The pattern itself

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.

Worth reading first: A disturbance with a memory · Counting the spirals · The sequence has a memory.

A memory in a plant’s disturbances manufactures nothing, and the reason is that a memory decays. The thing that would manufacture a comb is a disturbance that returns — one that comes back to the same value at a fixed separation instead of fading away from it. This essay builds that disturbance deliberately, drives it into a lattice with no rule anywhere in it, and reads the result with the same instrument.

It works. A comb appears, at the spacing it was given, well clear of the sampling band.

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.
Fig. 1 A kinematic lattice at the same divergence, rise and scatter as every control in this thread, with one change: the azimuth error repeats every eight organs. Seven tenths of each displacement is a fixed pattern of length eight, played over and over; the rest is a fresh draw. The teeth at 8, 16 and 24 are as tall as a real stem’s, on an arrangement in which nothing was ever placed by a rule.

So that earlier work’s control is not sufficient on its own. Independence was doing work in it, and a disturbance that is not independent in the right way puts up that work’s headline observable for free.

What saves the result — and it does survive, in a weakened form — is that the forgery is bad at the second number, and the site’s own habit of reading more than one stem is what exposes it.

What the forgery gets right, and what it invents

The readout returns a pair. It finds a spacing by looking for the residue class whose members average highest, and it finds a partner by looking for a second class at the same spacing. On the forged stem the spacing is 8, which is right, because 8 is what the disturbance was built to repeat at.

The partner is where it comes apart.

A periodicity reports a different partner every time. eight kinematic lattices, differing only in the seed of their disturbance, each read by the same instrument. The disturbance repeats every 8 organs at a weight of 0.7: it puts a strong comb at spacing 8 — 0.43 against a band of 0.07 — and the partner it names is 8/10, 8/12 across the 8 stems and never 8/13, which is what the position counter finds in every one of them. There is no placement rule in any of these arrangements.
Fig. 2 The eight forged stems, differing only in the seed of their disturbance. Four of them refuse. The four that report name 10, 10, 12 and 12 as the partner — and the position counter reads the same pair from every one of these arrangements, because they are the same arrangement. Not one of the four is right, and the two answers that do appear are answers to which pattern was drawn rather than to which lattice was built.

The partner is not a measurement at all. Here is why, and the arithmetic is short enough to check by hand.

A disturbance that repeats with period 8 has a covariance that depends only on the separation modulo 8. Whatever the fixed pattern happens to look like — and it is drawn once per stem, so it looks different on every stem — its circular autocorrelation at offsets 1 through 7 is a set of seven numbers of order one over the square root of eight, with signs that are an accident of the draw. Those seven numbers reappear at every lag congruent to them: the value at lag 2 is the value at lag 10 is the value at lag 18.

So a periodic disturbance does not put up one comb. It puts up eight of them, one per residue class, and seven are junk. The readout picks the tallest of the seven, and which one is tallest is a property of the pattern that was drawn for this stem. Change the seed and a different class wins.

That is why the answers run 10 and 12 rather than clustering near 13. There is nothing about 13 in the disturbance at all. The forger knows one number, and the readout demands two.

The junk classes are not small, either, which is why they get through. A random pattern of length eight has circular correlations at offsets 1 to 7 that scatter around zero with a spread of roughly one over the square root of eight — about a third — and the periodic part of the disturbance carries seven tenths of the variance, so the tallest of the seven arrives at the readout at something like a quarter. The threshold a second comb has to clear is three sampling bands divided by the square root of its own length, which for a four-member comb is about a tenth. A quarter clears a tenth comfortably. The forgery’s partner is not squeaking past a test; it is passing it by a factor of two, on a quantity with no information in it.

That is worth stating as a general caution rather than as a fact about this forgery. A threshold set by sampling noise does not protect against structure that is not the structure being looked for. The band in every figure here is what a correlation wanders by when there is nothing there; it says nothing about what a correlation does when there is something there that is not the thing being measured.

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 0.9. The largest comb mean is 0.860 against a sampling band of 0.073, and the readout returns 8/10.
Fig. 3 The same forgery, driven harder — nine tenths of the displacement is the repeating pattern. The main comb rises to 0.75, above the rule’s own 0.64, and the junk classes rise with it: now seven of the eight stems report, naming 10, 12, 10, 12, 12, 12 and 11. Making the forgery stronger does not make its partner more nearly right; it makes the wrong partner more confident, and it adds an eleventh answer to the list.

Driving it harder buys confidence and not accuracy

Set the two strengths beside each other, because the pair of them says something the stronger one alone does not.

At seven tenths, four of eight forged stems report and the partners are 10, 10, 12 and 12. At nine tenths, seven of eight report and the partners are 10, 12, 10, 12, 12, 12 and 11. So raising the disturbance’s weight moves the report rate from a half to seven eighths and leaves the accuracy at zero of eleven. Not one reading in either set names 13.

That is worth stating as a caution rather than as a fact about this forgery. The ordinary reading of a stronger signal is that it is more trustworthy — a comb clearing its band by a factor of two is better evidence than one clearing it by a tenth. Here strength and correctness come apart completely: the weight controls how much of the variance is periodic, so it controls the height of the true comb and of the seven junk classes in equal measure. Both rise together, the readout’s threshold does not, and what improves is only the fraction of stems on which the instrument is willing to speak.

A forger who wanted a convincing single stem would therefore turn the weight up, and a reader who took a confident reading as a good one would be exactly wrong. The quantity that separates the two is not how far a comb clears its band but whether two stems say the same thing.

The weak forgery is the hard one to convict

Which makes the multi-stem test’s power depend on the forgery’s strength in the opposite direction from the obvious one, and that is a number the specification needs.

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 0.5. The largest comb mean is 0.323 against a sampling band of 0.073, and the readout returns 8/10.
Fig. 4 The weak forgery: half the displacement periodic rather than seven tenths. Fewer of its stems report anything at all, which is exactly what makes it the hard one to convict.

The test convicts by disagreement, so it needs at least two stems that report. At nine tenths, seven of eight stems report, so three plants give between two and three readings and a disagreement is likely on the first try. At seven tenths, four of eight report — so three plants give one or two readings on average, and a single reading cannot disagree with anything. Half the time, three stems of the weaker forgery produce one confident pair and no way to check it.

Getting six reporting stems out of a species that refuses half the time takes about twelve plants, and getting them out of one that refuses seven times in eight takes fifty. So the cost of the test is set by the refusal rate, and the refusal rate is set by exactly the thing that makes the forgery least convincing on any one stem.

Two consequences for the survey, and both are cheap to adopt. Refusals have to be counted and reported, because the number of stems that declined is what says how much power the agreement between the ones that spoke actually has. And the number of plants cannot be fixed in advance; it has to be set by how many readings come back, which means the survey is run until it has enough agreeing readings rather than until it has enough specimens. A study that measured twenty plants, got four readings, and found them consistent would have run a test with almost no power and reported a clean result.

The check that catches it is a second specimen

Everything above is invisible from one stem. A botanist handed a single forged angle sequence would read a clean comb at spacing 8, a partner clearing the band, and would report a pair — with no way to know that a second plant of the same species, disturbed the same way, would report a different one.

So this essay changes the specification, and it changes it in the direction the survey thread has been pushed twice already. The pair must be read on several stems, and the readings must agree.

That is not a statistical nicety about averaging out noise. It is the only test that separates a lattice from a periodicity, because the lattice is a property the stems share and the periodicity is a property each stem draws for itself. The rule’s own stems agree with each other and with the position counter; the forgery agrees with nothing.

The site has had the habit for a long time and it was never justified this sharply. The reason to run five seeds was that any one of them might be unlucky. The reason now is that a single stem cannot distinguish a pattern from a disturbance that repeats.

Why the forger cannot simply be told the second number

The obvious repair, from the forger’s side, is to use a disturbance that repeats at 13 as well as at 8. That is possible and it is the subject of the next essay, where it is built out of something a plant plausibly has rather than out of two periods chosen by hand. But it is worth noticing first what happens with a periodicity at 13 alone, because the answer is not what a reader would guess.

A repeating error the search cannot name. 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 13 organs at period 13, weight 0.7. The largest comb mean is 0.117 against a sampling band of 0.073, and the readout refuses.
Fig. 5 A disturbance repeating every thirteen organs, at the same strength as the one that forged a comb at eight. The readout finds nothing to name. The spacing it settles on is 10 rather than 13, because 13 is outside the range it can return at all, and the comb it assembles there averages 0.117 against a clearance of 0.126. It refuses on all eight stems, and the average across them is 0.049 against a band of 0.073. The disturbance is exactly as strong and exactly as periodic as the one above. The instrument simply cannot see it.

The reason is the readout’s own arithmetic, and it is the same constraint that turns up from the other direction in the essays about the fine end of the ladder. A comb has to have three teeth inside the lag window before it is called a comb — two lags at a common spacing is a coincidence available at every spacing — and the window is thirty lags. So the largest spacing the search can return is ten. A period of 13 has teeth at 13, 26 and 39, and the third is outside the window; the search never proposes 13 as a spacing, so it never finds the comb that is sitting there.

This cuts both ways and both are worth having.

For the forgery: a disturbance repeating at the larger parastichy number is invisible, so a forger has to know the smaller one specifically. Knowing “some number in this plant’s arrangement” is not enough.

A lattice with an error that repeats every 5 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 5 organs at period 5, weight 0.7. The largest comb mean is 0.437 against a sampling band of 0.073, and the readout refuses.
Fig. 6 A disturbance repeating every five organs. Five is a parastichy number of coarser lattices than this one, and the readout’s arithmetic is what decides whether a period lands anywhere it can be mistaken for a family.

For the instrument: there is a whole class of real periodicity the readout would miss, and the miss is silent. A stem whose disturbance repeats at 13 would be reported as having no comb, which under that earlier work’s reading would have been evidence that no rule made it.

Could a plant’s disturbance have a period?

The forgery above was built by hand, and a forgery built by hand is only worth worrying about if something in the world could build it. Three candidates are worth naming, and the third is the one that matters.

A daily cycle beating against the plastochron. Organs are produced at roughly regular intervals, and the environment varies on a twenty-four-hour cycle, so a plant producing one organ every fraction of a day would receive a disturbance that repeats every few organs. This is real and it is not the threat: the period it produces is set by the ratio of two times, which has no reason to be a whole number and every reason to drift as the plant grows. A period that drifts is a memory with extra steps — the correlation smears across neighbouring lags and the comb loses its teeth.

An oscillation in the meristem itself. Some models of primordium initiation have a genuine clock in them, and a clock with a period of a few plastochrons would put structure at the lag of that period. Again the number is set by the clock rather than by the pattern, so it would have to coincide with the smaller parastichy number by accident — and it would go on repeating at the same lag while the pattern climbed the ladder and its parastichy numbers changed, which is a prediction a real stem could refute in one reading.

And the one that is not an accident: the neighbours. The organs that touch an organ are, by the definition of a parastichy pair, the ones m and n places back in the order of production. So any disturbance transmitted through contact — a physical push, a shared vascular connection, a local depletion of something — is correlated at exactly the lags the readout examines, and it is correlated at them because they are the lattice’s, not by coincidence.

That third candidate is the real threat and it is the subject of the next essay. It is worth seeing clearly what makes it different from the forgery here: this essay’s disturbance knows one number, chosen by whoever built it. A neighbour-transmitted disturbance knows both numbers, and does not have to be told them, because it lives on the arrangement whose numbers they are.

Which also disposes of the reassurance the previous section might otherwise offer. Reading several stems catches a forger who has to pick a period, because the period is a free parameter and the seven junk classes come out differently each time. It does not catch a disturbance whose period is a property of the lattice, because then every stem of the species has the same one.

A lattice is not a list of periods

The title is the point and it is worth stating as a claim rather than as a summary.

A phyllotactic lattice is a set of points on a cylinder with a metric on it. Its parastichy numbers are the index offsets whose two organs come out closest together on the surface — which is a fact about the geometry, and the reason the numbers are Fibonacci is that a nearly-golden divergence makes those the short hops. The offsets are related to each other: 21 is 8 plus 13 because the vector to the 21-neighbour is the sum of the vectors to the other two, and this site measured earlier in this collection that every contact family but the two smallest is a sum of two others.

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.
Fig. 7 The property a real lattice has and a list of periods does not: every contact family above the two smallest is the sum of two others, so the whole set is generated by two numbers. The forgery has one number and produces a set generated by nothing — eight residue classes with unrelated heights, of which the readout picks whichever won this time.

A periodic disturbance has none of that structure. It has one period and seven accidents. It reproduces the symptom the readout was built to detect and none of the arithmetic underneath it, and the arithmetic is what the second comb was found to be made of earlier in this collection: it sits at the difference of the pair because correlation travels between neighbours and a one-step chain reaches offsets a·m + b·n with b = 0 or ±1.

So the forgery in this essay is a forgery of a statistic, not of a lattice. That is a real weakness of the statistic and it is why that earlier work’s control was too weak. It is also why the repair is available: demand the part of the statistic that reflects the arithmetic, on more than one plant.

What the specification now says

Three requirements, each added by a measurement rather than by caution.

Several stems, and the readings must agree. From this essay. A single stem cannot separate a lattice from a period, and the disagreement between forged stems is visible with as few as three.

The angle reading must agree with a position count on the same stem. From the earlier work, and now doing much more work than it was asked to. The forged stems all have the same positions and therefore the same counted pair, so the counter contradicts four of the five that report.

And a refusal is not evidence of anything. From the figure at 13 above. The readout’s window means a periodicity at the larger parastichy number leaves no trace, so “no comb” cannot be read as “no periodicity”, and — under that earlier work’s interpretation — could not have been read as “no rule” either.

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.
Fig. 8 The forgery that does know both numbers, and does not have to be told them. It is the next essay’s, and it is what a periodicity would look like if the period were a property of the lattice rather than a parameter of the forger.

A periodicity at long scales

A disturbance that simply repeats manufactures a comb and is convicted by reading more than one stem. At long scales it does something else worth recording: its block means carry less variance than independent errors would, because a periodic component averages away exactly.

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 forgery needs a history — both name artefact, autocorrelation, counting blind, discrimination, ensemble, evidence, lattice offset, measurement, noise, null model, parastichy pair
  • A disturbance that is not passed on — both name artefact, autocorrelation, discrimination, ensemble, evidence, lattice offset, measurement, noise, null model, parastichy pair
  • The control a survey would need — both name autocorrelation, counting blind, discrimination, evidence, identifiability, measurement, null model, parastichy pair, specimen
  • What a forgery has to know — both name autocorrelation, discrimination, evidence, identifiability, lattice offset, measurement, noise, null model, parastichy pair
  • What the pair costs — both name autocorrelation, counting blind, discrimination, divergence angle, ensemble, identifiability, measurement, parastichy pair, specimen
  • A difference forgets a drift — both name artefact, autocorrelation, ensemble, evidence, identifiability, measurement, noise, null model

Named objects

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

ArtefactAutocorrelationCounting blindDiscriminationDivergence angleEnsembleEvidenceIdentifiabilityLatticeLattice offsetMeasurementNoiseNull modelParastichy pairSpecimen