Phyllotaxis
the spirals are counted, not admired
Stems and cones

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.
13 min read 20 figures State the claim, state the testForm is not mechanism

Worth reading first: Where a handover sits · The organ that was taken away · The angle the ladder returns to.

The step ordering is not what decides which family survives a removal. That was established on two bands, fifty-five wrecked cuts and two counted pairs, and it is the load-bearing negative of this whole thread.

There are six bands now. This essay cuts them all, and the claim survives in a narrower form than it was stated in — with the narrowing being the more interesting half.

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.
Fig. 1 Every band, with the cuts made on it and the families left standing anywhere across it.

What is cut

Each band is swept for its geometry at two parts in a thousand, which on the widest is 126 rises. It is cut at nine of them, evenly spaced in the logarithm of the rise, plus both ends and the handover itself.

Sixteen cut stems a rise, each grown three hundred organs past the cut with a control beside it, is why. Cutting every rise of every band is several hours of computation for a quantity the design predicts to be constant, and a constant is tested at the extremes and at the crossing rather than by sampling the middle more finely.

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.
Fig. 2 The six bands with their widths and the separation of their two steps at each end.

Which bands wreck

Four of the six. The golden 5/8 band wrecks 16 cuts, the golden 8/13 band 43, the Lucas 4/7 band 16 and the Lucas 7/11 band 39 — 114 wrecked cuts against fifty-five before.

The two that wreck at nothing are the golden 3/5 band and the Lucas 3/4 band, both at the coarse end of their branches, where a single removal cannot wreck a stem at any offset. That is the coarse rung behaving as measured rather than a failure of the design, and both rows are in the table.

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.
Fig. 3 The same table with the two bands that wreck at nothing listed rather than dropped.

The claim, stated over the crossing

At every offset of every band, take the nearest cut rise above the handover and the nearest below it. The two contact steps are the other way round at those two rises; the counted pair is the same; the settled divergence differs by less than a twentieth of a degree.

The family left standing is the same at both. On every offset of every band where both comparisons exist.

That is the claim the design is for, and it is now carried by four counted pairs — 5/8, 8/13, 4/7 and 7/11 — on two branches.

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.
Fig. 4 The two curves whose crossing is a handover, which is the rise the comparison is taken either side of.

What is not true

Two weaker-looking statements, both of which an earlier version of the check asserted, and both false.

The first: a band keeps one family. Three of the four do — the 5/8 band keeps only the 5, the 4/7 band only the 4, the 7/11 band only the 7 — and the 8/13 band keeps the 4 at one offset and the 8 at the others. That is the census’s oldest result: different offsets keep different families, and it has nothing to do with the ordering.

The second: a fixed offset keeps one family across a whole band. On the 8/13 band, offsets six, seven and eight change answer somewhere inside it.

Which lag survives, at every lattice and every offset. A row for each of twelve lattices and a column for each offset an organ is removed at, with the lag whose hop survived written in the cell. 30 cells never repair. The lag left standing is one of the two contact families at 29 of them, and at 17 it is the second shortest step on the lattice rather than the shortest, which is what refuses the reading that a rule holds its nearest neighbours. The ringed cells are the five the offset rule gets wrong. Two lattices carry no cells at all because no single removal wrecks them.
Fig. 5 The census’s grid of offsets against surviving families, which is where the first false statement comes from.

Where those changes are

Not at the handover.

At offset seven the survivor is 8 at every rise from 0.00682 down to 0.00601 and 4 from 0.00584 down. The handover is at 0.00605, so the change sits two sweep steps below the rise where the ordering reversed, at a place where the ordering has already been settled for two steps.

At offset six the change sits four steps below. At offset eight the answer does not change once at all: it goes 8, 8, 8, 8, 8, 8, 8, 4, 8, 4 down the band, alternating rather than switching.

An alternating answer is what a quantity near a boundary does, and it is not what a quantity following a single reversal does.

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.
Fig. 6 The cuts on each band, of which only the finest golden row keeps more than one family anywhere.

Why the widest band shows it

Because a band is a narrower experiment than its two ends suggest, and the 8/13 band is the one wide enough for that to matter.

It covers 72 per cent of its rung and spans a factor of 1.28 in the rise. Across it the divergence is held to 0.043° and everything else the rise controls moves: the neighbourhood size, the front’s depth, the lengths of both contact steps, how many organs are within reach of the placement rule.

The two bands the previous round built were swept at an absolute step of 0.0002 and covered about a fifth of that span in the rise. There was nothing wrong with them; they were narrow, and a narrow band cannot show that a wide one is not clean.

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.
Fig. 7 The six bands as stretches of their rungs, on which the finest golden one is most of a rung.

Which makes the design’s limits visible

A band holds two of the three quantities that move along a rung. It does not hold the rise, and the rise is what everything else here is a function of.

So “held the divergence and the counted pair” is not the same as “held everything but the ordering”. It is held everything the rung labels, which is two things, while the rise varies by up to twenty-eight per cent underneath.

That is a real limitation of the design and it is worth stating now rather than when something else fails to reproduce. The way to see it was to build a band wide enough to break, and the way to break it was a step that resolves fine rungs.

The divergence slides along 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.
Fig. 8 The divergence across a whole rung, of which a band is the flattest stretch and not a point.

The shortest-hop reading, again

Across the 114 wrecked cuts, the surviving family is the shorter of the two contact steps on 44 of them.

That is the reading this thread has been refuting for two rounds, scored on a set of cuts built specifically so that the ordering reverses underneath. Forty-four of 114 is not a score; it is what a reading looks like when the quantity it is stated over is not in the mechanism.

Read band by band it is 9 of 16, 14 of 43, 3 of 16 and 18 of 39 — which varies enormously and tracks how much of each band sits on which side of its handover rather than anything about the cuts.

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.
Fig. 9 One band’s cuts with the surviving family and the step ordering at each rise.

What the check now asserts

Three things, and the shape of them is the point.

That the family is the same either side of every handover, on every offset where both comparisons exist, with at least eight such offsets across the bands.

That the ordering really does reverse inside every band the claim is scored on — a constancy under nothing is not a constancy.

And, separately, that not one of the places where a fixed offset’s answer does change sits at a handover. That is asserted over the changes rather than asserted away, so a future sweep that produced a change at a crossing would fail rather than be absorbed.

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.
Fig. 10 The bands and their handovers, which is what the three assertions are indexed by.

What a first version would have reported

Five bands with a clean constancy and one refutation.

The version of the check that asked whether a whole band keeps one family fails on the 8/13 band, and the obvious readings of that failure are both wrong: either the ordering does decide the survivor after all, or the 8/13 band is defective and should be dropped. Neither is what happened. The band keeps two families because its offsets keep two families, which was known before the band existed.

That is the second time in this round a check written at the wrong granularity produced a wrong answer with a plausible face on it, and both times the repair was to ask the question the design actually poses.

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.
Fig. 11 The families kept per band, which is the column the wrong version of the check read.

What still varies across a band

Something does, on the widest one, at three offsets. What it is, this essay does not say.

The candidates are the ones a band does not hold: the rise itself, the neighbourhood size it sets, the front’s depth, the absolute lengths of the two steps. Any of those varies by tens of per cent across a 1.28× band, and separating them needs a design that holds the rise — which nothing here has.

It is worth noticing that the offsets where the answer changes are six, seven and eight, on a stem counted 8 and 13. Offset eight is a counted number; six and seven are not. That may be nothing.

The next organ moves for the last 13, and for no others. One row per organ removed, counted back from the tip of a stem at a rise of 0.005 whose counted pair is 8 and 13. Removing any of the last 13 moves the next organ by 2.6° to 167.6°; removing an older one moves it by at most 0.47°, which is under the azimuth grid. The boundary is at 13, and 13 is the larger parastichy number — so the experiment counts the spirals without measuring an angle.
Fig. 12 The offsets a fine stem wrecks at, which is the axis the changes here sit on.

What each band contributes

The golden 5/8 band wrecks at three offsets across ten cut rises, sixteen cuts in all, and keeps the 5 at every one of them. That is the band the original result was built on, swept here at a finer step and over a wider stretch, and it gives the same answer.

The Lucas 4/7 band wrecks at two offsets, sixteen cuts, and keeps the 4 throughout. It is the other original band, and it keeps a different family from the one the census’s 4/7 stems keep at other rises — which is the offset’s doing rather than the band’s.

The Lucas 7/11 band wrecks at five offsets across ten rises, thirty-nine cuts, and keeps the 7 at all of them. It is new this round and it is the largest clean contribution.

The golden 8/13 band wrecks at five offsets, forty-three cuts, and keeps the 4 and the 8. It is new, it is the widest, and it is the one that does not behave.

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.
Fig. 13 Each band’s cuts and what it keeps, with the two that wreck at nothing listed.

The cuts that keep nothing

None, across all 114.

That is worth stating because it is not guaranteed: a wrecked stem can keep no rigid hop at all, and when it does, a count of surviving families has a null in it that must not be folded into a family. Two-organ cuts produce nulls routinely and single-organ cuts on these bands produce none.

So every one of the 114 comparisons is between two families rather than between a family and an absence, which is the cleanest form the comparison can take.

Which hops survive one wall, the other, and both. One row per lattice. The last three columns are the lags whose hop the cut stem still holds, unchanged from a control that shares its history — the measurement that identifies what a wrecked stem has become. Removing a single wall always leaves something standing, which is what every single-organ cut in this collection does. Removing both leaves nothing at all on two of six lattices, including the coarse rung that no single removal can wreck. A stem that keeps no rigid hop is not a wrecked lattice with a slip in it; it is a stem that is no longer a lattice.
Fig. 14 Cuts that keep nothing, from a different design, which is what these 114 do not contain.

What a wider band would have shown sooner

The 8/13 band is 72 per cent of its rung and the two original bands were about a fifth of that in the rise. If the first version of this design had swept at a ratio, the 8/13 band would have existed in the previous round and the narrowing would have happened then.

That is not a criticism of the previous round; it is a consequence of a step size, and the step size was the repair this round made. What it says is that the cleanliness of a design can depend on a parameter of the sweep that has nothing to do with the design’s logic, and that the way to find out is to push the parameter until something breaks.

Here what broke is small — three offsets on one band — and it broke in a direction that leaves the central claim standing. It did not have to.

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.
Fig. 15 The widths the ratio step produced, of which the largest is the band that shows the limitation.

Read against the matched pairs

Two designs, two negatives, and they now rest on comparable weight.

The band design says the ordering does not decide the survivor: 114 cuts, four counted pairs, two branches. The matched-pair design says the settled divergence is not sufficient: four comparisons, four counted pairs, two branches.

The first has thirty times the cuts and the second has the cleaner logic — a matched pair varies one labelled quantity and a band varies one labelled quantity while the rise moves underneath. Neither is redundant, and what they leave standing is the counted pair.

The same angle, two rungs, two answers. Each block is one matched pair: two rises whose stems settle on the same divergence and whose counters return different pairs. Under each is the family every wrecked cut leaves standing. On three of the four pairs both rises wreck at some offset, and on every one of those the two stems keep different families — so the divergence, which is held, is not what decides the survivor. The two stems keep exactly the counted numbers their two pairs share, including the pair that shares none and keeps none.
Fig. 16 The other design’s comparisons, which are fewer and hold a different quantity.

What the comparison needs at each offset

Both a wrecked cut above the handover and one below it. Where an offset wrecks on only one side, there is nothing to compare and the row is skipped.

That happens often enough to be worth counting. On the golden 5/8 band, cutting three places back wrecks at three rises and all three are above the handover; on the Lucas 4/7 band, cutting five places back wrecks only below it. Both are skipped, and the check asserts that at least eight offsets across the four bands do bracket their handover — so a future sweep in which almost nothing brackets could not pass by comparing two rows.

Which offsets wreck at which rises is itself a measurement and not a nuisance: the set of offsets that wreck a stem is the front, and it moves down a rung, so a band that crosses enough of a rung will see offsets appear and disappear.

On a rung the response is a run of offsets; near a transition it has a hole. One row per rise, from 0.032 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 — 5, 8, 13 at the pairs shown on the left. Every rise shown here is in the middle of its rung, so every row is a run and nothing past it is felt — what happens between the rungs is a separate matter.
Fig. 17 Which offsets a removal is felt at, by rise, which is what decides whether a band’s comparison exists at a given offset.

What the alternation means

At offset eight on the 8/13 band the answer runs 8, 8, 8, 8, 8, 8, 8, 4, 8, 4 down ten cut rises. That is not a switch; it is two answers appearing alternately at the fine end.

The obvious reading is a boundary. If two families are nearly equally available at those rises, small differences between adjacent rises could tip a cut either way, and what looks like noise is a genuinely marginal decision being made ten times.

Testable, and not tested: cut at every rise of the band rather than at ten of them, and the alternation is either a pattern with a period or a scatter. That is sixteen cut stems times 126 rises for one band, which is the cost the sampling was adopted to avoid, and it is the one place on these six bands where paying it would buy something.

Which lag survives, at every lattice and every offset. A row for each of twelve lattices and a column for each offset an organ is removed at, with the lag whose hop survived written in the cell. 30 cells never repair. The lag left standing is one of the two contact families at 29 of them, and at 17 it is the second shortest step on the lattice rather than the shortest, which is what refuses the reading that a rule holds its nearest neighbours. The ringed cells are the five the offset rule gets wrong. Two lattices carry no cells at all because no single removal wrecks them.
Fig. 18 The census’s own view of which family is kept at each offset, against which the alternating rows can be placed.

What the two coarse bands would need

An experiment that can wreck a coarse stem, and there is one. Removing both walls of the slot wrecks a 3/5 stem where either alone heals, so a two-organ cut across the golden 3/5 band would produce the comparison a single removal cannot.

That is worth naming as available rather than as done. The band exists, it is a proper matched pair with its ends separated by four per cent, and the only reason it contributes nothing is that a single removal at the coarse end has nothing to sever. Cutting it with two organs is one change to a loop.

The Lucas 3/4 band would still contribute nothing, for a different reason, which is the distinction between the two empty rows.

Removing one wall of the slot, then the other, then both. The growing tip sits in a slot between its two chain-neighbours — the organ 3 places back and the organ 5 places back. Each bar is how far the next organ placed moves when those are removed, against a control sharing the same history. Either alone is a cheap removal: 41.7° and 20.6°, both inside the 45° that separates taking a neighbour from taking anything else. Together they move it 47.8°, against 62.3° for the two effects added, so the interaction is -14.5°. The slot is not two independent walls.
Fig. 19 The cut that wrecks a coarse stem, which is what the golden 3/5 band is waiting for.

The one line

A hundred and fourteen wrecked cuts across four bands, on four counted pairs and two branches, and at every offset of every band the surviving family is the same immediately above the handover and immediately below it with the ordering reversed between. The answer does change inside the widest band at three offsets — two to four sweep steps away from the crossing, and at one of them by alternating — which says a band a quarter as wide in the rise was a cleaner experiment than anybody had reason to know.

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.
Fig. 20 The whole table, which is what the claim now rests on.

Where this goes next

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. 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. 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.

What links here

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

Shares its objects with

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

  • The shortest hop was a coin flip — both name ablation, claim testing, control, falsifiability, lattice offset, negative result, parastichy pair, rigid hop, rise, rung, underdetermination
  • One rung, two answers — both name ablation, control, falsifiability, lattice offset, negative result, parastichy pair, rigid hop, rise, rung, underdetermination
  • The family that lost a member — both name ablation, claim testing, control, falsifiability, lattice offset, negative result, parastichy pair, rigid hop, rung, underdetermination
  • The organ that was nobody's neighbour — both name ablation, claim testing, control, falsifiability, lattice offset, negative result, parastichy pair, rigid hop, rung, underdetermination
  • The side the census sat on — both name ablation, claim testing, control, handover, lattice offset, negative result, parastichy pair, rise, rung, underdetermination
  • The front deepens down a rung — both name ablation, claim testing, control, lattice offset, negative result, parastichy pair, rigid hop, rise, rung

Named objects

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

AblationClaim testingControlFalsifiabilityHandoverLattice offsetMatched designNegative resultParastichy pairResolutionRigid hopRiseRungUnderdetermination