The third family

No cut-off on distance makes the two definitions of neighbour one relation on a seed head, and the pair a count returns names two thirds of a cell’s walls and no more. Both of those are negative, and both were measured on a disc.

On a cylinder the same comparison returns a zero, and the zero is exact.

What a stem lets a comparison do that a head does not #

A stem’s counted pair is the same the whole way up. A head’s is not: it changes with radius, so every reading on a disc is a reading in an annulus and every disagreement can be blamed on the annulus it was read in.

Removing that gives the comparison something a head cannot. If two definitions of neighbour disagree on a strip where the lattice does not change, the disagreement is a property of the definitions. If they agree there, the agreement is one number rather than a mean over bands.

The sweep #

One hundred and forty rises, from 0.60 down to 0.004, spaced by a ratio rather than by a step, with the unrolled strip’s seam closed by replication so that a node’s partners round the back of the stem are its real partners. Each rise gets both relations recomputed from the same points.

The reading is taken over the middle half of the strip, so that the two nodes at its ends — the only two whose neighbourhoods the replication cannot complete — are outside every measurement.

Three regimes, and the comparison means something different in each #

A stem is not one object across that range. At the coarse end a node has two walls, in the middle four, and below a rise of about 0.21 it has six.

Six rises above 0.5010 give a thread, twenty-three between 0.2187 and 0.4833 give a ribbon, and 111 below 0.2109 give a surface. Averaging the comparison across them would produce a number describing no stem at all.

The surface #

The surface regime is the one a plant is in and the one every result on this site about stems is about. Every node has six walls, the tessellation is an ordinary triangulation of a lattice, and the counted pair is a genuine pair.

That last clause is the finding. Three index differences, not two.

Exactly zero, at every one of 111 rises #

Cut the contact relation at three whole families — the three shortest index lags, each taken in both directions — and the two neighbour relations are the same set of edges. The symmetric difference is 0.000 per cent of their union at every one of the 111 surface rises, from 0.0040 to 0.2109.

Not small. Zero, at every rise, with no threshold anywhere in the construction. That is the strongest positive statement the comparison makes and it is the only place either definition is reproduced by the other.

The absence of a threshold is what makes the zero worth reporting. A cut at three families is a cut at a count, not at a distance: it keeps the three shortest lags whatever their lengths happen to be, so it cannot be tuned and there is no value of anything to choose. A result that holds across a factor of fifty in rise with no free parameter in it is a different kind of object from a residual minimised over a grid.

Exactly a third, at the cut a count makes #

Cut it at two families instead — which is what a parastichy count returns — and the disagreement is 33.333 per cent of the union, again at every one of the 111 rises, again exactly.

Six walls a node, three families sharing them, two families kept: a third of the walls left without a family. The number is not approximately a third and it is not a third on average. It is a third because a cut at two of three families throws away one of three families, and the arithmetic is that short.

Which is what names the missing third #

The head measurement found the same third missing and could not say what it was, because on a disc the shortfall is spread over several offsets and the composition changes with the band.

On a stem it has a name at every rise, and the name is an index lag the counting instrument found, ranked, and discarded for being third. The missing third is not a set of walls the contact relation cannot see. It is a set of walls it saw and dropped, and the cut that dropped them is the convention that a count returns two numbers.

Why that is a family and not a residue #

A residue would be a scatter — a few walls here whose offsets do not repeat. This is a whole family: the same lag at every node of the strip, carrying its share of the walls, and present at every rise of the surface regime.

It is also the sum of the other two. The three lags the graph carries at a given rise are a rung’s two members and their sum, which is why a head’s contact families are closed under addition and why the third one has never needed to be counted separately: it is determined by the pair. Determined is not the same as absent, and a description of the tissue that leaves it out is missing a third of the walls.

The three shares are a tie #

The obvious repair is to read the pair off the graph instead — take the two families carrying the most walls and call those the counted pair. It does not work, and the reason is a measurement rather than an argument.

Across all 111 rises the three shares never differ by more than 1.80 percentage points, and the largest margin between the second and third share is 3.51 per cent of the leading share. Three families dividing six walls a node divide them almost equally, because that is what a triangulation of a lattice does.

So the graph’s “two largest” is a tie-break #

A quantity separated by three per cent of itself is not a measurement that orders anything. Reading the two largest shares off a stem’s contact graph is choosing between three nearly equal numbers, and what decides it is whichever family happens to have the extra wall at that rise.

And the tie-break names a different pair at half the rises #

At 55 of the 111 surface rises the two families carrying the most walls are not the two the hop lengths rank shortest. That is 49.5 per cent — the fraction a coin would produce, which is what a tie-break with nothing to work on looks like.

So a parastichy pair cannot be recovered from the contact graph on topological grounds. The graph knows which three families exist and does not know which two of them a count would name. This is the same defect the head’s two rankings show more mildly, sharpened by a stem to the point where the agreement is exactly chance.

What survives that #

The counted pair is always in the graph. At every one of the 111 rises both of its members are among the three families the tessellation carries — the argument is over which two, not over which three, and the graph never carries a lattice the count did not find.

That is worth keeping because it is the half of the correspondence that does hold. A tessellation of a stem contains the counted pair and one more family, always, and the information a count discards is exactly one family and its share of the walls.

It also fixes the direction of the repair. Going from a count to a tessellation is arithmetic — take the pair, add its members, and the three families are known — while going from a tessellation to a count is the coin flip above. The instruments are not interchangeable and they are not symmetric: one determines the other and is not determined by it.

Two ladders, and both have closed forms #

The two instruments also change their answers at different rises, and both sets of rises can be written down rather than swept for.

The counted pair changes at the classical transitions, where the hops of the two families meeting at a rung become equal. The tessellation changes which third family it carries at the rise where the two counted families are perpendicular, and that rise is √(−⟨mδ⟩⟨nδ⟩ / mn) for a rung (m, n), where ⟨mδ⟩ is how far m turns of the divergence miss a whole turn by.

They alternate, strictly #

Interleave the two sequences over eight rungs and they alternate all the way down: a counted transition, a change in the tessellation, a counted transition, and so on for sixteen rises without one exception.

That means every rise at which one instrument reports a change is a rise at which the other reports nothing at all. The two have no interesting rise in common, over the whole range either has been swept on.

Merged, they step by φ #

Each ladder alone steps by 2.6215 between one member and the next, which is the familiar factor of φ² between consecutive transitions to four parts in a thousand. Merged, the sixteen rises step by 1.6180, against φ = 1.6180.

A sequence stepping by φ² interleaved with another stepping by φ² gives a merged step of φ only if each member of the second sits at the geometric middle of two members of the first. That is what the ratios say: the flip inside a rung divided by the transition out of it is 1.6140, 1.6196, 1.6174, 1.6183 and 1.6179 on the five finest rungs.

Which puts the tessellation’s change in the middle of every rung #

Not at a rung’s edge, where a count changes, but at the geometric centre of the rung’s interior. So the tessellation’s own ladder is a second ladder on the same lattice, offset from the counted one by half a rung in the logarithm of the rise.

The coarsest rung is the exception and it is the informative one: at rung (1, 2) the ratio is 1.7013 rather than about 1.618, which is where the closed forms stop being asymptotic. That is also the rung whose flip at 0.21233 the sweep brackets between 0.2109 and 0.2187, and it is the boundary between the ribbon and the surface.

What alternation costs an observer #

An observer watching a stem’s counts change up its length sees nothing at the rises where its tessellation reorganises, and an observer watching the tessellation sees nothing at the transitions. Neither instrument can be used as a warning that the other is about to change.

That is a stronger statement than the two instruments merely disagreeing. Two sequences that alternate share no member by construction, so no amount of finer sweeping will find a rise where both report.

It also settles a question the head could not put. A stem changing hands between two lattices and a stem reorganising its walls are two events, at two rises, a factor of about 1.62 apart — and on a disc those two rises are two radii, close enough together in the middle of a head that a single annulus can contain both. Anything read at one radius is therefore reading a mixture the cylinder separates.

Five of the eight, in the swept points #

The closed form for the flip is derived rather than measured, so it needs checking against points. Five of the eight predicted flips fall within 6 per cent of the rise on a swept rise where the tessellation’s third family does change; the other three sit below the sweep’s own resolution or outside its range.

Five of eight is a check and not a proof, and the honest form of the ladder claim is that the alternation is derived for one divergence and confirmed where the sweep is fine enough to confirm it.

The ribbon, where two families are exact #

Above the surface regime the arithmetic inverts. Between rises of 0.2187 and 0.4833 a node has four walls and the tessellation carries two families, so a cut at two families reproduces it exactly — 0.00 per cent — and a cut at three is wrong by a third.

So the number of families a cut should keep is not a constant of the definition. It is half the number of walls a node has, and that is a property of the rise.

Which retires the question of what the right cut is. There is no number of families that reproduces the tessellation on a stem — there is a number for each regime, two on the ribbon and three on the surface, and the regime is decided by a rise the instrument making the cut never measures. A count returns two families everywhere, and everywhere is one regime.

Two nodes, two walls #

The ribbon carries something no neighbour set can express. Two nodes meet twice — once directly and once the other way round the stem — so the tessellation gives them two walls where a set of neighbours can only record that they are neighbours.

That gap — four in the set against six in the tessellation — is not a disagreement about which nodes are neighbours. It is a disagreement about what a neighbour relation is, and it runs the whole width of the ribbon regime.

The thread, where a node meets its own image #

Above a rise of about 0.50 the strip is a thread: each node has two walls, and a node’s cell wraps far enough round the stem to share a wall with itself, one period along.

Self-walls appear at all six of the sweep’s rises above 0.5010 and at none below. The threshold is somewhere between 0.4833 and 0.5010 and the sweep does not place it more finely than that.

Where no cut-off works at all #

A wall from a node to its own image has no index difference, because there is no difference between a node and itself. A pair meeting twice has one index difference and two walls.

Both are walls that no cut on index lag can name, at any threshold, in any number of families — which is a failure of a different kind from the empty interval on a head. There the two relations are both well defined and no threshold aligns them; here one of them cannot express the answer the other gives. Between 0.219 and 0.483, and above 0.50, the contact instrument is not badly tuned. It has no vocabulary for what is there.

What this does not establish #

It does not establish anything about a plant. A rise of 0.35 is a stem whose internodes are a third of its circumference apart, which is a shoot rather than a capitulum, and rises above 0.50 are coarser than anything the counting work treats as countable. The two exotic regimes are properties of a model lattice at rises real phyllotaxis does not visit.

Nor does the exact zero on the surface transfer to a disc. A Vogel head is a family of cylinders rather than one, so a head’s cells sit at many rises at once and its bands are annuli of finite width; a third family reduces the head’s disagreement to 7.57 per cent and not to zero, and the gap between 7.57 and 0.000 is the whole of what a disc costs.

And the ladder result holds for one divergence. Every rise in it is computed at the golden angle, so the alternation is a statement about that lattice checked on its own points, not a theorem about lattices.

What would refute it #

A surface rise at which a three-family cut leaves anything in dispute. There are 111 of them, they span a factor of fifty in rise, and one non-zero reading would mean the exact agreement is a coincidence of the sweep’s own spacing rather than a property of the strip.

The cheaper refutation is on the other side. The tessellation’s ladder predicts a flip at a stated rise inside every rung; sweeping one rung finely enough to bracket its flip, in the way a rung has been walked at its own grid before, would either locate it or fail to find it. Three of the eight are currently outside the resolution the sweep has, and finding one of them where the closed form says it is would turn five of eight into six.