on the left. A filled cell is a cut that wrecks, and its tone is the family left standing; a pale cell is a cut that recovers. The golden 8/13 band above changes its answer at three of its six offsets, 19 times in all. The Lucas 7/11 band below changes it nowhere: every cut that wrecks on it keeps the 7 family, at every offset and every one of its 124 rises.

The second band, cut whole

One band was cut at every one of its rises and came back with a transition region — a stretch where three offsets change their answer, in short islands with uneven gaps. The obvious question is whether that is a picture of bands or a picture of that band. The other wide band answers it.

8 figures Which offsets wreck across the Lucas 7/11 band. One row per offset and one column per rise, with a mark where a single removal at that offset wrecks the stem. The set is not the same at every rise: on this band one offset wrecks at only 27 of its 124 rises, in several separate stretches, while others wreck at all of them. So a census taken at one rise of a band and a census taken at another are censuses of different sizes, and every claim of the form "at every offset that wrecks" is quantified over a set the rise decides.

The wrecking set moves again

Which offsets wreck a stem was assumed to be a property of the lattice. On one band it turned out to be a property of the lattice and the rise, changing on nearly a fifth of that band's steps. On the second band it changes more, and one offset's wrecking is broken into five separate stretches.

8 figures What nine rises find on each band, against what all of them find. Two bars per band: the changes of surviving family a nine-rise design finds, and the changes the full sweep finds. On the Lucas band the two agree exactly, at none and none. On the golden band they agree that something changes and disagree about how much — 5 against 19 — because one step of that design is 16 rises and the band carries features one to three rises wide. A sample was never wrong about whether; it was wrong about how many.

When nine rises are enough

A coarse design was shown to be misleading on one band and it has been criticised on that ground ever since. On the second band it is exactly right, and the difference between the two cases is a property of the band rather than of the design — which is the awkward part.

8 figures The golden 5/8 rung swept at 27 rises, with each removal's cost. How far the first organ placed after a cut moves, at every rise the sweep visits, coarse on the left. Removing both walls costs far more than removing the larger one alone above a rise of 0.00998, and exactly what the larger one costs below it. The change happens in one step of the grid the ladder is named on: 163.59 degrees at 0.00998 and 9.14 degrees at 0.00997, which is a fall of 154.5 degrees for a change of one part in a thousand in the rise.

Where a slot loses a wall

Two rungs were reported to go free at 84 and 85 per cent of themselves — the second removal stops costing anything over the larger one alone. Three samples a rung cannot say whether that is a transition or a slope, and twenty-nine more say it is a transition one grid step wide.

8 figures All three bands cut at every rise, offset by offset. One row per wrecking offset on each band cut whole, one cell per rise, coarse on the left. A pale cell is a rise at which that offset's cut recovers and has no survivor; a dark cell is a cut that wrecks and keeps one of the band's own counted pair; a warm cell is a cut that keeps a family off the pair. The vertical rule on each row is that band's handover, where its two contact steps change places. The golden 8/13 band changes the family it keeps 19 times, the golden 5/8 twice and the Lucas 7/11 not at all.

The third band, cut whole

Two bands cut at every rise disagreed about whether the family a cut keeps ever changes, and three explanations were available for a difference between two things. The cheapest third band settles which of them survives, and it settles it against the account nobody was betting on.

8 figures The 17 rows of the exchange table, gathered by the lag they kept. One bar per surviving lag, its length the number of rows the table holds at that lag, with the hop that lag's stems keep written beside it. The hop is nearly constant inside a lag, so a correction fitted over 17 rows is fitted over four hops — which is the denominator that matters and is much smaller than the row count suggests. Adding a lag to the table is worth more than adding rows at a lag already in it.

A family that is a multiple

When a wrecked cut keeps a family that is not one of the lattice's counted pair, the first case looked like a rule: it was half of one of them. The second case is four times the other, which makes the rule a coincidence and leaves a weaker statement that is probably the true one.

8 figures Four accounts of which bands speckle, scored on the three cut whole. Each candidate explanation of why one band's cuts change the family they keep and another's do not, against what the three bands cut whole actually do. A tick is an account that puts that band on the side the sweep does. The branch the band sits on is right on all three; the size of the counted pair, the number of wrecking offsets and how much of its rung the band spans are each wrong on two. Three bands can eliminate and cannot confirm, and this eliminates three of the four.

The branch is what is left

Four accounts of why one band's cuts change what they keep and another's do not were written down before a third band was cut. Three of them are now wrong on a band each, and the survivor is the one with no mechanism behind it.

8 figures What nine rises find on each band, against what the whole band holds. The coarse design cuts nine rises of a band, evenly spaced in the logarithm of the rise, and asks whether the family a cut keeps changes anywhere. On the golden 8/13 it finds five of nineteen changes and on the Lucas 7/11 it finds none of none, so it had never been wrong about whether anything changes. On the golden 5/8 there are two changes and it finds neither, both of them at single rises with the offset recovering on either side. Its record on that question is now 2 of 3.

Nine rises were not enough

A coarse sample of a band had never been wrong about whether anything changes inside it, and that record was the argument for trusting a negative from it. The third band cut whole makes it two of three, and the missed feature is one rise wide.

8 figures Offset 5 on the 5/8 band, and the change that cannot be placed. Every rise of the band, coarse on the left, with offset 5's cut drawn at each: pale where it recovers, dark where it wrecks and keeps 5, warm where it wrecks and keeps 20. It keeps the off-pair family at two rises above the handover and then does not wreck again for 34 rises, so its return is bracketed across a stretch that contains the handover. The flag that says a change sits at a handover fires here for the first time, and it is a statement about where the offset stops wrecking rather than about the handover.

A change with nowhere to be

The claim this whole thread rests on is that a survivor does not change where the two contact steps change places. Nineteen located changes never put one there. The twentieth is flagged at a handover, and it is flagged because the offset stops wrecking for thirty-four rises.

8 figures The Lucas 7/11 rung between 0.007 and 0.006, cut at every rise. One row per offset, one column per rise of the search, coarse on the left. A pale cell is a rise at which that offset's cut recovers; a dark cell is a cut that wrecks and keeps a lag of 7; a warm cell is one that keeps 11. The nine intermediate rises of the ten-thousandth grid were cut first and the nine of the hundred-thousandth grid inside the one bracket they opened. The change sits between 0.0064 and 0.00639, one step of the grid the ladder names its rises on, and above it no cut keeps 11 anywhere.

One step of the grid, again

A search of the fine end stepped from one rise to another a hundred grid steps away, and the family a cut keeps changed somewhere between. Cutting the rises in between puts the change inside one step, with the lattice identical on both sides.

8 figures The two shortest hops across the Lucas 7/11 rung, and where they cross. The lengths of the two shortest hops at every one of the 378 rises of the rung, fine on the left, computed from the settled divergence rather than measured off a cut. They cross exactly once, between 0.00804 and 0.00803, which is this rung's handover. The rise at which a cut first keeps the longer family is 164 steps of the grid further down and is marked separately; nothing happens to either length there.

The hops cross once

Walking a whole rung at the grid its rises are named on costs a few hundred stems and no cuts at all, and it answers a question nobody had asked: whether a rung has one handover or several. It has one, and the ladder had recorded it in the wrong place.

8 figures Two rises on the Lucas 7/11 rung, each located to one step of the grid. The whole of the Lucas 7/11 rung, coarse on the left, with the two rises this round located. The two contact steps change places between 0.00804 and 0.00803, found by walking all 378 rises of the rung and evaluating hop lengths, with no cut stems at all. The family a cut leaves standing changes between 0.0064 and 0.00639, found by cutting 89 stems. They are 164 steps of the grid apart, which is 45 per cent of the rung's whole span, so the geometry crossing is not what moves the survivor.

Two rises far apart

The whole handover thread rests on the claim that where the two contact steps change places is not where the survivor does. On one rung both rises are now located to a single step of the grid, and they sit forty-five per cent of the rung apart.

8 figures What each offset keeps, above and below 0.00639. Every offset that wrecks anywhere in the search, with the family it keeps above the transition and below it. Offsets 5, 6, 7, 9 keep the same family at every rise they wreck at, on both sides. What changes at 0.00639 is that offset 9 begins to wreck at all, and what it keeps is the larger of the counted pair. One offset does change its own answer, and it cannot be located, because it does not wreck at the rises in between.

An offset that arrives

The rise where a lattice first keeps a new family is located to one step of the grid, and what happens there is not what the question assumed. No cut changes its mind: a cut that was not wrecking starts, and what it keeps is the new family.

4 figures Where every wrecked run finishes, rung by rung. One row per rung of the ladder, one mark per cut cell of the slot design, placed at the divergence that run ended on. The open mark on each row is the divergence the intact stem of that rung settles to. On the Lucas 1/3 and golden 2/3 rungs every wrecked run finishes at the same value; on the golden 5/8 they finish at 13 values spanning 215 degrees. 27 cells recover, and each of those finishes at its own settled divergence to within 0.03 degrees.

The column nobody read

Every cell of the slot design carries where its run finished as well as how far its first organ moved. One rung's worth had been plotted and called unusable. Reading all eight says the endpoint is exact on two rungs, wanders on six, and is worst on the one it was read on.

8 figures Wrecked endpoints against the settling table's destinations. The upper lane is the 15 divergences the settling table reaches, grown from intact stems started at nine arbitrary angles across four falloff exponents and eight rises, with no cut anywhere in them. The lower lane is where the slot design's 63 wrecked runs finish, read without handedness so that a run ending at 209 degrees is placed at 151. 29 of them sit within 1 degree of a destination and 34 do not. The two measurements share no run and no design, so the agreement is not a construction.

A wrecked run goes somewhere

Where a wrecked stem finishes was called unstable. Half of them finish within a degree of a destination measured from intact stems started at arbitrary angles — two tables that share no run, no design and no question.

8 figures The twelve wrecked runs that finish at a half turn. Wrecked runs whose final divergence is a half turn, which is two files of organs rather than a spiral. Every one is on a rung at the coarse end of its branch, where the front is short enough that removing one organ reaches past it. The value is within 0.35 degrees of 180 on eight of the twelve.

Two files, and a way back

Twelve wrecked runs finish at exactly a half turn, which is a pattern with no spiral in it, and every one is at the coarse end of the ladder. Three finish at the divergence they would have had anyway, after being thrown a hundred degrees off it.

8 figures The Lucas 4/7 band cut at every one of its 86 rises. One row per wrecking offset on each band, one cell per rise, coarse on the left, with each band's width in proportion to the rises it holds. A pale cell is a rise at which that offset's cut recovers and has no survivor to report; a dark cell keeps one of the band's own counted pair and a warm cell keeps a family off it, so a change of answer is where the tone changes. The Lucas 4/7 is the test the branch account most needed: 86 rises, 774 stems, 144 wrecks across two offsets and not one change of the surviving family. Across the one bands drawn, 144 cuts wreck of 774 grown. The dashed rule on each row is that band's own handover, where its two contact steps change places.

The fourth band, cut whole

Three bands cut at every rise left one account of which bands change their answer standing, and the account was the one nobody had a reason to prefer. The band that would have killed it has now been cut, and it did not kill it.

7 figures The two bands that answer nothing, 586 stems and no wreck anywhere. The g35 and L34 bands cut at every rise, with one row per offset tried and one cell per rise, coarse on the left. A pale cell is a cut whose stem recovers and a mid cell is one that neither recovers nor leaves an orbit to count; the warm tone means a cut that wrecks, and it appears on the reference row below, which is offset 4 of the Lucas 4/7 wrecking at 83 of its 86 rises. g35 grew 490 stems across 70 rises at seven offsets and L34 grew 96 stems across 16 rises at six offsets, and wrecked none of them, so this is a zero that was measured rather than a band nobody cut. The dashed rule on each block is that band's own handover, where its two contact steps change places.

Two bands that wreck nothing

The census that reads a band refuses two of the six, and the refusal is correct: a band with no wrecked cut has no surviving family, so it has no answer to change. Cutting them anyway turns a refusal into a measurement, and the measurement has a third tone in it that the census cannot see.

7 figures Four accounts scored on the four bands of six that can score them. One column per band of the ladder and one row per candidate account of why a band's wrecking cuts change the family they leave standing. Each cell is what that account predicts of that band, in plain type where the band agrees with it and pale where it does not, and the header of each column is what the band actually does. The two bands with no wrecked cut are shown silent, because a band with no surviving family has no answer to change; over the four that can answer, the branch the band sits on is right 4 times of 4 and the other three are wrong twice each. A warm block marks each cell where the account and the band disagree.

Six bands, one table

Every rung of this ladder that carries a handover now has a band grown on it and cut at every rise it holds, and four accounts of which bands change their answer are scored on all six at once. The survivor is right on every band that can test it, and the same table read one cell differently kills it.

8 figures What nine rises find on six bands, against what those bands hold. The coarse design cuts nine rises of a band, evenly spaced in the logarithm of the rise, plus both ends and the handover, and asks whether the family a cut keeps changes anywhere on it. Scored on six bands it is right about whether on five and wrong on the 5/8, where it finds none of 2. On the one band where it finds anything it finds 5 changes of 19, so its record is a record about whether and never about how many. The figure under each band's name is that design's step on it, in rises.

The coarse design scored

Every claim this thread has made about an uncut band rests on a sample of nine rises, and its record was the argument for trusting it. Six whole bands close that record, and two of its six correct verdicts are correct only because there was nothing on those bands to find.

7 figures How broken the wrecking is on four bands, from 2.43 rises a stretch to 126. Every offset that wrecks on every band that wrecks, placed on a logarithmic axis of rises wrecked per separate stretch — the smaller the number, the more broken the wrecking. A filled mark is an offset that wrecks at every rise of its band and an open one comes and goes. The range runs from 2.43 on g58 offset 3 to 126 on g813, a factor of 52, and two of the four bands have no offset that wrecks everywhere at all. A band with no offset that wrecks at every rise is said so at the right.

A wrecking set with a range

Which offsets wreck a stem was taken to be a property of the lattice, and every band cut whole has found it to be a property of the rise instead. Six bands turn that replication into a measured range, and the range is a factor of fifty-one.

8 figures Six rungs walked at the grid, 10 crossings between them. Each row is one rung, drawn from its coarse end on the left to its fine end on the right and scaled to its own width so positions inside different rungs can be compared. The shaded stretch is the band the ladder grows around the rise it recorded as that rung's handover. five of the six rungs carry exactly one crossing, and on each of those the whole stretch a second one could sit in has been walked at the grid and closed at both ends. The 3/4 rung of the Lucas branch carries five, spaced 22, 12, 15 and 20 grid steps apart, and its band holds three of them. The rule on each row is a located crossing.

Five rungs walked

Six rungs of the ladder carry a handover and only one of them had ever been walked at the resolution its rises are named on. Walking the other five costs 1,224 grown stems and no cuts at all, and it returns a crossing count per rung — five ones and a five.

7 figures A reading that steps against an ordering that slides, on six rungs. Every hop length here is arithmetic on the settled divergence, and the settled divergence is a mean of angles read off a lattice placed on a fixed set of azimuths — so it is read in steps rather than continuously. Between two steps the ordering slides with the rise; at a step it jumps. The bar is how much a jump is worth in slides on each rung, measured inside the stretch that was walked at the grid so that the six are read over comparable regions. The account is a threshold at one and it is right on all six: the 3/4 rung sits at 12.71 and carries five crossings, and on three rungs the reading does not step inside the window at all.

One crossing or two

One rung of the ladder changes hands five times where the other five change hands once, and the difference is not in the geometry. It is a divergence read in steps against an ordering that slides, and the account is a threshold at one that is right on all six rungs.

7 figures Six handovers relocated, in steps of the sweep that recorded them. The ladder finds a handover by stepping at a ratio of one per cent, which never lands on the grid the rises are named on, so a recorded handover is the nearest rise the sweep visited to a crossing nobody had located. This is the difference, in units of the sweep's own step at that rise: 0.082 to 0.835, mean 0.374. All six are positive and all six are inside a single step of the sweep, and both of those are predictions rather than summaries: the ladder reports the first rise it visits at which the ordering has already changed, so the rise it records must sit on the fine side of a crossing and within one of its own steps. The dashed rule is one step of the sweep, which is the bound the sampling predicts.

The handovers corrected

Six recorded handovers, relocated to the grid against where a one-per-cent sweep put them: all six sit on the fine side of a crossing and all six inside a single sweep step. Nothing about the rung explains the size of the discrepancy, which is what a sampling artefact is supposed to look like.

7 figures 19 changes and 2 at the coarse step, 21 and 2 at half of it. The changes of surviving family each band makes, at the step the sweep has always used and at half of it. The golden 8/13 band goes 19 to 21 and the golden 5/8 band 2 to 2. Halving the step cannot lose a change, since the fine grid holds every coarse rise with the same answer; what it can do is find one, and it does so on one band and not on the other. So whether a tally is a floor or a total is a property of the band rather than of the sweep — a floor where there are islands narrower than the step, and exact where there are none.

A count or a floor

Nineteen changes of surviving family on the widest band have been quoted as a number since the band was cut, with nothing to say whether a finer grid would find more of them. Halving the step finds twenty-one there and nothing at all on the next band along.

8 figures The rise at 0.005845 the coarse grid steps over, under offset 6 at both steps. Offset 6's cut drawn at every rise between 0.00596 and 0.00573, coarse above and fine below, with a cell per rise coloured by the family the cut leaves standing and pale where the cut recovers. It changes its answer 3 times at the coarse step and 5 times at half the step, and the gain is the single rise 0.005845, where the 4 family is kept with 8 on both sides. That island is 0.86 parts per thousand of the rise wide, against 1.47 for the smallest step the coarse grid takes anywhere on the band, so no coarse rise could have landed on it.

New islands or old edges

Halving a band sweep's step found two more changes of surviving family, and there are two quite different things they could have been. Every coarse change and every coarse island turns out to be carried by exactly one fine one, so the extra pair is a rise the coarse grid stepped over rather than a boundary it misplaced.

7 figures The 125 steps of the golden 8/13 band, in units of the grid they are rounded to. The sweep's rises are held to five decimal places, so every step it takes is a whole number of units of that fifth place. Each bar is how many of this band's 125 steps are that many units: 99 of 1, 26 of 2. One unit is 1.653 parts per thousand of the rise at the handover of 0.00605, against a nominal step of 2.00, so a sweep asked for at one part in a thousand would land on the same rise twice and the finer grid has to be laid at 7 places instead.

The last unmoved setting

Halving a band sweep's step is only a halving if the sweep steps where it says it does, and this one does not: its rises are rounded to five decimal places, so at one handover the grain is 1.65 parts per thousand against a nominal step of two. Moving the last setting nobody had moved found the setting was never what it was called.

8 figures

Where the angle comes from

137.5° is not a constant of nature. It is where a rule settles — a rule that places each new element as far as it can from the others, and which settles somewhere else when one parameter changes.

The rule, 26 steps in, at a growth of 0.40. The next primordium goes where the repulsion is least — the marked minimum at 216°. Nothing in the rule refers to any particular angle.

The angle is an output

137.5° is not a constant of nature. It is where a rule settles — a rule that places each new element as far as it can from the ones already there, contains no reference to the golden ratio, and reaches the same answer from starting angles a hundred and sixty degrees apart.

8 figures What the model settles on, against how fast the meristem grows. A broad golden branch, a transition, and then the two-whorl regime at exactly half a turn. 9 of 27 converged settings land within 4° of the golden angle; 15 land more than 20° away.

The bifurcation diagram

Sweep the one parameter of the rule and the settled angle traces a diagram — a broad golden branch, a transition, and a two-whorl regime at exactly half a turn. The famous constant is one branch of it, which is a more useful thing to know than the constant.

8 figures Two runs of the same rule from unrelated starting angles. Both settle at 137.0°, within 0.5° of the golden angle, from seeds 166° apart.

Droplets with no biology in them

Douady and Couder dripped magnetised ferrofluid into a dish of silicone oil, and got spiral phyllotaxis with Fibonacci parastichy numbers out of a system containing no cells, no genes and no plant. That is the strongest evidence the pattern is physics — and the clearest warning about what a model can claim.

8 figures Where this implementation stops converging. Below about G = 0.18 the settled angle wanders over 113° however long the run. That is the model's limit, not a fact about plants.

Where the model stops

Below a growth parameter of about 0.18 this implementation does not converge — the settled angle wanders over a hundred degrees however long the run. That is the range where the literature says the interesting behaviour lives, and it is worth a figure rather than a quietly chosen axis.

8 figures What a two-organ cut does at each rise of the 2/3 rung. At every rise the coarse rung is a lattice on, all 36 arrangements of two organs removed, with the ones that never repair counted and split by where they end up. 22 of 324 cuts across the rung reverse the stem's handedness onto the mirror of the divergence they were cut from. 40 fall instead into a cycle whose mean is half a turn, which the lattice they came from has no number for. 4 rises give only the first, 4 give only the second, and at a rise of 0.075 both happen in the same table, which is what says the fate belongs to the cut and not to the rise.

Half a turn, four at a time

At four of the nine coarse rises no wrecked cut reverses. What those stems do instead is stop settling: they repeat 171.09°, 269.53°, 189.14°, 90.23° without end, which adds to two whole turns over four organs. The mean is exactly half a turn and a counter finds four files where the lattice had three.

8 figures Two paths down the same tree. Both start at the same first fork. Keeping the larger family every time reaches 137.473°; one different choice reaches 99.549°. Neither angle is in the arithmetic — both are limits of a path.

The tree and the attractor

The dynamical model settles on the golden angle over a range of one parameter and on the Lucas angle outside it, and it cannot reach the low-growth end at all. The lattice tree reaches everywhere, has no dynamics in it, and produces the same two angles as limits of two paths. Two routes, one pair of numbers.

7 figures A stem grown at 42 nodes per rung. 167 nodes, each placed where the repulsion from the ones below it was least, with the rise falling from 0.2 to 0.0045. Counted blind in a sliding window the pattern walks 1/2 → 2/3 → 3/5 → 5/8, and the marks are where its answer changed.

A pattern with a rate

Every lattice in the essays before this one is a static object indexed by a parameter, and a plant is not. Put the rise on a clock, place each node where the repulsion from the ones below it is least, and the object that comes out has a history — which is the first thing here that could disagree with the ladder.

8 figures The lag between two rates, measured against the rate. Each dot is one rate: the mean gap between where the grown pattern changed its count and where the static ladder puts that transition, in rungs. Over rates from 9 to 135 nodes per rung the worst is 0.087 of a rung. A lag of one rung would put a dot on the top line.

The lag that is not there

A pattern built out of its own history should hold its old parastichy pair past the point where a fresh lattice would have changed, and the gap should grow as the shoot is hurried. Over a fifteenfold range of rate it does not — every transition lands within a tenth of a rung of where the static ladder puts it.

7 figures One seed, two rates, two ladders. Both stems begin as forty nodes of Lucas lattice at a rise of 0.12. At 65 nodes per rung the divergence stays at 99.5° and the counts walk 1/3 → 3/4 → 4/7 → 7/11. At 131 it leaves for 137.7° and walks 1/3 → 2/3 → 3/5 → 5/8 → 8/13 instead.

The rate decides the branch

Forty nodes of Lucas lattice, carried down to the same fine rise twice. Hurried, the pattern holds the Lucas ladder through three more forks at 99.5°. Given room, it abandons it at the first fork it reaches and walks to 8/13 at 137.5°. Same seed, same rise, two ladders — with a threshold between them at about ninety nodes per rung.

8 figures Fibonacci at a rise of 4.8e-3, asked two ways. Choose a divergence at random and a static lattice at this rise gives a consecutive Fibonacci pair 10.8 per cent of the time. Start coarse at a divergence nobody chose, grow the stem down to the same rise, and it is 100 per cent of 16 runs. The geometry is not generous; continuity is.

Continuity from a coarse start

At a fine rise, one divergence in seven gives a Fibonacci pair. Grow a stem from a coarse start at a divergence nobody chose, down to that same rise, and sixteen runs out of sixteen end on 8/13. The earlier work's interpretation was that continuity does the work; this is the measurement it never had.

8 figures What becomes of a seeded branch at 65 nodes per rung. Each bar is 10 runs at one amplitude, divided by what the blind counter found at the top of the stem. With no noise this rate ends on the Lucas branch. Placement noise displaces the node after the rule has chosen; field noise perturbs the energy the rule chooses over. Across 160 runs, 1 reached the Fibonacci branch with the lattice intact.

Noise is not a slow rate

A stem seeded on the Lucas branch keeps it below ninety nodes per rung and abandons it above — which invites the objection that a real apex's fluctuations would knock it off regardless. Measured across a hundred and sixty runs of two independent kinds of noise, one escapes, at the amplitude where the pattern is already coming apart.

7 figures A lattice survives about 1.8° of scatter, whichever way the noise arrives. The largest divergence scatter at which a run is still a lattice, from each kind of noise at the largest amplitude that leaves one. The two routes share no code below the placement rule: one displaces the node after the choice, the other perturbs the energy the choice is made over. They agree to 0.33°.

Two degrees of scatter

A lattice tolerates about two degrees of wander in its divergence angle, and two kinds of noise sharing no code agree on the number to within a third of a degree. It is not a constant: carried finer, the same stem survives 0.8°, and the tolerance tracks the band of angles that produce its pair at all.

5 figures Only noise that arrives before the choice can change what is chosen. Intact runs only, from the whole amplitude sweep. Placement noise displaces the node after the rule has picked an azimuth: 44 runs, none of which changed branch at any amplitude that left a lattice. Field noise perturbs the energy profile the rule picks over, so it can move the minimum into a neighbouring gap: 1 of 66 did.

Where the noise gets in

Ninety runs of noise applied after the rule has chosen, and not one changes branch. Fifty-six of noise applied to the choice itself, and one does. Only a disturbance upstream of the decision can restructure which nodes are neighbours of which — which is what a branch is.

6 figures Which neighbours decide where an element goes. Each line is one exponent: how much each shell of neighbours makes the energy profile vary around the circumference, divided by what the nearest shell contributes. At p = 0.5 the nearest shell leads the next by a factor of 1.1 and a node is placed against the whole neighbourhood at once. At p = 3 it leads by 9.7e+3, and a node is placed against its immediate neighbours — which is what a lattice is.

How far a primordium reaches

The placement rule's repulsion falls as an inverse cube because that is what two magnetised droplets do, and nothing about a plant supplies the exponent. Asking what it controls produced one tidy wrong answer and one measured right one — and the difference between them is the difference between a total and a variation.

7 figures A rule too long-ranged makes no pattern; every shorter one makes the same pattern. Each dot is 4 runs from a coarse start at one exponent, separated by 0.2° of placement noise. Below p ≈ 1.1 the divergence scatters by tens of degrees, which is what an arbitrary sequence gives. From p = 1.25 to p = 8 — a sixfold range — every run ends on 8/13 with the scatter between 0.75° and 1.06°.

The exponent that barely matters

An unchecked claim, repeated since the first essays, held that the repulsion's falloff exponent hardly changes the answer. It could not be tested, because the function that would have taken it never passed one down. Tested at last, it is true on a disc — by three and a half degrees across a sixteenfold range — and on a stem it decides whether there is a pattern at all.

8 figures The same rule at p = 1, cut off at two distances. The top 220 nodes of two stems grown by an identical rule whose energy does not converge. Allowed to see 3/√h neighbours it produces 8/13 at 137.62° with 0.58° of scatter — a lattice no test here would question. Allowed 12/√h it produces 44° of scatter and no pattern. The truncation was doing the work.

A window that makes a pattern

A rule whose energy has no well-defined minimum produces a clean 8/13 lattice at 137.62°, with half a degree of scatter, when its neighbourhood is cut at three node spacings. Let it see twelve and the pattern is gone. Every simulation of this kind truncates something, and truncation manufactures exactly the result it is used to look for.

8 figures Three cut-offs at the same nominal width of 3 spacings. The weight the interaction is multiplied by, against distance. They halve at 2.08 (exponential), 2.50 (gaussian), 3.00 (hard) spacings — so a rule described as "cut off at 3 spacings" is three different rules until the falloff is named. Every later figure is read in half-weight radii for that reason.

A neighbourhood is a hypothesis

Every simulation of this kind stops summing somewhere. The earlier work found that where it stops decides what pattern comes out — so the stopping place is not a detail of the program but a claim about how far a primordium's influence reaches, and it should be written down as one.

7 figures The landscape the rule chooses over, at a cut-off of 3 spacings. One height of an ideal lattice, swept around the circle. The exponential cut-off hands the rule a smooth landscape; the hard one hands it a landscape with steps, because a neighbour enters the sum as the candidate slides past it. Halving the sample resolution multiplies the largest jump between neighbouring points by 1.99 on the smooth curve and by 1.19 on the hard one — which is the definition of the difference, since a smooth function's steepest step is bounded by its derivative and a discontinuity's is not. An argmin taken over steps is pinned to the steps.

A hard edge is not a falloff

The prediction was that cutting the neighbourhood at three spacings would reproduce the pattern truncation had manufactured. It does — if the cut is smooth. A hard cut at the same distance produces no pattern at any width, and the reason is that it is the only one of the three whose neighbour set depends on where the candidate is.

7 figures Two shapes, two ranges, one contrast. The exponential's lattice ends at 3.63 spacings and the gaussian's at 2.25 — ranges 47% apart — and at those two ranges the contrast is 5.74 and 6.09, 6% apart. The band is what the exponent route leaves: 3.98 at p = 1, where the uncut rule makes nothing, and 7.63 at p = 1.25, where it makes a lattice.

Two shapes, one threshold

Read in the same unit, an exponential falloff and a gaussian one disagree about where the lattice ends by half. The quantity they agree on turns out to be one the earlier work measured for an unrelated reason — and it agrees with a bracket left by a sweep of a completely different parameter.

7 figures Which lattices survive a fifth of a degree of noise. The share of runs that still have a lattice. The prediction was that a cut-off at this range would be as fragile as the truncation it replaces; it is not. Stating the neighbourhood as a function of distance did not merely make the old result honest — 100% of runs survive against 33%, at a scatter an inverse-cube rule cannot be told from.

The fragility belonged to the window

A pattern that exists only because the rule cannot see far was expected to be held together by that cut, and to fall over when nudged. It does — while the cut is a loop bound. Written down as a falloff at the same range, the same rule keeps every run under the same nudge, at a scatter an inverse-cube rule cannot be told from.

8 figures Placements that went to a different minimum, per thousand. The rule's own counterfactual, run beside it: where would this node have gone with the noise taken away and everything else left alone? Placement noise displaces the node after the argmin, so the answer is always "here" — 0.2°, 0.4°, 0.8° all give zero. A jostle and a field perturbation are upstream of the choice and change one or two placements in a thousand while the lattice is still intact.

Which minimum was chosen

The rule takes an argmin, so there are two completely different things noise can do to it: move the answer, or move the question. One of them can change what is chosen and the other cannot, ever — and the difference is exactly zero against one or two placements in a thousand, at amplitudes where every other measurement says the two are identical.

7 figures Where each kind's lattice gives way. The largest amplitude at which every run still has a lattice, and the scatter it produces there. The amplitudes are incomparable — field 0.015 (fraction of the barrier), jostle 1 (degrees of azimuth), placement 0.8 (degrees of azimuth) — and the scatters agree to 19%. The boundary belongs to the pattern rather than to the disturbance: a lattice fails at about a degree and a half of scatter, and which of three mechanisms produced it does not move where.

The boundary belongs to the pattern

Three kinds of noise, in three incommensurable units, destroy a lattice at the same place — about a degree and a half of divergence scatter. The earlier work measured that of two kinds and called it a scale rather than a constant. With a third it looks less like a coincidence and more like a property of what a lattice is.

7 figures The memory belongs to the rise, not to the lattice. The lag-one correlation of a noiseless rising stem, against how fast it climbs the ladder. Below about sixty nodes per rung it is negative; above it, 0.54, 0.74, 0.58, 0.58, 0.55 — flat across a fivefold change in rate. The horizontal line is the same rule with the rise held FIXED, where the correlation is -0.68. So the +0.74 the earlier work called the sequence's own memory is the pattern chasing an equilibrium that is moving under it.

A shoot too fast to remember

Sweep the rate at which a stem climbs the ladder and the correlation between one divergence and the next changes sign — negative below about fifty-five nodes per rung, positive above it, with the flip inside one step of the grid. The instrument the earlier work proposed is unavailable on a fast shoot, and nothing said so.

8 figures A window that fits inside a rung. Stems that climb the ladder at four rates, read over a window at the fine end. The condition is a ratio: the window has to be shorter than a rung. 250 internodes at 130 per rung is 1.92 rungs and agrees on 0 of 3; 400 internodes at 130 per rung is 3.08 rungs and agrees on 0 of 3; 250 internodes at 260 per rung is 0.96 rungs and agrees on 3 of 3; 400 internodes at 260 per rung is 1.54 rungs and agrees on 1 of 3; 250 internodes at 520 per rung is 0.48 rungs and agrees on 2 of 3; 400 internodes at 520 per rung is 0.77 rungs and agrees on 3 of 3; 250 internodes at 1040 per rung is 0.24 rungs and agrees on 3 of 3; 400 internodes at 1040 per rung is 0.38 rungs and agrees on 3 of 3. Read over the whole stem instead, every rate returns nothing — 0 of 3, 0 of 3, 0 of 3, 0 of 3 — because the quantity the comb is periodic in changes as the pattern climbs.

A window inside a rung

A stem that climbs the ladder has no comb in it at any rate, because the quantity the comb is periodic in changes as it goes. Read a window instead and it comes back, on one condition: the window has to be shorter than a rung — which makes the shoot's rate the thing that decides whether a plant can be asked.

8 figures Two windows on a shoot at 400 nodes per rung. A stem grown at 400 nodes to the rung with a disturbance of 0.25, its rise falling from 0.4 to 0.004 over 1914 nodes. The upper window is the last 250 internodes — where a count would be made on a real plant — and the lower is the same length shifted down 125. The upper reads 8/13; the lower reads 8/13. The verdict is agree, and the window holds 0.63 of a rung.

Two windows on one stem

A pair read off a climbing shoot can only be read through a window, and a window can straddle a transition. Read a second window half a length lower and the outcomes fall into four kinds — and agreement between them never happens on a shoot whose rung is shorter than the window, which turns the most awkward of the four refusal causes into something a reading can certify.

8 figures What the finer grid does to the rises already published. The two rises this site has argued from and the one it published as having no answer, each read at both azimuth grids, five stems apiece. A filled mark agrees with the position counter, a half mark contradicts it, an open mark is a refusal. At 0.013 and 0.005 the readings are identical at both grids, so nothing already written depends on the sample count. At 0.008 they are not: 384 azimuths gives 5/8, 5/8 and 1152 gives 8/13, 8/13, against a counter that says 5/8. That rise was chosen in the earlier work because the three shortest lattice offsets there are within a fifth of each other, and a stem with no answer answering differently on a different grid is the object behaving as it was said to.

What a sample grid decides

The rule takes its minimum over 384 sampled azimuths, and that number has been a constant since this site's first commit. Tripling it changes nothing at the two rises the collection argues from — five runs of five, identical readings — and changes which answer appears at the one rise published as having no answer. A parameter of the program, measured rather than assumed.

8 figures A cut eight back is never undone. The divergences of a stem whose organ eight places back was removed, against the same stem uncut. It never returns. What it settles into repeats exactly every 8 organs — 47°, 96°, 137°, 273°, 138°, 271°, 230°, 272° — and holds that cycle for the whole 300-organ run, with a mean of 186° and a spread of 83°. A rule that corrects a displacement does not correct a deletion.

A rule that cannot heal a hole

The placement rule corrects itself against a displacement — that is what the lag-one correlation of −0.6 has been saying since it was measured. It does not correct itself against a deletion. Which organ is removed decides whether the stem is back on its lattice in twenty-four organs or never, and the boundary between the two is sharp, reproducible and in the middle of the front.

8 figures A cut four back is never undone. The divergences of a stem whose organ four places back was removed, against the same stem uncut. It never returns. What it settles into repeats exactly every 8 organs — 139°, 137°, 138°, 138°, 138°, 271°, 231°, 271° — and holds that cycle for the whole 300-organ run, with a mean of 186° and a spread of 60°. A rule that corrects a displacement does not correct a deletion.

The pattern the cut leaves behind

A stem that never recovers from a removal is not disordered. Its divergences settle into a cycle of eight angles and repeat it exactly for the rest of the run, and a counter reading the positions calls the result 8/16 — a two-jugate lattice. The rule has a second attractor at the same growth parameter, and an ablation is how you get to it.

8 figures What a lattice survives depends on the colour of the disturbance, sixfold. five stems for each of seven disturbances at each of six displacements, every stream normalised by its own measured spread so that a displacement of half a degree is half a degree in every row. A filled mark is a stem that still has a lattice — a divergence scatter under 2° — and an open one is a stem that does not. Independent errors survive to 0.5°; errors that remember the last one to 1.5°; errors inherited from the contact neighbours only to 0.25°. The number beside each row is the scatter a protractor would record where the lattice is standing, and it is the quantity that explains the table: what destroys a lattice is not how far an organ moves, but how far it moves relative to the organs it is placed against.

A disturbance the organs share

This collection has put three kinds of noise into the placement rule and found the lattice fails at about the same recorded scatter whichever kind it was. None of them asked what happens when the displacements are correlated between organs. At equal displacement per organ, a lattice survives three times as much of a disturbance the organs share — and what a protractor records is the part they do not.

8 figures What a lattice survives depends on the colour of the disturbance, sixfold. five stems for each of seven disturbances at each of six displacements, every stream normalised by its own measured spread so that a displacement of half a degree is half a degree in every row. A filled mark is a stem that still has a lattice — a divergence scatter under 2° — and an open one is a stem that does not. Independent errors survive to 0.5°; errors that remember the last one to 1.5°; errors inherited from the contact neighbours only to 0.25°. The number beside each row is the scatter a protractor would record where the lattice is standing, and it is the quantity that explains the table: what destroys a lattice is not how far an organ moves, but how far it moves relative to the organs it is placed against.

The disturbance that travels

If a lattice survives three times the displacement when the organs share it, then a disturbance passed between the organs that actually touch should be the gentlest of all — it is correlated at exactly the offsets the rule places against. It is the harshest. Half the displacement destroys what independent noise leaves standing, and the reason separates two things that had been one.

8 figures Both edges of the front heal; the middle of it does not. The same removals, followed for 300 organs each. A cut one to three places back is undone within fifty organs and a cut ten to thirteen places back within sixty. A cut in between is never undone: the divergence sequence settles into an exactly repeating cycle of 4 or 8 angles and holds it for the rest of the run. The rule corrects a displacement and cannot correct a deletion.

A front with no middle

Take one organ out of a stem and the pattern sometimes never comes back — but that was measured on a front thirteen organs wide, where five of the thirteen offsets are beyond repair. Repeat it on a front of five and every single ablation heals. The band that cannot be undone is not a number the rule carries; it is what two fixed edges leave over.

8 figures On a rung the response is a run of offsets; near a transition it has a hole. One row per rise, from 0.02 at the top to 0.005 at the bottom, and one column per offset: the organ one place back at the left, 14 places back at the right. A cell is filled where removing that organ moves the next organ by more than 2.5°, and empty where it does not. On a rung the filled cells are a run from one to the larger parastichy number — 8, 12, 13 at the pairs shown on the left. Between a rise of 0.02 and 0.0065 the run ends at 5 and one more cell is filled at 7, with the offsets between them quiet to under a degree. That isolated column is one place inside the larger number of the pair the stem is climbing towards.

The response with a hole in it

Removing an organ is felt out to the larger parastichy number and no further — that is the intervention's headline, and it holds in the middle of a rung. Swept towards a transition the run of felt offsets stops early and one lone offset past it comes alive, with three quiet organs in between. The lone offset is one place inside the count the stem is about to have.

8 figures Two answers 138° apart, and one organ holding the second one up. The repulsion the rule minimises, around the circumference of a stem at a rise of 0.008, at the height the next organ will sit at. It has two low points 138.3° apart: the slot the next organ takes, and the slot the organ after it will take. The runner-up is 13.6% higher. The organ 13 places back carries 14.6% of the energy at the winning slot and twelve places back carries 16.3% at the runner-up — and that is more than the gap, so taking that organ away makes the runner-up win and the next organ appears a whole divergence away. Neither guard is a contact of the organ being placed; twelve is one place inside the larger number of the pair this stem is climbing towards.

The organ that guards the second slot

An organ twelve places back is the furthest of any from where the next one goes, and removing it moves the next one by a whole divergence. The reason is that the rule's profile has two low points rather than one, the second is the slot after next, and that organ is holding it up. The comparison between what it holds up and how far behind it is decides the whole thing.

8 figures The block a wrecked stem settles into is the count it was cut from. A stem is cut in the middle of its front and followed for 300 organs. It does not come back to its lattice; what it does instead is repeat a fixed sequence of divergences exactly, to the resolution of the azimuth grid. Each row shows that sequence twice over, one bar per organ, drawn to the same scale. At the 5/8 rung the sequence is five angles long and advances -36.1° per block; At the 8/13 rung the sequence is eight angles long and advances 22.5° per block. Every block is the smaller number of the pair that was cut, so the second attractor carries the first one's count — and every precession is one part in twice the block, which makes each of these a two-jugate arrangement with 10 and 16 rows.

The block is the count it was cut from

A stem that never recovers from an ablation settles into a repeating block of eight angles precessing by 22.7°. Eight was also the smaller parastichy number of the lattice that was cut, which left two possibilities and no way to choose between them. Cut a stem one rung coarser and the block is five.

5 figures A second cut moves the next organ, and does not move the boundary. Every pair of organs that can be taken out of a settled stem at a rise of 0.032, where the pattern is 3/5. A row is the offset of the nearer organ removed, counted back from the tip; a column is how many further places back the second one sits. The shade is how far the next organ ends up from where it would have been, from under a degree in the palest cells to a half-turn in the darkest. The run of shaded rows ends at 5, which is the larger parastichy number, and every row below it is blank across the whole width — the largest displacement anywhere past the front is 2.34°. So the nearer of the two organs decides whether the removal is felt at all, and the second one, wherever it is put, cannot make the pattern notice an organ it was not going to notice.

A cut of two organs

One organ removed from a stem is felt out to the larger parastichy number and no further, and at the coarsest arrangement the stem always repairs itself — so the one rung where the interesting prediction could be checked had no experiment that could reach it. Two organs can. The second cut brings a parameter with it, and that parameter turns out to be a control.

8 figures The wrecked stem is the lattice it was cut from, wound the other way. The divergence of each organ placed after two were removed from a settled stem at a rise of 0.032, over the 220 organs following the cut. The upper line is the divergence the undisturbed stem holds, 139.6875°; the lower is 220.3125°, which is 360° minus it and therefore the same lattice with the opposite handedness. The sequence is thrown by the cut, wanders for a few dozen organs, and settles on the second line to four decimal places — 220.3125° against 220.3125° — where it stays. A counter shown the positions afterwards returns 3 and 5, the pair the stem was cut from. Nothing about the pattern has been lost; its chirality has been reversed, which at this rung a single removal cannot do.

The stem that changed hands

A stem that never recovers from an ablation is supposed to end up somewhere worse than it started — a repeating block of angles, a pattern with the wrong counts in it. At the coarsest arrangement it ends up somewhere that is not worse at all: at 220.3125°, which is 360° minus the divergence it was cut from. The lattice is intact and its handedness is reversed.

8 figures One wrecked stem, lag by lag — golden, rise 0.005, 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 60 degrees. The lag-8 hop swings by 0.09 degrees and sits 0.19 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 8, which is the surviving lag and not a coincidence.

The hop that survived

A stem that never repairs after an organ is removed settles into an exactly repeating block of angles, and the period of that block is a spiral count of the lattice it was cut from. Nobody could say why. Read the wrecked stem by lags rather than by neighbours and the answer is one line: one family of the original lattice is still standing, organ by organ, and the block is its period.

8 figures 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.

One turn per survivor

If a wrecked stem keeps one family of its old lattice exactly, then the angle it settles at is not free. Over the period of the family that survived, the pattern has to come back to where that family left it — which means the whole change in the divergence is a whole number of turns spread over a small whole number of organs. Measured, it is one turn, at seventeen of nineteen.

8 figures More organs removed, more stems that never come back. The share of arrangements at the 5/8 rung that never return to the divergence they were cut from, against how many organs the cut removed. One organ wrecks 2 of 8 arrangements and five wreck 63 of 64. The number of arrangements differs from bar to bar because a cut of five organs has more ways of being placed than a cut of one, and it is printed on each bar for that reason. What the dose decides is whether a stem falls off its lattice; where it lands when it does is decided by something else.

Three organs and no mirror

A coarse stem cut of two organs can end up as its own mirror image — the same lattice wound the other way, counts unchanged, handedness reversed. Finer stems never do it, and two accounts of why were on the table: coarseness, or the share of the neighbourhood removed. A three-organ cut at the finer arrangements settles it, and the answer is the first.

8 figures The mirror belongs to the lattice, not to the dose. How close the closest arrangement came to the mirror of the divergence it was cut from, against the share of the front that was removed. The marked point at 40 per cent is the coarse 3/5 rung with two organs taken, which reaches the mirror exactly. Every other point is a finer rung: five sizes of cut at 5/8 running from 13 to 63 per cent, and three organs at 8/13. Taking a larger share of a larger front than the coarse rung needs gets nowhere near, so the quantity that decides it is not the fraction of the neighbourhood removed.

The share was not the thing

Two organs out of a front of five reverses a stem's handedness; three out of eight does not, and neither does five out of eight, which is a larger share of a larger neighbourhood. The hypothesis under test was that the dose decides the destination. It decides whether a stem falls off its lattice and nothing about where it lands.

8 figures The short list, and what is on it. Every distinct place a wrecked stem settles, for cuts of one organ through five at one rung, with what each one turned out to be. Three of the six are the lattice the stem was cut from with one lag left rigid and a whole number of turns inserted over its period — the description the single-organ work established. The rest have no rigid lag at any period up to twenty-four, and a counter shown their positions returns a pair the original lattice does not carry. So the list is short and it is not homogeneous: a large enough cut can put a stem onto a different lattice rather than onto a slipped version of its own.

A wreck has a short list

Cuts of one organ through five, over two hundred and forty-six stems that never came back, land on six settled divergences between them. Removing five organs instead of one wrecks nearly everything and reaches nowhere the single cut had not already found — and half the list turns out to be the old lattice slipped by a turn, while the other half is not the old lattice at all.

8 figures 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.

A survivor has to be a neighbour

A stem that never repairs after a removal keeps exactly one lattice hop rigid, and nothing predicted which one. Sweep every offset at twelve lattices and the answer narrows sharply: at twenty-nine of thirty the surviving hop is one of the two families a counter returns, and the one exception is a step six times too long to be one.

8 figures 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.

Not the shorter of the two

If a damaged stem keeps one contact family standing, the obvious guess is that it keeps the nearer one. Across thirty wrecked offsets that is true twelve times and false seventeen, and on one lattice the two steps differ by a quarter of a per cent — where the words shorter and longer are doing no work at all.

8 figures 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.

One offset, two answers

Which contact family a wrecked stem keeps is decided by where the cut landed, at twenty-five of thirty offsets, by the simplest rule anybody would write down. It is refuted by two runs: the same counted pair, the same offset, two different rises, and two different surviving families.

8 figures Which family survives, along the 5/8 rung. The family a wrecked stem keeps, at every offset that wrecks and every rise on one rung. A counter shown any of these stems returns 5 and 8 spirals, at all twelve rises, so nothing in the pair distinguishes the columns. The cells say otherwise: at an offset of 4 the stem keeps the 5-family at the coarse end of the rung and the other one at the fine end. The offsets that wreck at all grow from 1 to 5 as the rise falls, because the front deepens, and the extra offsets are the ones that keep the larger number. So the rule stated over the offset alone is a rule with the rise left out of it.

One rung, two answers

The offset accounts for twenty-five wrecked stems of thirty and is refuted by a single pair of runs that differ in nothing but the rise. Sweep one rung at a thousandth and the refutation stops being an anomaly: the same offset on the same lattice keeps one family at the coarse end and the other at the fine one.

8 figures A prediction and its opposite, scored on the same rows. The reading under test said the family whose member was removed is the one that breaks. Scored across every wrecked offset where the removed organ lies on exactly one contact chain — 9 of 30, the other 21 being silent because the organ lies on neither — it is right no times and its opposite is right nine. A chain that loses a member does not stop existing: the organs above the hole are still spaced at that lag and the rule that placed them is still minimising the same sum, while the other chain has lost the organ its members were positioned against.

The family that lost a member

The offset rule restated in the arrangement predicts that the chain whose organ was taken is the chain that breaks. Scored on the nine offsets where the question can be asked, it is right none of the time and its opposite is right all nine.

8 figures How many offsets wreck, along the 5/8 rung. The count of offsets that never repair, at each rise on one rung. It runs from 1 at the coarse end to 5 at the fine end, while a counter returns 5 and 8 spirals at every one of them. The front — the run of recent organs at which a removal is felt at all — deepens as the rise falls, so there are simply more places a cut can land and fail to heal. That is the mechanism under the grid: the offsets that appear at the fine end are the ones beyond the smaller contact number, and those are the ones that keep the larger family.

The front deepens down a rung

The offsets that never repair grow from one to five across a single rung, while a counter returns the same pair at every rise. The extra offsets are not a random extension of the ones already there: they are the ones past the smaller counted number, and they are the ones that keep the larger family.

8 figures The two contact steps change places inside the 5/8 rung. Measured at every rise on one rung, where a counter returns 5 and 8 spirals throughout. The settled divergence slides from 136.6406 to 137.8672 degrees. The ratio of the two contact steps falls to 1.0131 at a rise of 0.016 and the ordering changes hands at 0.015: above it the shorter step belongs to the 5-family and below it to the 8-family. Neither quantity is available to a counter, which is shown positions and reports a pair, and both of them move while that pair does not.

The shortest hop was a coin flip

The reading that a wrecked stem keeps its shortest hop was refuted at twelve of twenty-nine across the census. Re-scored along a single rung, where the counted pair is held and the step ordering reverses, it is right at sixteen of thirty-one — which is not a refutation but an absence of information.

8 figures Every wrecked offset, and whose neighbour was removed. Each row is a stem that never repaired, with the family of the organ that was taken and the family that survived. An organ five places back on a stem counted at 5 and 8 spirals lies on the tip's five-chain, so the question can be asked there; an organ four places back lies on neither chain and it cannot. Of 30 wrecked offsets in the census, 9 remove a member of exactly one family and 21 remove a member of neither. On every one of the 9 the family that lost a member is the family left standing, which is the opposite of what the reading predicted.

The organ that was nobody's neighbour

Twenty-one of the thirty wrecked offsets remove an organ that lies on neither contact chain through the tip. The reading that explains the other nine has nothing to say about them, and the honest thing is to say so rather than to widen the definition until it does.

8 figures The deeper rule against the disturbance's memory, at three contact scales. How many of six seeds agree that the deeper rule passed more drift, swept across the correlation length of the disturbance, at three rises. The rule, the amplitude and the run length are identical in every panel; only the rise differs, and with it the contact numbers — 3 and 5, then 5 and 8, then 8 and 13. The shapes are not the same: 3/5 is crossing, 5/8 is no corner, 8/13 is crossing. A corner that sat at a fixed number of organs would look the same in all three, and it does not.

The corner moves with the rise

The corner was either the contact scale or simply any memory at all, and nothing in the thread had ever varied the rise — the one knob that moves the contact numbers while leaving the rule, the amplitude and the run length alone. Swept over it, the comparison does not keep its shape.

8 figures Every transition as the rise falls. The pair climbs 1/2 → 2/3 → 3/5 → 5/8 → 8/13. Consecutive transitions are 0.381, 0.382, 0.382 of the previous rise — 1/φ² is 0.3820.

A stem too fine to settle

Below a rise of about four thousandths the counter stops returning contact families and starts returning pairs like 2/13 and 13/24. Lengthening the stem does not fix it. That is a ceiling on every sweep this collection runs up the ladder, and it has never been written down.

7 figures The 3/5 rung at a tenth of the ladder's step. Every rise of one rung, sampled ten times as finely as the ladder that found the locked band. All 23 are counted at 3 and 5 spirals and all 23 settle: the largest wander is 0.221 degrees, against the 0.5 degree threshold and against the 0.79 to 1.60 degrees the band on the coarse rung wobbles by. The divergence slides smoothly from 139.0625 to 136.7344 degrees with no rise stuck on a rational and none stuck on anything else. Whatever the band is, it is not something a coarser sampling was hiding here.

The band was not the sampling

Five rises in the middle of the coarse rung stick on three eighths of a turn, and the ladder that found them is swept at five thousandths — coarse enough that a band of the same kind could sit inside any finer rung unsampled. Swept at a tenth of that across a whole finer rung, nothing locks.

8 figures Which family survives, along the 5/8 rung. The family a wrecked stem keeps, at every offset that wrecks and every rise on one rung. A counter shown any of these stems returns 5 and 8 spirals, at all twelve rises, so nothing in the pair distinguishes the columns. The cells say otherwise: at an offset of 4 the stem keeps the 5-family at the coarse end of the rung and the other one at the fine end. The offsets that wreck at all grow from 1 to 5 as the rise falls, because the front deepens, and the extra offsets are the ones that keep the larger number. So the rule stated over the offset alone is a rule with the rise left out of it.

One rise per rung is a sample

Every census on this site takes one rise from each rung, because the question was always which pair. Any rule later scored on those rows inherits a variable that was never varied — and two of this collection's results turn out to be about the sampling as much as about the rule.

8 figures The two spiral families a counter finds between 0.68 and 0.92 of the radius. 34 spirals one way and 55 the other, found from the point positions alone — the counter is never told the divergence angle.

What a count cannot decide

A spiral count is the measurement this whole subject is built on, and it is deliberately blind to everything that varies inside a rung. Four results this collection now holds are results about that blindness rather than about the arrangements.

8 figures Where the deeper rule changes hands, and where it never does. One row per rise, with the contact numbers on the left and what the comparison does on the right. At the coarsest the deeper rule wins at both ends and loses in the middle; at the middle rise it wins throughout; at the finest it crosses once, from losing to winning. The contact scale runs over a factor of 2.6 across these three rises and the behaviour is not a translation of one curve — it is three different curves. That refutes a corner fixed at a short correlation, and it does not by itself establish one that tracks the contacts.

The panel with no corner

Sweeping the rise gave three shapes where one was expected, and the middle one is the informative panel: at the 5/8 contact scale the deeper rule wins at every correlation and there is no crossing to locate. That is either a fact about the lattice or a fact about the pair of exponents, and one measurement separates them.

8 figures The 3/5 rung at a tenth of the ladder's step. Every rise of one rung, sampled ten times as finely as the ladder that found the locked band. All 23 are counted at 3 and 5 spirals and all 23 settle: the largest wander is 0.221 degrees, against the 0.5 degree threshold and against the 0.79 to 1.60 degrees the band on the coarse rung wobbles by. The divergence slides smoothly from 139.0625 to 136.7344 degrees with no rise stuck on a rational and none stuck on anything else. Whatever the band is, it is not something a coarser sampling was hiding here.

The slide a counter holds constant

Inside one rung the settled divergence moves by more than a degree, monotonically, with no flat stretch anywhere — measured at a thousandth on one rung and at half a ten-thousandth on another. A rung is a plateau in one reported number laid over a geometry that never stops moving.

8 figures The column the census never carried. One row per lattice the ablation census was grown at. The bar shows where inside its own rung that rise sat, measured in the logarithm of the rise because the ladder is geometric, with zero the coarse transition and one the fine one. The mark on each bar is that rung's own handover, the rise where the two contact steps change places. Of the ten lattices that ever wreck, eight sit past their handover and one sit before it, with one sitting so close to one that the two steps differ by parts in a thousand. The rise was recorded in every table this collection has published; this fraction was in none of them.

The column that cost no stems

Every table in this collection records the rise a stem was grown at. None records where inside its own rung that rise sat, and the fraction turns out to be computable from numbers already written down — which makes it the cheapest column anybody here has ever added and the one that changes the most about how the tables read.

8 figures The column the census never carried. One row per lattice the ablation census was grown at. The bar shows where inside its own rung that rise sat, measured in the logarithm of the rise because the ladder is geometric, with zero the coarse transition and one the fine one. The mark on each bar is that rung's own handover, the rise where the two contact steps change places. Of the ten lattices that ever wreck, eight sit past their handover and one sit before it, with one sitting so close to one that the two steps differ by parts in a thousand. The rise was recorded in every table this collection has published; this fraction was in none of them.

The side the census sat on

Eight of the ten lattices the ablation census wrecks at were grown past their rung's handover, one before it, and one so close that the ordering it quotes differs by parts in a thousand. A reading scored over the step ordering was therefore scored against a quantity the census was nearly holding fixed.

8 figures The ordering changes and the survivor does not. Every offset that wrecks, at every rise of the band, with the family left standing written in the cell. The counted pair is 5 and 8 at all 18 rises and the settled divergence is held to a twentieth of a degree, so the one quantity moving across the columns is which of the two contact steps is the shorter — and it changes hands at the marked rise. The cells do not: the 5 family survives at all 24 wrecked cuts, on both sides. Scored on this band, the reading that a wrecked stem keeps its shortest hop is right 14 times out of 24, for an answer that never changed.

The ordering was not the actor

Cut an organ out of every rise of a band where the counted pair and the settled divergence are held and the two contact steps change places, and the family left standing does not change. Fifty-five wrecked cuts on two branches, and the ordering reverses underneath every one of them.

8 figures What a cut moves, organ by organ. A stem counted at 5 and 8 spirals with the organ five places back from the tip removed, compared against a control that shares its history to the last digit. Each mark is one organ placed after the cut and how far its azimuth ended up from where the control put the same organ. The quantity folds at half a turn, so 152 degrees is near the largest displacement there is; and it does not decay with height, which is what a stem that never repairs means. There is therefore no organ that the cut disturbed most in any useful sense, and a reading that needs one has nowhere to stand.

The organ that moved furthest

A reading that works on nine of thirty rows needs a reference organ, and the obvious repair is to measure one rather than to choose it. Measured, the disturbance turns out to have no far edge at all, so there is no organ that moved furthest in any sense the reading can use — and the generalisation that does work needs no reference organ.

8 figures The same comparison at three sizes of disturbance. Each row is one panel: how many of six seeds the deeper rule beats the shallower one on, at six correlation lengths, on the lattice at a rise of 0.013. At the size of jostle the thread used, cells across the middle sit at six of six — the largest number the panel can print — so no feature could have appeared there whatever the lattice did, and the reading that this rise has no corner was a reading of that ceiling. Raised, the cells come down, and at the largest disturbance the panel shows a clean crossing: the deeper rule loses at white noise and takes every seed once the disturbance remembers itself for a few organs. The corner is there, and the flat middle was the instrument.

Six of six is not a measurement

A panel comparing two rules seed by seed reported the deeper one winning every one of six seeds at four correlation lengths out of six, and was read as a lattice with no corner in it. Six of six is the largest number the panel can print, so the flat middle was a reading of the ceiling — and raising the disturbance brings a corner out of it.

5 figures How many starting angles reach a lattice, by rise and by falloff. Every cell is 9 stems grown from 9 starting angles spanning 40° to 180°, at one rise and one exponent in the placement rule's falloff, for 1200 organs each. The number is how many of them settle onto a lattice. Reading down a column, the basin closes as the rise falls, which is the wall. Reading across a row, nothing much happens: the exponent-three column is the published settling table digit for digit, and the other three are the same column inside their own sampling error. The account that a steeper falloff should wall somewhere else does not survive the table it predicted.

A steeper rule walls nowhere else

The account of the wall at the fine end was that the basin narrows because the neighbourhood deepens, which predicts a steeper falloff walling somewhere else. Grown at four exponents, the four columns settle 30, 31, 29 and 27 of 72 — a spread of 0.056 against an error of 0.058.

7 figures How many organs settling takes, by falloff exponent. One mark per stem that settles, placed at the organ from which every later divergence stays within a degree and a half of the run's own final value. At the exponent every other measurement here uses, the slowest is 290 organs — the number four rounds of this collection have carried as though it belonged to the rule. It belongs to the exponent: at 4 the slowest is 808 and at 5 it is 674. A table of shares cannot see this, because a stem that settles at organ 800 and one that settles at organ 8 are the same entry in it. And it is still not a shortage of time: of 72 pairs of runs grown to 1200 organs and then to 3200, 0 settle at the longer length after failing at the shorter.

The clock a share cannot see

Settling takes nothing to 290 organs, and four rounds of this collection have carried that as a property of the rule. At a steeper falloff the slowest is 808 — and not one of the 72 pairs of runs settles at 3,200 organs after failing at 1,200, so the fine end is still a wall.

4 figures Every divergence a stem settles on, by falloff exponent. One row per exponent, one mark per distinct settled divergence anywhere in that column, read to a tenth of a degree. This is the sharpest instrument in the sweep and the cheapest: it is already in every run, it does not average, and it is read on the same azimuth grid at every exponent. The steeper falloffs reach 42.3°, 47.9°, 148.1° — values that appear nowhere in the two shallower columns — while the traffic the other way is empty. So the exponent changes the map of where a stem can end up while leaving alone the share of starting angles that end up anywhere, which is a result a measurement of "did it settle" was never going to find.

Destinations only a steep rule reaches

A settled stem's divergence is the sharpest instrument in the sweep and the cheapest: it is already in every run, it does not average, and it is read on one grid at every exponent. Exponents 4 and 5 reach 42.3°, 47.9° and 148.1°, and the two shallower ones reach none of them.

8 figures Every destination in the settling table, counted. One row per destination, placed across at the divergence the stems settled on and labelled with the parastichy pair a counter returns at the top of the run. 117 settled runs land on 15 destinations when two values within half a degree are called one. five of them sit on the golden or the Lucas sequence and ten do not, and the ones that do not are reached at every falloff exponent including the shallowest. Nothing here refuses to count: the settling test does not admit arrangements this collection would decline to call lattices.

A counter on the settling table

The settling table has reported an angle for every run that reaches a lattice, and nobody has ever counted one. A hundred and seventeen settled runs, regrown and counted: every one of them has a parastichy pair, and sixty-five of them land on sequences the ladder does not carry.

7 figures What the 3 steep-only destinations count as. Each block is a divergence that only the two steeper falloff exponents settle a stem on, with the pair a counter returns and the sequence that pair belongs to. All three count. None is a rung of either ladder. Two of them are the only members of their sequence anywhere in the table; the third, at 148.1 degrees, is the coarsest rung of a sequence that supplies another destination at every exponent — so it is not a new kind of arrangement but a coarser one of a kind the table already reached.

What a steep rule counts as

Exponents four and five settle stems on 42.3°, 47.9° and 148.1°, and the two shallower ones reach none of them. All three count: 8/9, 8/15 and 2/5. None is a rung of either ladder, and one of them is the coarsest rung of a sequence the table already had.

7 figures The 10 sequences the settling table's destinations belong to. Every pair the counter returns is two consecutive terms of a sequence in which each term is the sum of the two before it, which is the ladder's own rule. two of these are the sequences plants are observed to follow. The rest are not, and they are not rare: the 1, 4, 5, 9, 14 sequence and the 2, 5, 7, 12, 19 sequence each supply several destinations, at every falloff exponent the table is grown at. The right-hand column is the divergences that land on each.

Ten sequences, two of them the ladder's

Every parastichy pair the settling table produces is two consecutive terms of a sequence in which each term is the sum of the two before it. Ten such sequences account for all fifteen destinations — nine, once one of the ten turns out to be a reading rather than a ladder — and the two the collection is built on are neither the largest nor the smallest.

8 figures The destinations a steep rule reaches, read two ways. Above: the list read to a tenth of a degree, which is what the settling table prints — 13 entries. Below: the same list read at half a degree, which is two steps of the grid the azimuths are placed on — 3. The 10 entries the finer reading adds are destinations a shallow exponent also reaches, a tenth or two of a degree away, and a tenth of a degree is a fifth of one step of that grid. The pale marks are every destination in the table, for scale.

A list that was a rounding

Three destinations only a steep falloff reaches, read at two grid steps. Thirteen, read at a tenth of a degree. And four arrangements, read by what the counter returns rather than by the angle — a different four, with one the angle reading hides.

7 figures Where a stem started at each of twenty angles ends up, at a falloff exponent of 3. One column per starting angle and one row per rise, coarse at the top. A filled cell is a run that reached a lattice, its tone the destination it reached; a pale cell is a run that never settles. Runs of one tone across neighbouring columns are basins, and the widest of them spans 7 consecutive angles. The angles are 6.25 to 10 degrees apart, so a basin narrower than that cannot be seen here and a single filled cell says nothing about how wide its basin is.

A basin has a width

A destination reached from one starting angle is a presence. A destination reached from seven consecutive starting angles spanning forty-five degrees is a basin with an extent, and nine angles could not have measured one — they were too far apart to have two of them land in the same place.

8 figures The settling share at every rise, sampled two ways. How often a stem reaches a lattice, against the rise, at each of the four falloff exponents. The horizontal rule is the half share the wall is read at. At nine starting angles the four exponents cross it at rises spanning a factor of 1.96 and order themselves 2, 5, 3, 4; at twenty they span a factor of 1.18 and order themselves 5, 4, 2, 3. The ordering reverses and the spread collapses, so the conclusion that the exponents do not separate is confirmed and the numbers that had suggested otherwise were the coarser sampling's own error.

A wall that stopped moving

Four falloff exponents were reported not to move the rise below which stems stop reaching a lattice. Their measured walls spanned a factor of two and ordered themselves 2, 5, 3, 4. At twenty starting angles they span a fifth of one and order themselves 5, 4, 2, 3 — so the conclusion was right and its arithmetic was noise.

4 figures How often each group of starting angles reaches a lattice. The share of runs that settle, for the nine angles the table was grown from, for eight angles placed halfway between them, and for three angles below the nine's lowest. The nine settle far more often than either. So it is the refinement rather than the extension that drags the share down, and the settling share this collection reports is biased upwards by the choice of angles rather than by their range.

Round numbers are not a sample

The nine starting angles the settling table was grown from reach a lattice four times in ten. Eight angles placed exactly halfway between them reach one a quarter of the time. The difference is not noise and it is not the range — several of the nine sit next door to somewhere a stem could settle.

8 figures Where a stem started at each of twenty angles ends up, at a falloff exponent of 3. One column per starting angle and one row per rise, coarse at the top. A filled cell is a run that reached a lattice, its tone the destination it reached; a pale cell is a run that never settles. Runs of one tone across neighbouring columns are basins, and the widest of them spans 7 consecutive angles. The angles are 6.25 to 10 degrees apart, so a basin narrower than that cannot be seen here and a single filled cell says nothing about how wide its basin is.

Twenty angles instead of nine

Every claim in this collection about where a stem ends up rests on nine starting angles a cell, and the file that uses them says so — it computes a binomial error of 0.17 and declines to read a spread against it. Eleven more angles halve that error and change what several of the numbers were.

8 figures How often a stem settles, at three samplings of the starting angle. The share of runs that reach a lattice, over the whole table of four falloff exponents by eight rises. The upper three bars are the pooled share at nine, twenty and forty starting angles; the lower three are the share for each group of angles on its own. The pooled figure falls 40.6 to 32.3 to 30.0 per cent, by 8.3 points and then 2.3, so it is converging. The eleven angles added at twenty and the twenty added at forty settle at 25.6 and 27.7 per cent, which differ by less than their own error.

Forty angles, and a limit

Nine starting angles turned out to be a biased sample of the circle, and doubling to twenty said by how much. Doubling again says the estimate is converging — to a smaller correction than one doubling extrapolated to.

8 figures The settling share against the rise, at 40 starting angles. One line per falloff exponent: how many of the 40 starting angles still reach a lattice at each rise, coarse on the left. The wall is where a line crosses a half, and the shaded band is the range of rises consistent with that crossing at one standard error — from the finest rise whose share is confidently above a half to the coarsest whose share is confidently below it. Every one of those bands is wider than the factor of 1.62 the four walls differ by, and at exponents 4 and 5 no rise in the table has a share confidently above a half at all.

A wall that was never measured

Three samplings of the starting angle give three orderings of the four falloff exponents' walls and a spread that does not shrink, while every error bar behind them halves. The reason is that a wall is a crossing of a nearly flat curve, and nobody had asked how well it is located.

4 figures Destinations found at nine, twenty and forty starting angles. How many distinct divergences the settling table reaches at each sampling, split into those sitting on the golden or Lucas sequence and those off both. The on-ladder count is 5 at every sampling, over a fourfold refinement of the starting angle — so the rule's on-ladder targets were enumerated by the first nine runs. The off-ladder count goes 10, 13, 14, and the one the last doubling found is off both ladders.

A list that can only shrink

The destinations only a steep falloff reaches grew from three to five when the sampling doubled, and everyone read it as a list filling in. Doubling again takes two off it, which is the only direction a list defined by an absence can ever move.

8 figures The widest basin, at twenty starting angles and at forty. The starting angles of one cell of the table, drawn as ticks, with the run of consecutive angles that all reach one destination filled. At twenty angles it holds 7 angles and spans 43.75 degrees; at forty it holds 14 and spans 47.5. Halving the spacing doubled the count and left the width where it was, which is what a real basin does and what a run of angles produced by the sampling does not. The other half of the same table goes the other way: 170 of 233 runs are a single angle, against 116 of 151.

A basin that doubled

A run of consecutive starting angles reaching one destination is a basin, and its width is a lower bound. Halving the spacing doubled the angles in the widest one and left its width alone, which is what a real basin does and a sampling artefact does not.

7 figures

The pattern itself

A seed head is a set of points, and nearly every claim about one is really a claim about how many spirals run through it. The counting can be done from the points alone, and the answer is not what the captions say.

A head of 200 primordia at a divergence of 137.51°. Nothing is placed by hand: the nth point sits at n·137.51° and radius √n. The closest any two points come is 1.60 of the mean spacing.

A head is a set of points

The nth primordium at n times an angle, and a radius of root n. Two lines of arithmetic produce a sunflower head, which is either remarkable or suspicious depending on how carefully the claim is stated — and stating it carefully is most of the work.

8 figures The two spiral families a counter finds between 0.68 and 0.92 of the radius. 34 spirals one way and 55 the other, found from the point positions alone — the counter is never told the divergence angle.

Counting the spirals

Almost every claim about phyllotaxis is a claim about how many spirals run through a pattern, and the count is almost never done. It can be done from the points alone, by a count that is never told what angle built them — and then a count of 34 is evidence rather than a restatement.

8 figures The spiral counts, band by band, in one head. The same flower gives 13/21, 21/34, 34/55, 55/89 at different radii. A caption saying "34 and 55 spirals" is a statement about one annulus.

The counts change with radius

The same head gives 13 and 21 near the centre, 21 and 34 further out, 34 and 55 beyond that, and 55 and 89 at the rim. The transitions are at computable radii, and a photograph captioned with one pair is a statement about one annulus rather than about a flower.

8 figures A round trip on four heads of 400 primordia: the divergence angle recovered from each. The counter is shown the points and nothing else. The worst recovery across the four is 0.035°.

Recovering the angle from the counts

Build a head at a stated divergence angle, forget the angle, and get it back from the spiral counts alone. Four angles, worst error twelve thousandths of a degree — and the only thing that crossed between the two halves was a list of coordinates.

8 figures three stems: 1, 2, 3 primordia at a time. 1-jugate at 137.51° counts 2 and 3; 2-jugate at 68.75° counts 2 and 4; 3-jugate at 45.84° counts 3 and 6. Every count shares the factor k, and dividing it out leaves an ordinary lattice: what a k-jugate stem is, exactly, is k copies of an ordinary one wrapped k times round.

Two at a time

Every counter in these essays asks how far it is from element i to element i plus m, and that index is a claim that the elements arrived one at a time. For teasel, for Cephalaria and for a real minority of plants the claim is false — and what those patterns turn out to be is an ordinary lattice, wrapped twice.

8 figures Tracing one family: 4 chains. Every node is joined to the node one repeated displacement away, and the chains that result are drawn separately. There are 4 of them, which is the parastichy number of this family. No index of arrival was used anywhere, which is what lets the same count be made on a pattern where 2 primordia appear at once.

Counting without an index

A person counting spirals on a cone puts a finger on one scale, follows a family round, and counts how many distinct chains there are. That needs no order of arrival — and building it turns out to be strictly more general than the counter that reads the order of arrival, and to find a bug in the counting of a bijugate stem that nothing had caught.

8 figures The 2-jugate forks converge on 68.7539°. Every fork sits at a rational divergence, with denominator 4(m² + mn + n²) — 10/28, 30/76, 74/196 and so on. The limit is 68.7539°, which is 137.5078 divided by 2, and it is at none of them.

Half the golden angle

The forks of the van Iterson tree converge on 137.5078° and sit at none of them. Divide the whole tree by two and the same thing happens at 68.7539° — which is where teasel is, and where a bijugate sunflower counted 42 and 68 has to be.

7 figures The angles against the positions, rise by rise. three rises, five seeded stems each. A filled mark is a run whose angle readout returned the pair the position counter finds in the same stem; an open mark is a refusal. At 0.032 the counter says 3/5 and the angles agree on 0 of 5, refusing 5. At 0.013 the counter says 5/8 and the angles agree on 5 of 5. At 0.005 the counter says 8/13 and the angles agree on 5 of 5. The two instruments share no code path: one is given a list of angles, the other a list of coordinates.

A counter that sees no positions

This site has counted spirals two ways, and both were handed coordinates. A third counter is handed a list of angles and nothing else. It returns one number instead of two, it refuses more often, and where it refuses it would have been wrong every time.

8 figures Every family but two is the sum of two others. Four heads and the contact families each actually has. An arc arrives at every family that is the sum of two smaller ones; the two with no arc arriving are the generators, and on every head they are the two smallest. whorled, 144°: 2, 3, 5 — golden, 137.508°: 8, 13, 21, 34, 55, 89 — Lucas, 99.502°: 11, 18, 29, 47, 76 — rational, 137.5°: 8, 13, 21, 34, 55, 89 — 137.0°: 8, 13, 21, 29, 50, 71, 92, 113. So once a pair is counted the rest is arithmetic, and a third counted family is a prediction rather than a second measurement.

Every family but two is a sum

A seed head has six spiral families and everybody reports two. That looks like a convention hiding information and it is the opposite — every family but the two smallest is the sum of two others, so a third count is a prediction rather than a measurement, and a check that catches a wrong pair.

8 figures The angles against the positions, rise by rise. five rises, five seeded stems each. A filled mark is a run whose angle readout returned the pair the position counter finds in the same stem; an open mark is a refusal. At 0.032 the counter says 3/5 and the angles agree on 0 of 5, refusing 5. At 0.013 the counter says 5/8 and the angles agree on 5 of 5. At 0.01 the counter says 5/8 and the angles agree on 5 of 5. At 0.005 the counter says 8/13 and the angles agree on 5 of 5. At 0.008 the counter says 5/8 and the angles agree on 2 of 5, refusing 3. The two instruments share no code path: one is given a list of angles, the other a list of coordinates.

The angles name the branch

Seed the same rule at the Lucas angle and the readout returns 4 and 7, then 7 and 11 — the pairs the position counter finds, and not Fibonacci numbers. So a list of divergence angles carries not only how many spirals there are but which family of ladders the plant is on.

8 figures A lattice with an error that repeats every 8 organs. The autocorrelation of 759 divergence angles from a kinematic lattice at a divergence of 137.8261° and a rise of 0.005, with 0.5° of scatter on each azimuth. There is no placement rule anywhere in it: node i is put at exactly i times the divergence and then displaced. The only thing that differs between this figure and the control is the structure of the displacement — here, an error that repeats every 8 organs at period 8, weight 0.7. The largest comb mean is 0.592 against a sampling band of 0.073, and the readout returns 8/10.

A periodicity is not a lattice

Give a lattice's errors a period of eight and a comb appears at spacing eight, on an arrangement with no rule in it. But the partner it names is 10, then 12, then 11, then nothing — an accident of the disturbance rather than a measurement of the pattern. The forgery is caught by reading a second stem, and by nothing else.

8 figures One wrecked stem, lag by lag — golden, rise 0.005, organ 6 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 55 degrees. The lag-4 hop swings by 0.12 degrees and sits 0.01 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 4, which is the surviving lag and not a coincidence.

A period that is not a count

Eighteen wrecked stems settle into a block whose period is one of their own spiral counts, and one settles into a block of four on a lattice counted 8 and 13. The odd one is not noise. It is the case that shows what the rule is actually conserving, and it is the reason this thread is about lattice steps rather than about spirals.

8 figures Everywhere a cut of one to five organs can send a 5/8 stem. Every settled divergence reached by any arrangement of up to five organs removed from a stem at the 5/8 rung, on one axis. There are six of them and no more. three are slips of the lattice the stem was cut from: each keeps the lag-5 family intact and sits a whole number of turns of it from the next, which is the ladder marked below the axis with rungs 72.0 degrees apart. The other three keep no lag at all and sit near a fraction of a turn, marked above: 175.0 degrees near 1 of 2 turns, 190.0 degrees near 1 of 2 turns, 235.0 degrees near 2 of 3 turns. A stem that is cut either slides along the ladder it was on or leaves it for a lattice with files in it.

A count with a factor in it

Cuts of several organs send a stem to six settled divergences and no more. Three of them are the old lattice with a family left standing. The other three are counted 2/6, 4/6 and 3/6 — and a pair whose numbers share a factor is the classic signature of a pattern that arrives several organs at a time.

8 figures Two patterns a counter cannot tell apart — counted 2/6 against 2/6. On the left, the top 90 organs of a spiral stem that never repaired after two organs were removed, settling at 175.01 degrees. On the right, a stem grown by a rule that places two organs at a time on every node. A counter shown the positions returns 2/6 for the first and 2/6 for the second, and a pair whose numbers share a factor is the usual signature of a whorled pattern. Rotate each pattern and the answer separates them at once: the whorled stem maps onto itself at a half turn and the wrecked stem maps onto itself at no fraction of a turn at all. Its shared factor is a fact about where its divergence landed, 4.99 degrees from 1 of 2 turns, and not about how it grew.

The symmetry that is not there

A wrecked stem counted 2/6 and a stem grown two organs at a time counted 2/6 are indistinguishable to a spiral counter. Rotate them by half a turn and one lands on itself and the other lands nowhere. A cut does not make a whorled pattern; it makes a pattern that counts like one.

8 figures Two readings of what a stem is, and one of them is wrong. The same runs read twice. On the left is the pair counted from the point positions by machinery that is never shown a divergence angle; on the right is the pair read from the divergence sequence, which is the column the depth thread has been quoting. Every exponent's stems are counted at 5/8 spirals from the points. Read from the angles, the deepest rule's stems come back as something else at every seed, and a pair with a 8 replaced in it is not a near miss but a family whose step is several times as long. The settled divergence moves by 0.118 degrees across all five exponents, so the rules are on one lattice and the disagreement is a defect in one of the two instruments.

The pair read from the angles

Checking that a jostled stem is still the lattice its panel is about turned up a disagreement. Counted from the point positions every rule's stems return five and eight spirals; read from the divergence sequence, the deepest rule's stems come back as five and seven at every seed. The points are right, and this site's founding rule is why.

8 figures All 19 destinations, counted at 17 windows from 20 to 800 organs. One row per destination, one cell per counting window, coarsest on the left. A dark cell is a window returning the same pair as the standing 200; a warm cell is a window returning a different one; a pale cell is a window that reaches below the run's own settling organ and was dropped. 5 of the 19 rows differ anywhere, and every difference sits left of the floor its own pair sets — 20 organs for a pair under 14 parastichies and the larger count plus 6 above it. The census read at 25 organs and at 800 is the same census.

The window nobody varied

Every parastichy pair reported for the settling table was counted over the top two hundred organs of a run, and that number had never been moved. Moved seventeen ways across a factor of forty, on all three hundred and eighty-four settled runs, it changes nothing at all.

8 figures What a 200-organ window allows against what the offset ceiling allows. Two settings decide what the counter can return, and at this window the offset ceiling is the tighter of them. A window of 200 organs holds a family of at most 194, since the counter scores an offset only where it has 6 hops of it inside the band; the largest offset it looks at is 60, a factor of 3.2 between them. The ticks are the rungs of the two ladders this collection is built on, and the largest counted number anywhere in the settling table is 19 — inside both bounds, which is what makes the table inert.

A plateau the instrument should have had

The counting window decides nothing on the settling table, and it has two bounds that decide everything outside it. A window one organ too narrow returns the previous rung of the same ladder, a family past the offset ceiling is reported as a coarser rung at every width there is, and neither failure produces a refusal, noise or a wide error bar.

8 figures The window 16 known pairs need, against the 20 organs the counter refuses under. Cylinders built at a stated divergence and rise, counted at every window from 20 organs to 200, so the pair is known before the counter sees it. The smallest window that reads it is 20 organs for every pair up to 14 parastichies and the larger count plus 6 above that — 13 of 16 measured exactly, on both branches, with nothing fitted. A pair of 76 and 123 needs 129 organs where 2 and 3 needs 20; the open marks are pairs the counter's own offset ceiling refuses at every window, read only once that ceiling is raised.

How many organs a pair needs

A count taken over too few organs does not fail. It returns the rung below, which is a perfectly good pair, and nothing anywhere says so. The window that avoids it is not a constant but the counter's own arithmetic, and 384 settled runs sit exactly where that arithmetic puts them.

8 figures The five readings of 384 whose two counted families wind the same way. Each row is an additive sequence with the term the counter stepped over drawn open between the two it returned. 4/9 is 1, 4, 5, 9, 14 with 5 skipped, and the pair a crossing requirement would have returned is 4/5 — a sequence the same destination's other runs already sit on. Dropping these readings takes the census at 40 starting angles from 31 arrangements and 15 sequences to 28 and 12, on 5 of 384 runs.

A sequence that was a reading

One of the settling table's additive sequences rested on a single count that skipped a term, and the account of it was that the count had been taken over the wrong patch. It was not. The counter works out whether its two families wind opposite ways and the reading throws the answer away, on five runs in three hundred and eighty-four.

8 figures The pairs grown stems of one, two, three and four organs a whorl count as their rise falls, against the folded rise. Four stems grown by the same rule over the same range of folded rise, 0.4 to 0.0012 — the rise between whorls times the jugacy, which puts every jugacy on one lattice axis — each whorl folded to one node and counted in a sliding window. With 1 organ a whorl the counted pairs run 1/2, 2/3, 3/5, 5/8, 8/13, 13/21; with 2 organs a whorl the counted pairs run 2/4, 4/6, 6/10, 10/16, 16/26, 26/42; with 3 organs a whorl the counted pairs run 3/6, 6/9, 9/15, 15/24, 24/39, 39/63; with 4 organs a whorl the counted pairs run 4/8, 8/12, 12/20, 20/32, 32/52, 52/84. Every jugacy passes the same transitions at the same folded rises, with its pairs multiplied by its jugacy.

A grown stem halves its ladder

A bijugate stem at 68.754° was derived to pass the ordinary transitions at half their rises, because a lattice wrapped twice round is the ordinary lattice at twice the rise. No bijugate stem had been grown through them. Grown by the same rule that grows an ordinary shoot, stems of two, three and four organs a whorl walk the ordinary ladder with every pair multiplied by the jugacy, change pair at the ordinary rise divided by the jugacy between whorls — and by its square per organ — lag behind the static ladder by the same hundredth of a rung, and settle within three hundredths of a degree of 137.5078 over the jugacy.

7 figures Which rates keep a Lucas seed, on stems of one, two and three organs a whorl, placed by whorls a rung. Every rate each configuration was grown at, drawn at the number of whorls the stem places between one transition and the next: a dark cell keeps the Lucas ladder to the end of the rise, a light cell gives it up. 1 a whorl, 512: kept to 86.6 whorls a rung, lost from 87.6; 1 a whorl, 256: kept to 86.6 whorls a rung, lost from 87.6; 1 a whorl, 1024: kept to 86.6 whorls a rung, lost from 87.6; 2 a whorl, 1024: kept to 87.6 whorls a rung, lost from 88.1; 2 a whorl, 1024, imposed: kept to 87.6 whorls a rung, lost from 88.1; 2 a whorl, 512: kept to 87.6 whorls a rung, lost from 88.5; 2 a whorl, 2048: kept to 87.6 whorls a rung, lost from 88.5; 3 a whorl, 1536: kept to 87.9 whorls a rung, lost from 88.5; 3 a whorl, 1536, imposed: kept to 87.9 whorls a rung, lost from 88.5; 3 a whorl, 768: kept to 86.6 whorls a rung, lost from 88.2. On this axis every edge falls within about two whorls a rung of every other.

A Lucas seed counts whorls

An ordinary stem seeded on the Lucas lattice keeps that ladder when its rise falls fast and gives it up when it falls slowly, with the edge near ninety nodes a rung. A stem that grows two organs a whorl is an ordinary stem folded twice round, so the identity predicts its edge — once it says whether the edge counts placements or whorls. Grown across the edge, stems of one, two and three organs a whorl keep a Lucas seed to 86.6, 87.6 and 87.9 whorls a rung: one number to within a per cent in whorls, and one, two and three times as far out in organs. No grid moves it and imposing exact whorls moves it not at all, even where free trijugate whorls come apart completely as the seed is lost.

7 figures How far grown whorls of two to eight members miss exact symmetry, at four lattices. Whorls grown one member at a time at a fixed rise, on a grid of 1,680 azimuths that every jugacy divides, read at folded rises of 0.27, 0.18, 0.09, 0.036. With 2 members the largest miss is 0.00°, 0.00°, 0.00°, 0.00°; with 3 members the largest miss is 0.64°, 1.93°, 0.21°, 0.00°; with 4 members the largest miss is 0.00°, 0.00°, 0.00°, 0.00°; with 5 members the largest miss is 1.29°, 1.29°, 0.21°, 0.21°; with 6 members the largest miss is 0.86°, 0.86°, 0.43°, 0.21°; with 7 members the largest miss is 0.86°, 2.14°, 0.21°, 0.00°; with 8 members the largest miss is 0.00°, 0.00°, 0.00°, 0.00°. Whorls of two, four and eight members are exact at every lattice; three, five, six and seven miss.

A whorl that misses its share

A whorl of k organs is defined by its symmetry, and the rule that grows one places its members one after another, each against the members already there. Nothing tells it to put them a k-th of a turn apart. Grown that way, whorls of two, four and eight members sit exactly on their shares of the turn at every lattice measured, and whorls of three, five, six and seven do not — the pattern a mirror argument predicts, since only a power of two leaves every new member a position that mirrors every member already placed. A trijugate whorl misses by 6.5° at a coarse rise and not at all at a fine one, by an amount the rise sets almost everywhere, and none of it moves a single transition.

7 figures How often three counters read a band's own pair, against the width of the band, on an ordinary stem. Bands of 40 to 320 organs slid up an ordinary stem grown at T = 300 over a rise from 0.05 to 0.0005, each read three ways. The counter without an index reads the pair at 65%, 66%, 87%, 88%, 88%, 88%, 89%, 100%, 100%, 92%, 54%, 23%, 8%, 2%, 1%, 0%, 0%, 0%; the counter families required to cross reads the pair at 65%, 66%, 87%, 88%, 88%, 88%, 89%, 100%, 100%, 92%, 54%, 22%, 8%, 2%, 1%, 0%, 0%, 0%; the counter with the index reads the pair at 100%, 100%, 100%, 100%, 100%, 100%, 100%, 100%, 100%, 100%, 100%, 100%, 100%, 100%, 100%, 100%, 100%, 100%. The index-free counter reads nine bands in ten or more only from 80 to 100 organs.

A counter that cannot be slid

The counter that needs no order of arrival follows each family into chains and counts them, and on an ideal lattice it agrees with the counter that does. On a stem whose rise falls it works only inside a band of widths, and outside the band it returns a pair rather than refusing: the rung below when the band is too narrow for the larger count, a pair on no rung when the band spans more than about a third of a rung of rise. The upper edge moves with the rate, so a width that is right on one stem is wrong on another, and on the fastest bijugate stem measured no width works at all.

7 figures The band the counter chooses for itself, against how slowly the stem grows. The counter is given positions and no width. It fits the decay of the rise through the spacings of its own band's whorls and takes the band that spans a third of a rung of rise: 49 organs on the stem that falls over 150; 97 organs on the stem that falls over 300; 193 organs on the stem that falls over 600. A stem grown four times as slowly chooses a band 3.94 times as wide, because a third of a rung is a statement about rise and a width is a number of organs. That is exactly why no one width fits the seven stems, and why the previous reading found a stem for which none worked.

A band that follows the rise

The index-free counter reads a growing stem only inside a band of widths, and the upper edge is a third of a rung of rise rather than a number of organs — so a width right on one stem is wrong on another. A counter that fits the decay of the rise through the spacings of its own band's whorls, and takes the band spanning a third of a rung, chooses 49, 97 and 193 organs on stems falling over 150, 300 and 600. Given no width at all it reads within eight points of the best of eighteen fixed widths on six of the seven stems, matching it exactly on one and beating it on two. Refusing any band too narrow to have shown the rung above the pair it counted removes every reading of the rung below, on every stem — and costs between five and fifty-one points of correct reading to do it.

8 figures Read as whorls, the three jugacies fall on one line. The rate at which a stem seeded on the Lucas lattice gives that ladder up, against how long the seed is, for one, two and three organs a whorl. The three lie on top of each other at every seed length: 15 whorls gives 30.80, 30.80, 30.80; 20 whorls gives 42.35, 42.35, 40.42; 30 whorls gives 63.52, 63.52, 63.52; 40 whorls gives 85.66, 87.58, 85.66. Every departure is one or two steps of the grain the edge is bisected to, which is two whorls of rate — 1.925 whorls a rung at every jugacy and every edge drawn.

A seed measured in whorls

A Lucas seed's length counts whorls, and the number published earlier for the rate edge counts nothing at all. Grown from seeds of fifteen to eighty whorls at one, two and three organs a whorl, stems of every jugacy lose the seed at the same rate in whorls a rung for the same seed length in whorls — 30.80 at fifteen, 63.52 at thirty, identically across the three — and at the same seed length in organs they differ by a factor of 3.24. So the seed is measured in whorls, as the edge is. The other half is worse for the earlier reading: the edge is not a constant but 2.24 times the seed less three, straight to within 2.7 whorls a rung over a fivefold range, so the eighty-seven whorls a rung reported everywhere is a property of the forty-whorl seed nobody varied.

8 figures The rate at which a Lucas seed is lost, read through four counting windows. Eighteen stems — one, two and three organs a whorl, seeded with 15 to 80 whorls of Lucas lattice — each with the edge found four times, through counting windows of 20, 26, 39, 52 folded nodes, the four dots at each seed drawn side by side. 17 of the 18 stems give the same organ through every window. The exception is three organs a whorl with a 30-whorl seed, where the widest window finds the edge at 207 organs against 201 for the others. A window twice as wide as the one every earlier reading used moves no other edge by an organ, so the window is not what makes the ratio of edge to seed change with the seed.

The window was not carrying it

The ratio of a Lucas seed's rate edge to its length rose from 2.05 at fifteen whorls to 2.20 at eighty, and the suspect was the counting window, which is most of a short seed. Read through windows of 20, 26, 39 and 52 folded nodes, seventeen of the eighteen stems lose the seed at exactly the same organ of rate, so the window carries almost none of it. Part of the rise was the grain of rate the edge was found on, worth up to six hundredths of the ratio. What is left rises by a tenth below thirty whorls and has stopped by sixty, at a level an ordinary stem reaches about one and a half per cent lower than a bijugate or trijugate one — and a trijugate edge is not always a line.

6 figures What the round trip returns as the organs are displaced: golden angle, 300 organs. Thirty heads at each size of displacement, from none to two and a half spacings, each run through the round trip and sorted by what came back. The undisplaced head counts 21 and 34. At 0.25 spacings, 97% the undisplaced pair, 3% a neighbouring pair; at 0.5 spacings, 50% the undisplaced pair, 50% a neighbouring pair; at 0.75 spacings, 30% the undisplaced pair, 70% a neighbouring pair; at 1 spacing, 13% the undisplaced pair, 80% a neighbouring pair, 7% a straddling pair; at 1.25 spacings, 7% the undisplaced pair, 57% a neighbouring pair, 7% a straddling pair, 30% refused; at 1.5 spacings, 40% a neighbouring pair, 3% a straddling pair, 57% refused; at 2 spacings, 10% a neighbouring pair, 90% refused; at 2.5 spacings, 3% a neighbouring pair, 97% refused. 10 recovered intervals in all exclude the true angle.

A head displaced before it is counted

The round trip from a head's spiral counts back to its divergence angle was tested on heads whose every organ sat exactly where the rule put it. Displaced by a normal error of up to two and a half spacings, heads of 900 organs keep counting a pair from their own sequence and return intervals holding the true angle to a spacing and a half; heads of 300 organs move to the neighbouring pair by half a spacing and then refuse, nine in ten of them by two spacings. Every moved count brings in the family whose chord was third shortest. Of 898 heads recovered, 19 intervals miss the true angle and 17 of those by about a tenth of a degree — displacement makes the reading coarser and then silent, not confidently wrong.

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What a plant might be doing

Turing's last work was on this, and it was unpublished when he died: a ring of cells, two diffusing substances, and a spacing that selects itself. The modern account uses auxin and a pump that works uphill. Both are made to predict a number and then to produce it.

Which patterns grow on a ring of circumference 1.40. Modes 4 to 15 have positive growth rates and mode 8 is fastest. Integrating the full equations from a disordered start gives 8 peaks.

Turing's last problem

Turing's final work was on phyllotaxis and it was unpublished when he died. Its core is a ring of cells and two diffusing substances, and the thing it does is select a number of peaks — which can be predicted from the equations before anything is integrated, and then counted from what the integration produces.

6 figures The angle between the first two peaks, for six starting disorders. The same equations, the same ring, six different starting perturbations: 177°, 47°, 109°, 151°, 47°, 133°. A spiral needs the angle between successive elements to be the same one each time, and a stationary ring does not supply that.

A ring cannot make a spiral

The peaks on a Turing ring do not all appear at once — there is a first and a second. But which two lead is decided by the starting disorder, so the angle between them comes out at 177°, then 47°, then 109°, then 151°. A divergence angle is a relationship that repeats, and this one does not.

8 figures 48 cells with a carrier that pumps auxin up the gradient. Each short line is one cell's polarisation — the neighbour it pumps towards, which is always the richer one. 10 peaks come out, at a contrast of 93%, from a start that was uniform to within 6%.

A pump that works uphill

The mechanism that actually has molecular support behind it does not use a diffusing inhibitor at all. Cells move auxin towards whichever neighbour already has more of it, which is the opposite of what transport is supposed to do — and it produces a spacing from a field that started uniform to within six per cent.

8 figures Which patterns grow on a ring of circumference 0.80. Modes 2 to 8 have positive growth rates and mode 4 is fastest. Integrating the full equations from a disordered start gives 4 peaks.

What a mechanism would have to show

Every model of phyllotaxis comes with the caveat that reproducing a pattern is not explaining it. That is easy to repeat and hard to make precise. Here it is made precise — a list of what an account of phyllotaxis would have to establish, with each item marked according to whether the models drawn in these essays establish it.

8 figures Three disturbances, three places to get in. The rule reads its neighbours, builds a profile of the energy at every azimuth, takes the least of it, and records a position. field noise enters at the profile; jostle noise enters at the neighbours; placement noise enters at the record. Two of the three are upstream of the choice and can change which minimum is taken; the third is downstream and never can.

The noise that arrives through the neighbours

The two kinds of noise this site had were idealisations that bracket the rule's choice. The realistic disturbance is neither: a primordium is placed exactly, and then the organ grows, so by the time the next one forms its neighbours have moved. That is a third kind, and it is invisible in every measurement a plant offers.

8 figures Three disturbances, three places to get in. The rule reads its neighbours, builds a profile of the energy at every azimuth, takes the least of it, and records a position. jostle noise enters at the neighbours; placement noise enters at the record. Two of the three are upstream of the choice and can change which minimum is taken; the third is downstream and never can.

A growing organ is part of the rule

Every model in the earlier essays places primordia on a surface and then treats the surface as furniture. But the surface grows between one placement and the next, and that growth reaches the rule through the only channel it has — where the neighbours are. What looks like a boundary condition turns out to be a term in the model.

7 figures The same lattice with no rule in it. A cylindrical lattice at a divergence of 137.826° and a rise of 0.005, built by placing node i at exactly i times the divergence and then displacing each azimuth independently by 0.5°. Its photograph is the photograph of the stem in the figure beside it and its parastichy pair is the same pair. The largest comb mean in it is 0.03 against a sampling band of 0.07, and the readout refuses.

A comb is evidence of a rule

Build the same lattice kinematically — every node at an exact multiple of the divergence, an independent error on each azimuth, no feedback anywhere — and the spectrum is empty. The photograph is identical and the parastichy pair is identical. The comb is not a property of the arrangement.

8 figures A lattice with independent errors. The autocorrelation of 759 divergence angles from a kinematic lattice at a divergence of 137.8261° and a rise of 0.005, with 0.5° of scatter on each azimuth. There is no placement rule anywhere in it: node i is put at exactly i times the divergence and then displaced. The only thing that differs between this figure and the control is the structure of the displacement — here, independent errors — that earlier work's control at 0.5° of independent scatter. The largest comb mean is 0.029 against a sampling band of 0.073, and the readout refuses.

A disturbance with a memory

That earlier work's control assumed that a plant's errors are independent from organ to organ, and nobody had tested it. Give the errors a memory — each one a fraction of the last, up to a coefficient of 0.97 — and the comb does not appear. The obvious threat to the result turns out to be empty, and the algebra says why before the measurement does.

8 figures A lattice with an error inherited from the two contact neighbours. The autocorrelation of 759 divergence angles from a kinematic lattice at a divergence of 137.8261° and a rise of 0.005, with 0.5° of scatter on each azimuth. There is no placement rule anywhere in it: node i is put at exactly i times the divergence and then displaced. The only thing that differs between this figure and the control is the structure of the displacement — here, an error inherited from the two contact neighbours at coupling 0.7. The largest comb mean is 0.514 against a sampling band of 0.073, and the readout returns 8/13.

Errors that pass between organs

An organ's neighbours are the ones eight and thirteen places back — that is what a parastichy pair is. So a disturbance transmitted by contact is correlated at exactly the two lags the readout examines, and it does not have to be told them. Driven into a lattice with no rule in it, it returns the counted pair on eight stems out of eight.

8 figures The two combs, in the proportions the rule gives them. The ratio of the second comb to the main one, for a kinematic lattice whose errors are inherited from its two contact neighbours, against how unevenly that inheritance is split. The horizontal line is where the placement rule's own stems sit, at 0.65. Weighted by distance — the coupling a d⁻³ interaction would give, which at this rise favours the 13-neighbour by 1.26 to one because the 13-hop is the shorter — the forgery sits at 1.46, well above the rule. It reaches the rule's value only at about 3 to one the other way, which is a factor of 4 against what distance supplies and in the opposite direction.

What a forgery has to know

A lattice with transported errors reproduces the comb and the pair, so one quantity is left: the two combs' relative strength. Weighted by distance the forgery puts more in the second comb than the first; the rule does the opposite. It matches only if the coupling is turned three to one towards the further neighbour, which no falloff supplies.

8 figures Take away the organ eight places back, and the next one goes into the hole. The last 34 organs of a stem at a rise of 0.005, unrolled. The open circle is the organ removed — eight places before the tip. The ring at the top is where the rule puts the next organ with every organ present; the filled mark beside it is where the rule puts it with that one missing. The two are 16.4° apart, against a local spacing of 25°, and the vacancy itself is 22.7° from the undisturbed answer. Nothing else differs between the two runs: same rise, same history, same rule.

The organ that was taken away

Every observable this site has is read off an arrangement that was finished before the reading began, and earlier work here showed what that costs. So remove one primordium from a settled stem and place the next one against what is left. The rule has to answer. The rival account cannot, because in it no organ's position was ever computed from its neighbours.

8 figures