Phyllotaxis
the spirals are counted, not admired
Stems and cones

The rung was not the instrument

The previous phase 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.
19 min read 12 figures State the claim, state the testCounted, not admiredThe round trip

Worth reading first: The sequence has a memory · The Fibonacci ladder · Two numbers out of the points.

The previous phase closed with four leavings and one of them was a ceiling:

The reading needs three multiples of the smaller parastichy number inside the lag window and needs the window short enough that the larger number is not itself a candidate spacing. At 5/8 and 8/13 that is satisfiable; at 13/21 there is no window that satisfies both, so the instrument as written has a ceiling one rung above where it works. That is arithmetic rather than statistics and no amount of stem fixes it.

Every clause of the first sentence is right. Everything after it is wrong, and the arithmetic that was supposed to prove the ceiling disproves it in two lines.

Every rung has a window it can be read inFor 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 previous phase used everywhere: it sits inside the 8/13 band and outside the 13/21 one, which is what was mistaken for a ceiling.runglag window, in internodes0153045605/8rise 0.0131524read at 218/13rise 0.0052439read at 3013/21rise 0.00193963read at 45the default window: 30three teeth in, the larger number outgenerated from a stated rule, not drawn to look right
Fig. 1 The two requirements, drawn. A comb needs three teeth, and its teeth are at multiples of the smaller parastichy number, so the window must reach 3m. The larger number must not become a rival spacing, so the window must stop short of 3n. The band is [3m, 3n) — [15, 24) at 5/8, [24, 39) at 8/13, [39, 63) at 13/21 — and it is never empty, because it is empty only when the larger number is no larger than the smaller. The dashed line is the window the previous phase used everywhere.

The band is never empty

Write the two requirements as inequalities. Three teeth of the main comb inside the window: L ≥ 3m. The larger number not itself a candidate: L < 3n. So the window has to satisfy 3m ≤ L < 3n, and since m < n at every rung there is always room. At 8/13 the band is [24, 39) and thirty sits in it, which is why the readout worked. At 13/21 the band is [39, 63) and thirty does not, which is a statement about the default rather than about the rung.

Widen the window to 45 and the requirement is met. That is not a repair to the instrument; it is the instrument being used at the setting its own arithmetic prescribes.

The mistake is a small one and it is the kind that gets written into a plan file at the end of a phase: the two constraints were both correctly identified, neither was written down as an inequality, and the conclusion that they are incompatible was reached by reasoning about one example. It survived a phase because a leaving is not gated.

Every transition as the rise fallsThe 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 0.3820.0.2500.5000.75011.25-2.50-2-1.50-1-0.500log₁₀ of the rise between nodes (falling to the right is the plant growing)log₁₀ of the larger parastichy number2/33/55/88/1313/21500 rises, shortest vectors recomputed at eachratio 0.3820 against 1/φ² = 0.3820
Fig. 2 The rung the argument is about. As the rise falls the parastichy pair climbs the ladder — 2/3, 3/5, 5/8, 8/13, 13/21 — so the window that a reading needs climbs with it. A single default window can only ever be right for one or two rungs, which is the thing that should have been noticed rather than the thing that was concluded.

What was actually stopping it

Widening the window is necessary and it is not sufficient. Two things about the stem had to change with it, and both are more interesting than the arithmetic.

The seed has to be a piece of lattice. Every fixed-rise run on this site starts from eight nodes of ideal lattice at a stated divergence, which at a rise of 0.005 spans four hundredths of a circumference — a stretch of pattern the rule can continue. At a rise of 0.0019 the same eight nodes span fifteen thousandths, which is a cluster: eight points nearly on top of each other with no lattice structure in them at all. The rule has nothing to continue, the run wanders, and the recorded divergence scatter comes out at 115° — no pattern anywhere in it.

Forty seed nodes at that rise span the same height eight do at 0.005, and the run settles to a scatter of half a degree.

And the azimuth grid has to be finer than the pattern. With the seed fixed, the readout still failed on every stem at every disturbance from 0.08 to 0.18. The rule samples 384 azimuths when it takes its minimum — a step of 0.94° — and at this rung the pattern needs finer resolution than that. What happens is worth describing precisely, because it is not a rounding error: the minimum lands on the same grid point run after run, every divergence comes out at exactly the same value, and what is left to autocorrelate is a deterministic repeat. The readout refuses it as a cycle rather than a sample, which is a refusal the previous phase built for exactly this failure and did not connect to this rung.

The 13/21 rung, at two azimuth gridsFive stems at each of five disturbances, all at a rise of 0.0019, read through a lag window of 45 from a seed of 40 nodes. A filled mark is a stem that returned 13/21, which is what the position counter finds in it; an open mark is a refusal. The only difference between the two rows is how many azimuths the placement rule samples when it takes its minimum: 384, which every run on this site has used since the foundation phase, against 1152. At the coarse grid the rule locks onto its own sample points and the readout refuses 22 of 25 stems; at the fine one it reads all 25. The ceiling was a parameter of the program.disturbance0.080.10.130.150.18384 azimuthsstep 0.94°1 of 25 read 13/21scatter 44.9°1152 azimuthsstep 0.31°25 of 25 read 13/21scatter 0.4°rise 0.0019 · seed 40 nodesgenerated from a stated rule, not drawn to look right
Fig. 3 The measurement. Five stems at each of five disturbances, at a rise of 0.0019, read through a lag window of 45 from a seed of 40 nodes. The only difference between the two rows is how many azimuths the rule samples. At 384 the whole sweep returns the pair once in twenty-five; at 1,152 it returns it twenty-five times in twenty-five, on stems whose scatter is 0.36° to 0.53° and whose position counter agrees.

So the claim goes

The instrument has no ceiling at 13/21. It reads the rung on every stem at every disturbance tried, with the pair confirmed by a counter that is shown coordinates and no angles.

And the reason the previous phase could not read it was neither arithmetic nor statistics. It was two settings — a seed length and a sample count — that had been constants since the foundation phase and had never been asked whether they were binding.

That is the third time this site has recorded a defect of that shape. reach and maxWindow were one number in the rising model until a sweep of a non-binding parameter reported the same answer at every setting and was read as robustness. A recency cut-off in the interaction-range work turned out to manufacture a lattice the model did not otherwise have. And now a sample grid sets which rungs an instrument can reach.

The common form: a parameter of the program, mistaken for a property of the model. Each was found by changing the parameter, and each had passed every gate the fleet has, because a gate checks that a computation is correct and none of these computations was incorrect.

Which offsets give short hops, at a rise of 0.0019The two lowest points are at 13 and 21, 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.0123102030index offsetmedian hop between node i and node i+m23300 nodes, 34 offsets triedshortest at 2 and 3
Fig. 4 The rung being read, from the lattice’s side. At a rise of 0.0019 the three shortest offsets are 21, 13 and 8 — 0.0409, 0.0487 and 0.0533 of a circumference — and the counted pair is 13 and 21, one rung finer than everything else in this thread. The lag window a reading needs is set by these numbers, and the numbers are set by the rise.

The failure is bimodal, which is how it hid

One detail of the measurement is worth pulling out, because it explains why the problem looked like a ceiling rather than like a setting.

A run at the fine rung does not degrade gradually as the seed is shortened or the grid coarsened. It either settles onto a lattice — scatter half a degree, pair read cleanly — or it never finds one at all, and comes back with a scatter of over a hundred degrees. There is nothing in between. At twenty-four seed nodes, four runs of five settle and one does not, and the mean scatter across the five is 23°, which is a number describing no run that happened.

So a sweep that averages over seeds sees a quantity that moves smoothly from 0.5° to 115° as the setting changes, and reads it as a pattern degrading. What is actually happening is a fraction of runs falling off a cliff. The two look identical in any summary and are completely different failures: one says the instrument is reaching its limit, the other says some runs are broken and the rest are fine.

That is worth a general note, because this site averages over seeds everywhere. A mean over an ensemble whose members are bimodal is a number describing none of them. Where a run can fail outright, the thing to report is the fraction that succeeded and the distribution of what the successes gave — which is what the figures in this essay do, one mark per stem, and it is why the cliff is visible in them.

What a leaving is worth

Two of this phase’s four leavings from the previous one turned out to be wrong in the same way: a claim written into a plan file at the end of a phase, on the basis of reasoning rather than of measurement, and carried forward as though it had been established. The other was the control that this phase withdrew.

That is not an argument against writing leavings down. A leaving is how a phase hands its unfinished business to the next one, and unfinished business that is not written down is unfinished business that is forgotten. It is an argument for marking which kind each one is:

A leaving that reports a measurement — the dip’s coefficient runs 1,400 to 3,300 and three denominators do not support a q-dependence — is a fact with a number attached and it held up.

A leaving that reports a conclusion — there is no window that satisfies both, and no amount of stem fixes it — is an argument, and an argument that has not been run against the machinery is a hypothesis wearing a result’s clothes.

Both kinds belong in a plan. Only one of them should be quoted in the next phase’s essays without being checked first, and this phase checked two of them and found both wrong.

What it cost, and why the default did not change

Tripling the sample count triples the cost of every noisy run, and this site grows a great many of them. So the finer grid is used where it buys a rung and the default stays at 384, which is a decision that needs the evidence that it is safe.

The evidence is in the next essay: at the two rises the site has already argued from, the readings are identical at both grids, five runs of five. Nothing already written depends on the sample count. What does change at the finer grid is the deliberately marginal rise where the two instruments were reported as disagreeing — and a stem with no answer answering differently on a different grid is the object behaving as it was said to.

The angles against the positions, rise by risethree 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.risefive stems, read from the angles alonethe position counter0.032refusedrefusedrefusedrefusedrefused3 and 50.0135/85/85/85/85/85 and 80.0058/138/138/138/138/138 and 13seeded at 137.3°, 900 nodes per stemfilled where the two instruments agree
Fig. 5 The published rungs, read at the site’s own settings. These are the numbers the collection’s arguments rest on, and this phase’s finding leaves every one of them where it was — which is the first thing to establish when a setting turns out to have been doing something invisible.

Why nobody had looked

The rung was never attempted, and the reason is worth recording because it is not laziness.

Reading at 13/21 costs more than reading at 8/13 in three ways at once. The stem has to be longer, because a comb at spacing 13 needs its teeth out to lag 39 and the sampling band falls as one over the square root of the sequence length, so the same clearance needs more angles. The seed has to be longer, which means more of the run is transient rather than measurement. And the sample grid has to be three times finer, which triples the cost of every node.

Multiplied together that is roughly an order of magnitude in computation per stem, against a rung that no essay needed. So the previous phase asked whether the instrument could reach it, reasoned about the lag window, concluded it could not, and moved on — which is a perfectly ordinary thing to do with a question that would cost a day to answer and does not block anything.

What made it worth answering now is that the same three settings turned out to be the whole content of the answer. A ceiling that is really three constants is a different object from a ceiling that is arithmetic: the first says the instrument is fine and the defaults are stale, and the second says a whole class of plants is unreadable. A survey designer told the second would have excluded every specimen at a fine rung, which is most of the interesting ones — the fine rungs are where sunflower heads and pine cones live.

What is visible in the outer part of a 4000-element organBoth 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.024680.2000.4000.6000.8001counting only outside this fraction of the organ's length or radiustransitions inside the counted partdisc: 2 beyond 50%cone: 1 beyond 50%flare 0.12 · 4000 elements1 against 2 in the outer 50%
Fig. 6 And why the fine rungs matter. A real head passes through several of them as it grows, and the finest one it reaches is the one a counter sees at the rim. An instrument that stops at the eight-and-thirteen rung stops one rung below where most of the counting in this subject is actually done.

What the reading needs, stated as a rule

The instrument now has three settings that depend on the rung rather than on the site’s history, and each has an arithmetic reason:

The lag window must satisfy 3m ≤ L < 3n. In practice: read at a short window first, take the spacing it returns as an estimate of m, and re-read at 3m + 6.

The seed must span a fixed height rather than a fixed number of nodes — about 0.076 of a circumference, which is 8 nodes at a rise of 0.01 and 40 at 0.0019. A seed shorter than that is a cluster and the run never finds a lattice.

And the azimuth grid must resolve the pattern, which at the fine end means about 1,150 samples rather than 384. The requirement scales with how finely the rule has to discriminate between competing minima, which tightens as the rise falls.

The first is a consequence of the readout’s own construction. The second and third are consequences of the model. All three were constants, and none of them should have been.

The plane of stems: divergence across, rise upEach 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°.-2-1100120140160180divergence angle (°)log₁₀ of the rise between nodes1,2,32,3,554 × 150 lattices, each solved747 runs drawn
Fig. 7 Why the requirement tightens down the ladder. The rungs are geometric — each a factor of φ² finer than the last — so the region of parameter space a pattern has to be resolved within shrinks by the same factor each rung. A fixed sample count therefore buys fewer and fewer rungs, and where it runs out is a property of the count rather than of the ladder.
What a divergence picked at random gives, at a rise of 0.002Fibonacci pairs take 6.6% of the circle at this rise, and the share falls as the rise does. The claim that Fibonacci counts are what nature "prefers" needs the preference to come from somewhere, and it is not from the geometry being generous.Fibonacci6.6%Lucas0.5%whorled42.0%other50.9%177 distinct pairs over 1200 divergencesrise 0.002Fibonacci 6.6%
Fig. 8 The census at the fine rise, which is what the position counter is reading against. The thirteen-and-twenty-one pair occupies a band of divergence angles here, and the stems settle inside it — so the angle readout’s answer and the counter’s answer are both answers about the same lattice.
Six stems built, forgotten and recoveredEach 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°.137.51°, rise 0.092.8e-14°counted 2/3137.51°, rise 0.031.7e-13°counted 3/5137.51°, rise 0.0122.8e-14°counted 5/899.50°, rise 0.083.0e-13°counted 1/3151.14°, rise 0.071.1e-13°counted 2/399.50°, rise 0.027.1e-14°counted 4/7error in the recovered divergence anglecounts and hop lengths onlyworst 3.0e-13°
Fig. 9 The site’s strongest instrument, for comparison with the one this essay repairs. The round trip recovers a divergence and a rise from counts alone to fifteen digits, and it has never needed a rung-dependent setting — because it works on positions, which do not have to be resolved by a sampling grid. Everything fragile in this essay is fragile because the pattern is being grown rather than read.
The order of the angles carries the countThree 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.rise 0.032counted 3/5angles say 336912150.5rise 0.013counted 5/8angles say 55101520250.5rise 0.005counted 8/13angles say 8816240.5151015202530lag, in internodescorrelation between a divergence and the one that many internodes later3 runs per rise · 320 internodes eachthe counter is never shown a position
Fig. 10 The single-number readout, which is where this thread started and which has the same window arithmetic underneath it: a comb of three teeth at spacing m needs 3m lags. It was never tested above 8/13 either, and it reaches the fine rung under the same three conditions.
Two combs, at a rise of 0.005The 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.-0.50000.500125810131621242629lag, in internodescorrelation between a divergence and the one that many internodes later816245132129spacing 8 · offset 5pair 8/13 — counter says 8/13sampling bandone stem · 760 divergences · disturbance 0.25generated from a stated rule, not drawn to look right
Fig. 11 And the rung below, where all of this was worked out. The instrument that reads this stem is the same instrument; what changed is that three of its settings are now functions of the rise instead of constants left over from the phase that first wrote them down.
What the finer grid does to the rises already publishedThe 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 previous phase 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.risefive stemsthe position counter0.0133845/85/85/85/85/85/80.01311525/85/85/85/85/85/80.0053848/138/138/138/138/138/130.00511528/138/138/138/138/138/130.0083845/85/85/80.00811528/138/135/8the previous phase's settingsgenerated from a stated rule, not drawn to look right
Fig. 12 The check that makes this finding safe to have made, and the subject of the next essay: the two rises this collection argues from give identical readings at both sample counts, so nothing already written depends on the setting that turned out to be doing the work here.

Where this goes next

-0.50000.500125810131621242629lag, in internodescorrelation between a divergence and the one that many internodes later816245132129spacing 8 · offset 5pair 8/13 — counter says 8/13sampling bandone stem · 760 divergences · disturbance 0.25generated from a stated rule, not drawn to look right 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. risefive stemsthe position counter0.0133845/85/85/85/85/85/80.01311525/85/85/85/85/85/80.0053848/138/138/138/138/138/130.00511528/138/138/138/138/138/130.0083845/85/85/80.00811528/138/135/8the previous phase's settingsgenerated from a stated rule, not drawn to look right 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. 0.2500.5000.75011.25-2.50-2-1.50-1-0.500log₁₀ of the rise between nodes (falling to the right is the plant growing)log₁₀ of the larger parastichy number2/33/55/88/1313/21500 rises, shortest vectors recomputed at eachratio 0.3820 against 1/φ² = 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.

What links here

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

Reads more easily once this is understood

Essays that name this one as worth reading first.

Shares its objects with

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

  • A periodicity is not a lattice — both name artefact, autocorrelation, counting blind, divergence angle, ensemble, identifiability, lattice offset, measurement, parastichy pair
  • Two windows on one stem — both name autocorrelation, divergence angle, ensemble, honest limits, identifiability, ladder, measurement, parastichy pair, rung
  • A refusal with a reason — both name autocorrelation, divergence angle, ensemble, honest limits, identifiability, measurement, parastichy pair, rung
  • A window inside a rung — both name autocorrelation, divergence angle, honest limits, identifiability, ladder, measurement, parastichy pair, rung
  • The angles name the branch — both name autocorrelation, counting blind, divergence angle, ladder, lattice offset, measurement, parastichy pair, rung
  • What the pair costs — both name autocorrelation, counting blind, divergence angle, ensemble, honest limits, identifiability, measurement, parastichy pair

Named objects

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

ArtefactAutocorrelationCounting blindDiscretisationDivergence angleEnsembleHonest limitsIdentifiabilityLadderLattice offsetMeasurementParastichy pairRiseRungTransient