A counter that cannot be slid

Counting without an index built the counter a person uses on a cone: find the displacements the points repeat, follow each one into chains, and count the chains. It needs no order of arrival, which makes it the only counter a stem that grows two organs at a time can be read with, and it was licensed by agreeing with the index counter on ideal lattices at every divergence and jugacy tried.

An ideal lattice is one lattice from top to bottom. A growing shoot is not: its rise falls, so any band of organs a person marks off is a little of one lattice at its bottom and a little of the next at its top. Sliding a counting window up a real shoot means choosing how wide to make it, and the question is what the index-free counter does as that choice varies.

Three counters on one stem #

Stems were grown by the ordinary placement rule over a rise per organ falling from 0.05 to 0.0005 — ordinary and bijugate stems at rates of 150, 300 and 600 organs, and a trijugate stem at 300 — and bands of eighteen widths, from 40 organs to 320, slid up each with a stride in proportion to the rate. Every band was read three ways.

The index-free counter, as it stands. The same counter with its two families required to wind opposite ways, the check it computes and does not make. The index counter, on the band itself for an ordinary stem and on the band folded whorl by whorl for a jugate one. Each reading is then classed against the static ladder at the band’s own rises: the pair at the band’s middle or at one of its ends is right; another rung of the same ladder is a wrong rung; a pair on no rung is off the ladder; and a refusal is a refusal.

The index counter at every width #

On the ordinary stems the index counter reads the band’s own pair at every band of every width from 40 organs to 320. On the bijugate stems it reads at least 95 bands in a hundred at every width, the rest being the rung below at the narrowest bands on the fastest stem. Width is not a setting it needs.

On the trijugate stem it refuses every band of fewer than twenty whorls — 42 to 57 organs, folded to 14 to 19 nodes — because it will not count a band that short, and it reads every wider band right. That is a refusal, stated, at a floor of its own; it is the behaviour the index-free counter does not have.

The index-free counter’s band #

On the same ordinary stem at T = 300 the index-free counter reads 65 and 66 per cent of bands right at 40 and 44 organs, 87 to 89 per cent from 48 to 70, all of them at 80 and 90, 92 per cent at 100, and then falls away: 54 per cent at 120, 23 at 140, 8 at 160 and nothing from 240 up. It reads nine bands in ten or more only from 80 organs to 100.

So the instrument that agreed with the index counter on every ideal lattice has, on a growing stem, a working range of widths a quarter as wide as its lower edge, and nothing on either side of that range tells a person it has been left.

Below the band, the rung below #

The lower edge has exact arithmetic. The counter keeps a family only if its chains link more than 55 per cent of the band, and a family of m chains leaves its top m nodes unlinked, so a band of N organs cannot keep a family of more than 0.45·N chains. The narrowest band that reads a pair whose larger count is m is therefore ⌊m/0.45⌋ + 1 organs, and at least forty.

On ideal lattices in the middle of each rung that is exactly where the counter starts reading. For an ordinary lattice the pairs 3/5, 5/8 and 8/13 read from 40 organs, 13/21 from 47 and 21/34 from 76 — which is 34/0.45 = 75.6, plus one. A bijugate lattice counted 26/42 reads from 94 and 42/68 from 152; a trijugate one counted 39/63 from 141. Every floor is the arithmetic, rounded up to whole whorls. Below its floor the counter returns the rung below — 13/21 read too narrowly comes back 8/13 — and never a refusal.

Why the band starts before the floor #

On a growing stem the floor is not one number, because the pair changes along the stem, and that is why the ordinary stem’s band starts at 80 organs rather than at the largest floor it could need. At T = 300 the rise per organ reaches 0.00101, where the pair becomes 21/34, at organ 300 × ln(0.05/0.00101) = 1,171 of the stem’s 1,382. So 211 organs, 15 per cent of the stem, carry a pair whose floor is 76 organs; the rest carry pairs whose floors are 47 or less.

A band narrower than 76 organs misreads that last stretch and reads the rest correctly, which is why bands from 48 to 70 organs are right 87 to 89 per cent of the time: the misread share of the stem is a little over a tenth. Only at 80 organs does every stretch clear its floor, and the band reads everything. The lower edge a person meets therefore depends on how far up the ladder the stretch being counted has reached, which is a second thing about the specimen the width has to be chosen against.

Above the band, a pair on no rung #

Too wide a band fails differently, and the difference is the whole danger. On a bijugate stem at T = 300 every band is classed by what it returns. At 40 organs, 75 per cent are the band’s own pair and 25 per cent another rung. At 70 organs, 99 per cent are right. At 100, 92 per cent are right and 8 per cent are pairs on no rung; at 160, 7 per cent are right and 93 per cent are pairs on no rung; at 240, every band is.

Refusals are nearly absent throughout: on every stem, at every width, the counter refuses fewer than three bands in a hundred, whatever it returns instead. It fails narrow by being one rung coarse and wide by inventing a pair, and in both cases it returns a number.

How a wide band invents a family #

The counter’s own record of the families it keeps shows where the invented pairs come from. In a band of 60 organs on the bijugate stem, spanning rises from 0.0480 to 0.0396, it keeps families of 4, 4, 2, 2, 6 and 6 chains — each real family twice, at lengths a few per cent apart — and reads 2/4, the band’s pair.

In a band of 140 organs from 0.0388 to 0.0245 it keeps families of 4, 4, 16, 2, 2 and 6 chains, and reads 4/16 where the band’s pair is 4/6. Across a band that wide the lattice vectors drift enough that a single displacement no longer describes one family; the clustering splits the drifted copies, and a displacement between two families links 89 per cent of the band into sixteen chains that belong to neither lattice. It passes the 55 per cent test, and it is shorter than the real partner, so it is chosen.

The edge moves with the rate #

The upper edge is not a property of the counter alone. On ordinary stems the counter reads nine bands in ten from 48 to 48 organs at T = 150, from 80 to 100 at T = 300, and from 80 to 180 at T = 600. The lower edge is set by the largest pair a stem reaches, and it barely moves; the upper edge doubles as the rate halves.

The reason is that a band’s width means nothing by itself. What spoils a band is how much of the ladder it spans, and a band of W organs on a stem whose rise falls as $e^{-i/T}$ spans a factor of $e^{W/T}$ in rise. A slower stem spans less of the ladder in the same number of organs, and tolerates more of them.

A third of a rung #

Placed at the share of a rung each band spans — its width over the rate, over $\ln \varphi^2 = 0.9624$ — the stems line up. The widest good band is 0.33 of a rung on the ordinary stem at T = 150, 0.35 at 300 and 0.31 at 600; 0.35 and 0.31 on the bijugate stems at 300 and 600; 0.31 on the trijugate stem. Past about a third of a rung of rise, on every stem that has a working band, the counter’s success falls away.

Worked at T = 300: a third of a rung is 0.32 in log-rise, and 0.32 × 300 = 96 organs, against the measured edge between 100, the last good width, and 120, the first bad one. At T = 600 the same arithmetic gives 192 organs against an edge between 180 and 200.

When the two edges cross #

The lower edge is a number of organs set by the pair; the upper edge is a number of organs set by the rate. They are independent, so they can cross, and on the fastest stems they do. On the ordinary stem at T = 150 the counter reads nine bands in ten at exactly one width, 48 organs. On the trijugate stem at T = 300 at exactly one, 90 organs.

On the bijugate stem at T = 150 it reads nine bands in ten at none. The pairs it reaches at its fine end need bands wider than a third of a rung of its rise can supply, so every width is either too narrow for the pair or too wide for the rise. The best it manages is 85 per cent, at 60 organs. There is no setting of the instrument that reads that stem, and nothing in its output says so.

Requiring the families to cross #

The counter computes whether its two families wind opposite ways and does not require it, and a pair on no rung might be expected to fail that check. Mostly it does not. At the widest band, 320 organs, requiring the families to cross turns about a fifth of the wrong readings into refusals on the fastest stems — 21 and 22 per cent — about one in twenty-five at T = 300, and none at T = 600.

So the check that a second family winds the other way is real but weak exactly where it is needed: the invented family of sixteen chains above winds the right way, and a wide band’s wrong pair passes it. It converts some silent errors into refusals on the stems least likely to be counted with wide bands, and almost none on the stems most likely to be.

Three organs a whorl #

The trijugate stem shows both counters’ floors at once. The index counter, folded, refuses every band under twenty whorls and reads every one above. The index-free counter reads 67 to 68 per cent at 42 to 48 organs, 90 per cent from 54 to 72, 92 per cent at 90 — its one width at nine in ten — 87 per cent at 102, and 56, 28, 10 and 4 per cent at 120, 141, 162 and 180.

On a stem with no order of arrival the index-free counter is the only one available without folding, and folding needs the jugacy known in advance. A grown jugate stem can be read either way once k is known; a specimen whose k is the question can be read only by the counter with the band.

What this asks of a person counting a shoot #

The same arithmetic tells a person with a real shoot what the counter needs. The band must hold more than 2.2 times the larger count it is expected to find, or it will return the rung below. It must span less than about a third of a rung of rise, which on a real shoot means knowing roughly how fast its internodes shorten, or it will return a pair on no rung. And the second requirement can defeat the first: on a shoot that shortens fast and carries high counts, no band satisfies both.

That is the same lesson the counting window taught for the index counter on a settled stem, with one difference. There, a window too small returned the rung below and a plateau of good windows existed above it. Here the plateau has a ceiling that moves with the shoot, the window nobody varied becomes the window that has to be chosen for each specimen, and a counter that sees no positions is not available to fall back on when the arrival order is gone.

The same problem on a head #

A seed head has the same structure laid out in radius rather than height. Its counts change with radius, 13 and 21 near the centre and 55 and 89 at the rim, and the transitions sit at radii a fixed factor apart. An annulus counted for its spirals is a band, and an annulus wide enough to span a good part of the factor between two transitions holds two lattices at once, which is the condition under which the chain counter here kept a family belonging to neither.

What differs on a head is that the counter reads positions with an index available, since every seed has its place in the spiral. The index-free counter is needed where that index is gone — two organs at a time — and it is there, on a jugate stem or a jugate head, that the band has to be chosen and cannot be slid.

A plateau with a ceiling #

The counting window on a settled stem had a floor and then a plateau: every window above the floor read the same pair, because a settled stem is one lattice all the way up. The same index counter on these growing stems still has no ceiling, reading every width to 320 organs. The index-free counter has a floor, a short plateau and a ceiling, and the ceiling is where the stem stops being one lattice within the band.

So the plateau was a property of the specimen as much as of the instrument, and the index counter’s lack of a ceiling is a property of its index: it asks about organs i and i + m, and a band that spans two lattices still has the same index offsets as its shortest hops in most of its length. The chain counter asks what displacements the band repeats, and a band spanning two lattices repeats three or four. Whether a trijugate stem’s pushed whorls move either counter’s edges is not measured; its bands were read only folded, where each whorl is one node.

What this does not establish #

That a person counting a real shoot uses this counter’s neighbour count, its clustering or its 55 per cent coverage. The shape of the failure is a property of any counter that keeps a family only when it links most of a band; the widths at which it happens are this counter’s. The stems are grown by one rule with an exponentially falling rise, and a shoot whose rise falls unevenly would move the upper edge along the stem.

What would withdraw it #

The index counter misreading more than a few bands in a hundred at any width on an ordinary or bijugate stem. A static floor that differs from ⌊m/0.45⌋ + 1 in whole whorls. An upper edge that does not grow with the rate, or that sits far from a third of a rung. The index-free counter refusing, rather than miscounting, more than three bands in a hundred at any width. Each is checked every time the measurement runs.

Still open: a band that follows the rise #

The upper edge is a third of a rung of rise, not a number of organs, which suggests the fix: a band whose width is set from the rise it is read at rather than chosen once. The next test is a counter that estimates the local rise from its own band — the spacing of the points already gives it — sizes the band to a third of a rung, and slides; whether it reads every stem here, including the bijugate stem at T = 150 that no fixed width can, is the measurement.