A count set by a delay

L-systems describe, they do not explain ended on a precise complaint. Two rewriting rules and a turtle produce something indistinguishable from a plant, and the trouble is not that the grammar is a model but that it has no parameter anything could disagree with: the turn angle, the shrink ratio and the rules were all chosen to make the picture look right.

There is a branching grammar whose one parameter is not chosen to look right, and it sits under the most repeated numerical claim about branching plants — that a stem which branches in seasons carries 1, 2, 3, 5, 8 growing points, and so on through Fibonacci’s numbers. The parameter is how long a new bud waits before it branches itself, and a person with a notebook and a plant can watch that.

A grammar over the age of a bud #

The symbols are growing points labelled by age. A mature apex, M, makes one new bud every season and stays mature. A new bud waits: with a delay of d seasons it passes through d stages and then becomes mature itself. Rewriting every symbol once is one season of growth, and counting the symbols is counting the growing points.

There is no geometry in it and no turtle. That is deliberate, because the claim it is about is a count, and a count needs only the lineage: which growing point produced which, and when. Lindenmayer’s grammars began as exactly this kind of object, a model of cell lineage in a filament, before a turtle turned them into drawings.

Fibonacci’s rabbits are a delay of one #

With a delay of one season the grammar is M → M B and B → M. It is the rule Fibonacci wrote down for pairs of rabbits in 1202, a new pair taking one season to mature, and the numbers of symbols after each rewrite are 1, 2, 3, 5, 8, 13, 21: each season’s count is the last season’s plus the one before, because every point that was mature a season ago has made a new bud since.

With a delay of two seasons the rule gains a stage, $M \to M\,B_0$, $B_0 \to B_1$, $B_1 \to M$, and the counts run 1, 2, 3, 4, 6, 9, 13. Each season’s count is now the last season’s plus the count three seasons back. Nothing else in the grammar has changed.

Every delay has its own rate #

Written generally, the count obeys N(t) = N(t − 1) + N(t − 1 − d), and a recurrence of that shape settles into growing by a fixed factor each season: the positive root of $x^{d+1} = x^d + 1$. With no delay every point branches at once and the count doubles. With a delay of one the root is the golden ratio, 1.6180. With two it is Narayana’s constant, 1.4656; with three, 1.3803; with four, 1.3247; then 1.2852, 1.2554 and, at eight seasons, 1.2132.

The rewritten strings and the recurrence agree symbol for symbol at every delay checked, from none to five seasons, and after fifty seasons each delay’s successive counts stand in the ratio of its own root to within half a per cent. The roots are all different, so a rate measured well enough names a delay.

The plastic number at four #

The delay of four is worth a line of arithmetic, because its root has a name for a reason. Its equation is $x^5 = x^4 + 1$, and $x^5 - x^4 - 1$ factors as $(x^2 - x + 1)(x^3 - x - 1)$. The first factor has no real root, so the growth rate is the real root of $x^3 = x + 1$, which is the plastic number, 1.324718.

The discarded factor does not vanish from the counts. Its two roots are complex with modulus one, so they add a part to the sequence that repeats every six seasons without growing, and the counts 1, 2, 3, 4, 5, 6, 8, 11, 15, 20 do not obey any shorter recurrence; only their ratio closes on the plastic number, as the repeating part shrinks against a total growing by a third each season.

Padovan’s sequence, 1, 1, 1, 2, 2, 3, 4, 5, 7, 9, 12, obeys N(t) = N(t − 2) + N(t − 3) and grows at the plastic number too. So a four-season wait and Padovan’s rule share a long-run rate and differ in every count along the way — the grammar’s own small instance of a pattern that does not select its mechanism, and the reason a rate is a weaker reading than a run of counts.

Worked at a delay of two #

The two-season recurrence can be checked by hand from the counts already given. Each count is the one before plus the one three seasons back: 4 = 3 + 1, 6 = 4 + 2, 9 = 6 + 3, 13 = 9 + 4, 19 = 13 + 6, 28 = 19 + 9. Its growth root, 1.4656, satisfies $x^3 = x^2 + 1$ to the digits printed: $1.4656^3$ is 3.1479 and $1.4656^2 + 1$ is 3.1479.

The gap it opens on Fibonacci’s rule is not subtle once it starts. After six seasons the two rules have 21 and 13 growing points; after ten, 144 and 60; after fourteen, 987 and 277, fewer than a third as many. A season’s extra wait costs a plant nothing it can see for three seasons and then costs it more every season after, which is why the counts agree so long and then part so completely.

A delay forbids every count but one #

This is the property the turtle grammar lacked. Fix the delay and the whole sequence of counts is fixed with it: a one-season rule forbids 1, 2, 3, 4, forbids 1, 2, 4, and forbids every sequence but Fibonacci’s. The grammar is still a description, but it is a description that an observation can contradict.

And what contradicts it is the one thing the parameter is about. A count of 4 after three seasons says a bud that should have branched did not, which is a statement about waiting that a person could check on the plant itself by marking buds and watching them. The mechanism checklist asks for a parameter that corresponds to a measurable quantity in the organism; the delay is one.

Four counts name a delay of one #

Laid side by side, every delay agrees with every other for its first three counts, 1, 2, 3 — they are the counts before any new bud has had time to branch — and then they part. A one-season delay reaches 5 at the fourth count, and a delay of two, three, four or eight seasons reaches 4 there. A delay of none has already parted at the third count, with 4 against 3.

Telling longer delays apart takes longer. Two seasons and three agree through 1, 2, 3, 4 and part at the fifth count, 6 against 5; three and four part at the sixth, 7 against 6. A delay of d parts from a delay of d + 1 in season d + 2 — the count numbered d + 3 — because the first bud either delay makes is formed in season one, and the shorter delay’s bud makes its own first bud one season before the longer’s.

Buds that do not keep time #

A real apex is unlikely to keep an exact schedule, so the next question is what an imperfect one does. Let each new bud wait one season with probability 1 − q and two with probability q, independently. The counts are no longer fixed, and a plant’s sequence may or may not match Fibonacci’s for as long as it has been counted.

The chance that it matches has a closed form, and the argument is short. A bud that waits two seasons never produces more than it would have: it produces its first bud a season late, and the total two seasons after it formed is one lower than Fibonacci’s — a delay only withholds, and nothing later makes the withheld bud up. So the counts through season T are all Fibonacci’s exactly when none of the buds formed up to season T − 2 waited two. Along the Fibonacci path those buds number 2, 4, 7, 12, 20, 33, 54 — each one less than a Fibonacci number — and the chance is (1 − q) raised to that number.

The chance of an unbroken run #

At q = 0.1, one bud in ten waiting two seasons, the counts through season four match with probability $0.9^2 = 0.81$. Through season six, $0.9^7 =$ 0.478: seven counts in a row come out Fibonacci’s less than half the time. Through season eight, $0.9^{20} = 0.122$; through season ten, $0.9^{54} =$ 0.0034; through season twelve, $0.9^{143}$, about three in ten million.

Slower drift makes the run last longer but not much longer. At q = 0.05 eleven matching counts, through season ten, turn up 6.3 per cent of the time. At q = 0.2 seven counts match 21 per cent of the time and nine counts one per cent. At q = 0.5 seven counts match 0.8 per cent of the time.

Checked against simulated plants #

The closed form rests on one claim — that the first mismatch comes exactly two seasons after the first long wait — and a claim like that is worth a test it could fail. Twenty thousand simulated plants at each q, grown from a stated seed, give shares that sit on the exact values: 0.8110 against 0.8100 through season four at q = 0.1, 0.4816 against 0.4783 through season six, 0.1230 against 0.1216 through season eight and 0.0033 against 0.0034 through season ten. At q = 0.3, 0.4923 against 0.4900 and 0.0819 against 0.0824.

Every comparison is within four standard errors of the exact probability, and the six quoted here are within one.

Where the counts end up #

A plant that drifts off Fibonacci’s sequence does not drift far at first and then drifts steadily further. After twelve seasons a one-season rule has 377 growing points and a two-season rule 129. With one bud in twenty waiting two, the simulated plants’ median is 362 and ninety per cent of them lie between 317 and 373 — below 377 at the top of the range. With one in ten the median is 343; with three in ten, 272; with half, 217.

So after a dozen seasons a count of exactly 377 is almost never what an imperfect one-season rule produces, and a count in the three hundreds is what it produces nearly every time. A count in the neighbourhood of a Fibonacci number is common; a count equal to one is not.

The rate barely moves and the run collapses #

Two instruments read the same plants, and they respond to the imperfection in opposite ways. The mean growth rate under a mixture of waits is the root of $x^3 = x^2 + (1 - q)\,x + q$, which runs from the golden ratio at q = 0 to Narayana’s constant at q = 1 and is 1.6011 at q = 0.1 — within 1.1 per cent of φ. The chance of eleven matching counts at the same q is a third of a per cent.

So a growth rate close to the golden ratio is weak evidence of a one-season rule, since a rule that misses one bud in ten reproduces it to one per cent; and an unbroken run of Fibonacci counts is strong evidence of an exact one, since that same rule almost never produces it. Which of the two a report quotes decides what it can claim.

How exact a plant would have to be #

The closed form can be turned round. For the counts through season ten to be all Fibonacci’s more often than not, $(1 - q)^{54}$ must exceed a half, which needs q below $1 - 0.5^{1/54} = 0.0128$: fewer than one bud in 78 waiting a second season. Through season twelve the bound is one bud in 207.

That is the practical content of the claim. A plant shown with eleven Fibonacci counts in a row is either keeping its buds’ schedule to better than a per cent or was chosen for being the one that did. How often a count comes out Fibonacci is a separate question about the geometry of seed heads, and how many plants it would take to settle it is a separate question about samples, and both are sharpened by knowing which kind of count a report took.

How long a diagram has to be #

The same arithmetic prices the diagrams that carry the claim. A drawing whose counts run 1, 2, 3, 5, 8, 13 shows six counts, through season five, and the buds formed early enough to spoil it number four. It is more likely than not whenever $(1 - q)^4$ exceeds a half, which holds for $q$ up to $1 - 0.5^{1/4} = 0.159$: a plant whose buds wait a second season one time in six still produces that drawing more often than it fails to.

Eight counts, through 21, need twelve buds to keep time, and a drawing of them is more likely than not only below one bud in 17.8 waiting two. So a short Fibonacci run is evidence of very little about a plant’s schedule, and each further count roughly multiplies the precision it demands by the golden ratio, because the number of buds that must keep time grows as Fibonacci’s numbers do.

Fibonacci twice over #

The same shape of result holds for the Fibonacci numbers counted in seed heads, which is worth setting beside this one. There, Fibonacci counts come from one branch of the placement model, reached at one range of growth rates, and a different branch gives Lucas numbers instead. Here they come from one waiting time, and a different waiting time gives Narayana’s numbers instead.

In both, the sequence is a consequence of a parameter taking one value rather than a property of plants, and in both the useful question is the parameter’s value. And in both the count is almost never made in the way the claim needs: through enough seasons, or across enough of a head, to tell the named sequence from its neighbours.

A description that can be wrong #

The complaint about L-systems was that a model which forbids nothing says nothing. The delay grammar forbids a great deal: every count sequence but one for each delay, and every growth rate but one. What it predicts about a plant is not the plant’s shape but the waiting time of its buds, and a waiting time is observable.

That puts it in a different class from the turtle grammar without making it an explanation. It says a one-season wait produces Fibonacci’s counts; it does not say why a bud would wait one season, any more than the cube law’s derivation says why a tissue’s upkeep costs what it does. It is a description whose single parameter a plant could contradict, which is what separates a count worth making from a picture worth admiring.

What the grammar leaves out #

Growing points in the grammar never die, never stop branching and never branch twice in a season. Real shoots do all three, and each changes the counts in its own direction: a death lowers every later count, a shoot that stops branching lowers them more slowly, and two buds in one season raise them. The grammar also counts growing points, and a count on a plant is often of something else — branches at successive heights, flowering stems — whose relation to growing points has to be stated before the arithmetic here applies.

Nor does anything here say that any plant’s buds wait a fixed time. The claim that a species branches in Fibonacci numbers is a claim about waiting times; the grammar says what it would take for the claim to be true and how quickly it would stop being true, not whether it is.

What would withdraw it #

A delay whose rewritten strings and recurrence disagree at any season. A growth rate that is not the root of $x^{d+1} = x^d + 1$. A one-season delay that agrees with a longer one past its third count. A share of simulated plants with an unbroken Fibonacci run that differs from (1 − q) to the number of earlier buds by more than four standard errors. Each of these is checked every time the measurement runs.

Still open: a count that loses its growing points #

The grammar has no deaths, and a stem that loses shoots is the common case rather than the exception. The next test adds a stated chance that a growing point dies each season, and asks whether a death leaves the growth rate a root of some recurrence, how it moves a Fibonacci run’s chance, and whether a count of scars restores the sequence the way counting scars restores a tip count on a branching tree.