Phyllotaxis
the spirals are counted, not admired
The claims, measured

A slip that remembers is caught by a second pass

A pinned count errs only by slipping between spirals on the way round, and on 34/55 only a count two out is silent, which independent slips make rare. An eye that has lost its place once is likelier to lose it nearby, and runs of slips ought to make that silent error common. Holding the mean rate of slipping fixed, runs do raise it, by more than five times when each slip makes the next hundred times likelier. It still never reaches the whole error of a count taken at a chosen place at the rim, which is fifteen in a hundred against under half of one. Bunching the same slips together leaves fewer counts wrong in all. And two passes that must agree take the wrong readings kept from 5 per cent to a tenth of one, with runs or without.
14 min read 6 figures Counted, not admiredState the claim, state the test

Worth reading first: What a count is worth.

A pinned count errs where the spirals crowd modelled the careful way of counting a seed head. A person who picks one organ as the start of a family and returns to that organ has no closing error, unlike one who simply starts and stops at a chosen place. What is left is losing the way round, skipping a spiral or counting one twice where the eye passes between neighbours. Each crossing of a spiral has a stated chance of that, which falls with the square of how far apart the spirals are at that place, and two slips the same way are the error nothing can notice: on a reading of 34/55, a count two out still has no factor in common with the other, and passes as a report of a pair. A count that can be wrong by one had established what a count one out gives away; on this pair, a count one out shares a factor with the other and announces itself. A count that drifts by two had already shown how slippery two-out errors are.

Independent slips make the second accident rare, so the pin’s silent share came out small. That essay said so, and it also said what was wrong with the assumption. Eyes do not lose their place independently. A stretch of damaged florets, a moment of distraction, a place where the spirals look briefly wrong: any of those makes the next slip more likely than the last, and if slips come in runs, two out becomes far commoner than a product of small chances allows. Its closing question was how much clustering it would take for a pin’s silent rate on 34/55 to reach a place’s error rate, and whether counting each family twice catches what the clustering adds. Both have definite answers, and neither is what a person might have expected.

An eye with a memory

The change to the model is small. Crossing by crossing, a count’s net error can be computed exactly, by following it and a second number, how many crossings are left in the current run. After a slip, the next few crossings are hot: at each, the chance of slipping is multiplied by a stated boost, up to a hundred times. Runs lasting two, five and ten crossings are all read.

It matters how the baseline is treated. If the boost were simply added on, the eye would be making more slips in all and the comparison would be unfair, because a higher overall rate of slipping is a different eye. So the chance at a quiet crossing is lowered until the mean number of slips per crossing is exactly what the independent eye had. Clustering here moves slips together and adds none. The geometry is what the earlier essay used: each family’s crossing distance on the counting circle at the rim of the band that reads 34/55, an ideal Vogel head, an eye that slips about once in five hundred crossings at one organ’s spacing, and the larger family crossing closer.

What runs do to one count

How far off a pinned count of 55 spirals can be, with slips independent and with each making the next 20 times likelierFifty-five crossings at the rim of the band that reads 34/55, the mean chance of a slip the same on both eyes. One out: 2.1% against 1.8% each way. Two out: 0.02% against 0.08% each way. A count two out is the error a reading of 34/55 cannot notice. Runs move weight from one out to two out and beyond.10⁻⁷10⁻⁵10⁻³10⁻¹-4-3-2-1+1+2+3+4how far the larger count of 34/55 is outchance, logarithmicslips independenteach slip makes the next 20× likelierrun of 5 · a slip per 1274 crossings on averagegenerated from a stated rule, not drawn to look right
Fig. 1 How far off a pinned count of fifty-five spirals can be, with slips independent and with each making the next likelier. The vertical axis is logarithmic. The dial sets how much likelier each slip makes the next.

The picture of a single count shows what a run does. With slips independent, the larger count is one out in either direction with a chance of 2.1 per cent each, and two out with a chance of 0.02 per cent. If each slip makes the next twenty times likelier, for five crossings, the one-out chance falls to 1.8 per cent and two out rises to 0.08. At a hundred times, one out is 1.2 per cent and two out 0.16. So runs move weight from one out to two out, and from there on to three and four, at about the rate a person would guess.

Two things in that picture are worth keeping. The first is that the two-out share rises by a factor of four to eight. The second is that the one-out share falls, because slips that fall in a run cost the same number of slips but spoil fewer counts.

The silent share

How often a pinned 34/55 is wrong and gives no sign of it, as slips cluster. The rim of the band, the mean chance of a slip held fixed at 0.002 of a crossing at one organ's spacing. Hot for 2 crossings: 0.07% at 1×, 0.07% at 2×, 0.07% at 3×, 0.08% at 5×, 0.09% at 10×, 0.12% at 20×, 0.19% at 50×, 0.27% at 100×. Hot for 5 crossings: 0.07% at 1×, 0.07% at 2×, 0.08% at 3×, 0.10% at 5×, 0.13% at 10×, 0.18% at 20×, 0.29% at 50×, 0.38% at 100×. Hot for 10 crossings: 0.07% at 1×, 0.08% at 2×, 0.09% at 3×, 0.12% at 5×, 0.17% at 10×, 0.25% at 20×, 0.36% at 50×, 0.42% at 100×. Clustering raises the silent share several times over.
Fig. 2 How often a pinned 34/55 is wrong and gives no sign of it, against how much likelier a slip makes the next, for runs of three lengths. The mean rate of slipping is the same at every point.

The silent share responds in the same way. For runs of five crossings it goes from 0.07 per cent to 0.13 at a boost of ten, 0.18 at twenty, 0.29 at fifty and 0.38 at a hundred. For runs of ten crossings, the same boosts reach 0.17, 0.25, 0.36 and 0.42. So strong clustering multiplies the pin’s silent share by about five. That is what the earlier essay feared.

It should be read with the scale of the axis in mind. These are shares of a few tenths of a per cent. A count at this eye’s slip rate is wrong about one time in twenty in total, and only about one wrong count in ten is the silent kind even at the strongest clustering. So the error that was described as a pin’s special risk stays a small part of a pin’s error, and clustering makes it a less small part without making it the main thing.

Fewer wrong counts in all

How often a pinned 34/55 is wrong at all, as slips cluster. The rim of the band, the mean chance of a slip held fixed at 0.002 of a crossing at one organ's spacing. Hot for 2 crossings: 5.2% at 1×, 5.1% at 2×, 5.1% at 3×, 5.1% at 5×, 5.1% at 10×, 5.0% at 20×, 4.7% at 50×, 4.4% at 100×. Hot for 5 crossings: 5.2% at 1×, 5.1% at 2×, 5.1% at 3×, 5.1% at 5×, 4.9% at 10×, 4.7% at 20×, 4.2% at 50×, 3.7% at 100×. Hot for 10 crossings: 5.2% at 1×, 5.1% at 2×, 5.1% at 3×, 5.0% at 5×, 4.8% at 10×, 4.4% at 20×, 3.7% at 50×, 3.0% at 100×. The same slips moved together leave fewer counts wrong, since a run's slips can only spoil one count at a time.
Fig. 3 How often a pinned 34/55 is wrong at all, against how much likelier a slip makes the next. The same slips, moved together, leave fewer counts wrong.

The total error goes the other way, and that could easily be missed. With independent slips 5.2 per cent of readings are wrong. For runs of ten crossings that falls to 4.8 at a boost of ten, 4.4 at twenty, 3.7 at fifty and 3.0 at a hundred. No slips have been removed. A count ruined by three slips and a count ruined by one are both one wrong count, so when slips cluster, more of them fall inside the counts that were already wrong, and more counts get through with none at all.

So it is too simple to say that a tired eye counts worse. For a given number of slips, a tired eye that bunches them produces more accurate readings overall and a larger share of the errors it does produce are unnoticeable. A survey that estimated its error rate by recounting a sample would see the first and not the second, and might conclude that clustering made its counters more reliable. That is true in one sense, and it isn’t the sense that matters for a census.

A further effect sits among the errors too: a slip in a run that goes the opposite way undoes the one before it. The net count is what matters. At high boosts, that cancels a share of multiple slips, and it is why the silent share, which grows quickly at first, flattens out by a hundred times. Above that, the hot crossings are so slippery that a run’s net error is more or less a coin-toss walk over its length, and more boost adds little.

Two numbers that are one

A run has two settings, how much likelier a slip makes the next and how many crossings that lasts, and it would be awkward if both had to be measured. They very nearly reduce to one. A boost of ten held for ten crossings gives a silent share of 0.17 per cent. A boost of twenty for five gives 0.18. A boost of fifty for two gives 0.19. Each of those has a product of a hundred. A boost of five for ten gives 0.12, a boost of ten for five gives 0.13, and a boost of twenty for two, a product of forty, also gives 0.12.

The reason is simple arithmetic. What a run contributes is the extra slips it brings, and the chance of one more slip during a run is about the boost times the length times the quiet chance, so long as that is small. So an eye can be described, for this purpose, by one number: how many more slips follow a slip than would have happened anyway. On this eye, that number has to be well above one before anything much happens. Up to a boost of five for two crossings the silent share is 0.08 per cent, indistinguishable from independence. It is a useful simplification for anyone trying to measure this, since it says that a test does not have to separate how long an eye stays flustered from how flustered it is.

Which pair

How often a pinned count is wrong and silent, for four pairs, as slips cluster. Hot for five crossings, at the rim of each band. 13/21: 1.5% at 1×, 1.4% at 10×, 1.0% at 100×. 21/34: 0.02% at 1×, 0.04% at 10×, 0.14% at 100×. 34/55: 0.07% at 1×, 0.13% at 10×, 0.38% at 100×. 55/89: 1.7% at 1×, 1.7% at 10×, 1.9% at 100×. On 13/21 and 55/89 a count one out is already silent, so runs have little to add; on 21/34 and 34/55 it takes two out.
Fig. 4 How often a pinned count is wrong and silent, for four pairs, with runs of five crossings. The vertical axis is logarithmic.

All of this depends on the pair, and in a way that follows from the arithmetic of which miscounts announce themselves. On 21/34 and on 34/55, a count one out shares a factor with the other and the reader notices, so only two out is silent, and clustering raises the silent share several times over: on 21/34 from 0.02 per cent to 0.14, and on 34/55 from 0.07 to 0.38.

On 13/21 and 55/89 a count one out is already silent in some directions, and there the silent share is high to begin with, 1.5 and 1.7 per cent, and it barely moves. At a boost of a hundred, 13/21 is actually down to 1.0 per cent, since there are fewer one-out counts then, and 55/89 is up only to 1.9. So clustering hardly matters where the error is already silent, and it matters most on the pairs that the coprime check protects. It wears away a protection that was never as strong as it looked.

It also means that the silent share of 34/55 and of 21/34 together does not reach that of the other two pairs at any boost read here. Whatever is dangerous about clustering, the worst-placed pairs for a silent error remain the ones that were worst to begin with.

Against a chosen place

A pinned 34/55's silent share as slips cluster, against the whole error of a placed count. Hot for ten crossings. At 0.0005 a crossing: 0.00% independent and 0.06% at 100×; At 0.002 a crossing: 0.07% independent and 0.42% at 100×; At 0.005 a crossing: 0.39% independent and 1.0% at 100×. A placed count at the rim is wrong 15.3% of the time and a third of the way in 1.1%.
Fig. 5 A pinned 34/55’s silent share as slips cluster, at three eyes, against the whole error of a count that starts and stops at a chosen place, at two places on the head. Runs of ten crossings.

The question as posed was about reaching a place’s error rate. A count that starts and stops at a place in the rim of this band is wrong 15.3 per cent of the time from the closing alone, however careful the counter. A third of the way in, that is 1.1 per cent, since the closing error is smaller against the wider family spacing there. So there are two targets.

The silent share never gets near the first. At a boost of a hundred and runs of ten, at an eye slipping once in five hundred crossings, it is 0.42 per cent: thirty-six times smaller than the place’s error at the same circle. Against the second, it reaches 0.42 of 1.1, and an eye slipping one in two hundred gets there, at 1.0 per cent. That eye is a poor one, with a total error of more than one in ten.

So the answer is that it cannot be done by clustering at a sensible slip rate. A pin’s silent share reaches a place’s error only for an eye so careless that the pin’s whole advantage has already gone, and the comparison needs the place a third of the way in and not the place at the same circle. Pinning remains the better choice at any boost, for anyone who can count at all. That recommendation from the essay on the pin stands. So does the first count anyone is taught to distrust, which was never pinned at all.

Two passes

What counting each family twice, from two pins, leaves of a pinned count's errors. Two independent passes at each family, a reading kept only when both agree. At 1×: one pass wrong 5.2%, silent 0.07%, two passes agreeing and wrong 0.10% of what is kept, 90.0% kept; At 5×: one pass wrong 5.1%, silent 0.10%, two passes agreeing and wrong 0.09% of what is kept, 90.2% kept; At 20×: one pass wrong 4.7%, silent 0.18%, two passes agreeing and wrong 0.08% of what is kept, 90.8% kept; At 100×: one pass wrong 3.7%, silent 0.38%, two passes agreeing and wrong 0.04% of what is kept, 92.8% kept.
Fig. 6 What counting each family twice, from two pins, leaves of a pinned count’s errors, with runs of five crossings. A reading is kept only when both passes agree. The vertical axis is logarithmic.

The second question has an answer that is even more favourable. Count each family twice, from two different pins, independently, and keep a reading only when the two agree. With slips independent, 5.2 per cent of single-pass readings are wrong. Of the readings a two-pass counter keeps, 0.10 per cent are wrong, and 90 per cent of readings are kept. That is a fifty-fold reduction in the error for doubling the work and giving up one reading in ten.

Clustering makes that better. A count that is wrong is more scattered when slips come in runs, since it is spread over more possible errors. Two passes agreeing on the same wrong count then needs both to have made the same set of slips, and the chance of that falls. At a boost of twenty, 0.08 per cent of what is kept is wrong, at a hundred, 0.04 per cent, and 92.8 per cent is kept. Two passes catch almost all of what clustering adds, and more besides.

This is a stronger result than it looks. The silent error is unnoticeable to anything that examines a reading, by definition: it is a perfectly good pair. Nothing about a single reading can catch it. Two passes catch it without knowing anything about pairs at all, since the error has to happen twice in the same way. It is the only check these essays on miscounts have found that does not depend on a reading looking wrong. The price is a second pass, which is a person’s time rather than anything clever.

What runs cannot do across passes

The independence of the two passes is doing all the work there, and the essay would be misleading if it did not say so. A slip that remembers has been modelled as an eye with a state, a run within one pass. Its other cause was offered as a damaged patch of florets. A damaged patch doesn’t go away on the second pass. It lies at the same place on the head, and a second pass, starting from a different pin, will cross it in the same way. If what caused the first pass’s run is on the specimen rather than in the eye, then two passes are correlated, and they could agree on the same wrong count much more often than these numbers say.

The model can’t speak to that at all. Nor can it say how much of real clustering is the eye and how much the specimen, and that split is what matters for a survey. An experiment could measure it directly, by having one person count one head several times and another person count it again, and comparing how often they disagree within and between themselves. Nothing here has any such data, and the essay’s number is a limit from one side.

What this does not establish

The slip law is the earlier essay’s stated choice: a chance that falls with the square of the spacing and one free number. A hot crossing’s boost is another free number, with its run length a third. So every result is stated across a range of those and none is a measured property of an eye. The head is ideal, and a real one has spacings that vary. Families are treated as independent, so a run in one family does nothing to the other, and an eye that has lost its way counting the larger family might well lose it again in the smaller. Only the band of the pair is read, at its rim, and the count of a whole capitulum would have other things happen at other radii.

How it could be wrong

If a measured run of slips had a boost and a length that took the silent share of 34/55 above a few per cent at an otherwise reasonable total error rate, this essay’s account of the scale would be wrong. If two independent passes by one person agreed on a wrong count at much more than the product of their separate error rates, the independence behind the recount’s number would be wrong. And if a count whose slips were bunched turned out to have a larger total error than one whose slips were spread, the most surprising result here would be wrong, because the argument for it is about how many counts a given number of slips can spoil.

Still open: a slip the specimen causes

Clustering here lives in the eye and it starts afresh on each pass. A slip that a damaged patch causes lives in the head, at a place. The next measurement puts a stated stretch of each family’s crossings, at the same azimuth for both passes and both families, under a raised chance of slipping, and asks how far two passes then still reduce the wrong readings kept, whether taking the two passes from pins on opposite sides of the patch helps, and whether both families being disturbed at one azimuth produces the joint errors that independence ruled out. If the patch defeats the second pass, then a survey would want its second pass counted in the other direction, and maybe by another person, and it would be possible to say by how much.

Where this goes next

What a pinned count of 34/55 reads for an eye that slips 1 in 500 crossings at an organ's spacing, beside a placed mark. Shares of readings exact, wrong and announced by a shared factor, and wrong and silent: pin, outer edge, 94.8% exact, 5.1% announced, 0.07% silent, both counts wrong on 0.04%, a count two or more out on 0.05%; pin, a third in, 91.0% exact, 8.8% announced, 0.21% silent, both counts wrong on 0.13%, a count two or more out on 0.15%; place, a third in, 98.9% exact, 1.1% announced, 0.00% silent, both counts wrong on 0.00%, a count two or more out on 0.00%. The placed mark is read from 2000 places on its circle and is never two out. A pinned count errs where the spirals crowd A counter who closes each family on a chosen place is wrong one time in eleven from the geometry alone. Pin the starting organ instead and the closing costs nothing; the only error left is a slip on the way round, and a slip is likeliest where neighbouring spirals cross the counting circle closest. They cross evenly round any one circle and much closer at the inner edge of a pair's band, and the larger family's always cross closer by the ratio of the counts — so it slips about four times as often per count. A pin is best at the outer edge, where a place is worst, and beats a place there for an eye that slips less than once in 156 crossings at an organ's spacing; to beat a place a third of the way in it must slip less than once in 2,400. And a pin makes an error a place never does: a count two out, silent on 34/55. Where a chosen mark closes: the organ nearest one place, and where its two spirals cross the counting circle. A stretch of a golden-angle head around one place on a circle inside the band of radii where 34 and 55 are the two shortest families of spirals, 75 per cent of the way out. The ringed organ is the one nearest the place, and both counts start on spirals through it. Its 34-spiral crosses the circle 0.48 of that family's spacing to one side of the place and its 55-spiral 0.42 of a spacing to the other, because the organ sits outside the circle and the two families wind in opposite senses. A count that closes on the place, rather than on the organ, is off by those amounts, which round to nought here. A chosen mark errs by a third of a spiral A counter who starts every family of spirals at the organ nearest one place on the head, and stops when the count comes back round to that place, makes a closing error the head decides: the gap between the place and where that organ's spiral crosses the counting circle. It is three tenths of a spiral spacing in every family of every pair, so in degrees it falls as one over the count — 3.2° on a 34-family, not the 7.2° assumed — and the two families' errors are correlated by where on the head the place is, from +0.35 at the inner edge of a pair's band of radii to −0.5 at the outer. From the geometry alone a reading is wrong one time in eleven and almost never in both counts, so 34/55 is silently wrong once in six thousand. An eye that misplaces the mark by a tenth of a spacing is enough to turn the correlation positive. What a report of 34 and 55 allows as a count is allowed to drift by one, two and three. The divergence axis from 20° to 180°. Dark: the band a report allows if both counts are right. Pale: the bands of every pair within the tolerance whose counts share no factor and which are the two shortest families somewhere in that range. 34/55 read with a tolerance of 1 allows 3 bands, 0.354° in all, the nearest wrong one 35/54 at 113.27°; 34/55 read with a tolerance of 2 allows 13 bands, 1.550° in all, the nearest wrong one 35/54 at 113.27°; 34/55 read with a tolerance of 3 allows 29 bands, 3.445° in all, the nearest wrong one 37/54 at 126.60°, and 2 more pairs whose bands lie below 20°; 21/34 read with a tolerance of 3 allows 28 bands, 8.767° in all, the nearest wrong one 18/31 at 139.55°, and 2 more pairs whose bands lie below 20°. A count that drifts by two A reported pair of 34 and 55 that may be wrong by one allows three sharp bands; allowed to drift by two it allows thirteen, and by three, twenty-nine — and the information lost is exactly the logarithm of that count, because every band is as narrow as the true one. The nearest wrong band stays twenty-four degrees away until a drift of three brings one to eleven. What does not survive is the protection: 21/34 and 34/55 could not be miscounted silently by one, but every Fibonacci pair can be by two, so a counter who drifts by two as readily as by one reports 34/55 silently wrong 9.5 per cent of the time rather than 0.13.

What links here

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

Shares its objects with

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CoprimeDiscriminationFibonacciHonest limitsJoint distributionMeasurement errorParastichy pairSilent failure