A pinned count errs where the spirals crowd

A chosen mark errs by a third of a spiral modelled the simplest count by hand. The counter fixes a place on a circle round the head, starts each family on the spiral through the organ nearest it, runs round counting spirals, and stops at the spiral whose crossing is nearest the place again. The organ is never exactly at the place, so the count closes a little off, and the error is the head’s: three tenths of a spiral spacing in every family of every pair, one reading in eleven a count out, and almost never both.

The careful counter does something else. They push a pin through the starting floret, or mark it with ink, and close each family on that organ rather than on a place. Then the closing contributes nothing — the trace ends where it began by construction — and the only error left is losing the way round: skipping a spiral, or counting one twice, as the eye passes from one to the next. That essay ended by asking what such slips do: whether they are commoner on one family, whether they ever move both counts, and whether a pinned count at the rim is better or worse than a placed count a third of the way in.

Where the eye can slip #

The picture from the placed count is the right one to start from, because the eye does the same thing in both: it travels round the counting circle and registers each spiral as it crosses. What changes with a pin is only the end of the journey. A slip happens in the middle of it, at a crossing, and the natural guess about where a slip is likely is the one the earlier essay made: where neighbouring spirals cross close together, so the eye passing from one to the next has least room to tell them apart.

That is a quantity the head fixes, and it can be measured before any assumption is made about eyes. On the counting circle at a given radius, each family’s spirals cross at angles that follow from the Vogel head, and the spacing between neighbouring crossings, divided by the distance between neighbouring organs at that radius, says how crowded the crossings are in the lattice’s own units.

Even round the circle, uneven across the band #

Round any one circle the spacing hardly changes: the widest gap between neighbouring crossings of a family is within two per cent of the narrowest. A slip is no likelier at one part of the head than another on the same circle, so where the counter starts round the circle does not matter, which was not true of a place.

Across the band it changes by half. On 34/55, the 34-family’s spirals cross 1.73 organ spacings apart near the band’s inner edge and 2.58 at its outer edge, and the 55-family’s 1.07 and 1.60. The same holds on every pair, at almost the same numbers in organ spacings: 1.85 and 1.15 at the inner edge of 13/21, 1.70 and 1.05 at the inner edge of 55/89. At the inner edge of any band the larger family’s spirals cross the circle barely more than one organ apart, and at the outer edge about one and a half.

The larger family is always closer #

At every circle on every pair, the larger family’s crossings sit closer than the smaller’s by the ratio of the two counts, to within two per cent — 0.618 on 34/55 — since n spirals share one circle that m share. A larger family therefore offers a slip in two ways. It has more crossings to make, n against m, and each is closer to its neighbours.

How much that costs depends on how an eye’s slips grow as spirals crowd, and nothing on a head says. The measurement here takes one simple law and states it: the chance of a slip at a crossing is a fixed number divided by the square of the spacing there, so halving the room between two spirals makes a slip four times as likely. The fixed number, written $q_1$, is the chance of a slip where spirals cross one organ spacing apart. It is the eye’s own, and every result below is either stated against it or given as the $q_1$ at which two counts break even.

A count wrong by its net slips #

A slip skips a spiral or counts one twice, each equally often, and a family’s count is wrong by the net of its slips round the circle. The two families are traced separately, each from the pin round to the pin, so their slips are independent. Under the law, the 55-family on the outer edge of its band slips 2.6 times as often per crossing as the 34-family, and with 55 crossings against 34 its count is wrong about 4.1 times as often: at $q_1$ of 1 in 500, the 34-count is wrong on 1.0 per cent of readings and the 55-count on 4.2.

That answers the first question, and the answer is the opposite of the placed mark’s. A placed mark erred by the same three tenths of a spacing in both families, so neither was wrong more often. A pinned count’s errors belong mostly to the larger family.

Both counts together, by coincidence only #

The second answer follows from independence. The share of readings with both counts wrong is the product of the two shares, and nothing more: at 1 in 500 on the outer edge, 0.042 per cent. The placed mark made both counts wrong twice in six thousand places from the geometry alone, about 0.03 per cent, because both of its families started from one organ’s offset — a correlation the essays on counts that slip together and two marks chosen by one eye spent their length on. A pin removes the shared offset, and with it the only mechanism that tied the two counts’ errors together.

Two out, and silent #

A pin buys an error a place never makes. A placed count closes within a third of a spacing, so it is never two out. A pinned count can slip twice the same way, and then it is.

On 34/55 that matters more than its rate suggests. Every single miscount of 34/55 shares a factor with the other count and announces itself, as a count that can be wrong by one showed, and that protection is why a placed mark reads 34/55 silently wrong under two times in ten thousand. Two out is not protected: 32/55, 36/55, 34/53 and 34/57 are all coprime. At 1 in 500 a pin on the outer edge reads 34/55 silently wrong on 0.067 per cent of readings, and a pin a third of the way in on 0.21; at 1 in 100, 1.4 and 3.8 per cent.

The protected pairs were protected against one kind of error. A pin trades it for another.

Better at the outer edge, worse a third in #

The third question was where each count should be made, and the two counts want opposite places. A placed mark was worst at the band’s outer edge, wrong on 15.3 per cent of readings of 34/55 there, and best about a third of the way in, wrong on 1.1, where the two families’ offsets cancel. A pin is best at the outer edge, where the spirals cross widest, and worse a third of the way in at every slip chance: 5.2 per cent wrong against 9.0 at 1 in 500, 2.6 against 4.7 at 1 in 1,000.

So the right comparison is not a pin and a place on the same circle but each on its own best circle — a pin at the outer edge against a place a third of the way in. That comparison is far harder for the pin to win.

The slip rate at which a pin wins #

On the same circle, a pin beats a place for any eye slipping less than the break-even chance, and that chance is largest where a place is worst. At the outer edge of 34/55 it is 1 in 156: an eye that loses a spiral less than once in a hundred and fifty crossings at an organ’s spacing does better with a pin. A third of the way in, where a place is almost never wrong, it is 1 in 4,300.

Against a place a third of the way in, a pin at the outer edge must slip less than once in 2,400 crossings at an organ’s spacing on 34/55. On 13/21 it is once in 170, on 21/34 once in 590, on 55/89 once in 3,800. The larger the pair, the steadier the eye a pin needs, because a larger count has more crossings to slip at while a place’s error, three tenths of a spacing, does not grow with the count.

Reading the break-even #

Whether 1 in 2,400 is an easy rate or an impossible one is a fact about eyes, and the head cannot say. What can be said is what it means in the head’s own terms. On the outer edge of 34/55, where the spirals cross wider than one organ apart, an eye slipping at the break-even chance loses its place once in about 6,100 crossings of the 55-family and once in 16,000 of the 34-family — about one slip in ninety complete counts of the pair, which is the placed mark’s one wrong reading in ninety a third of the way in.

That is the arithmetic of what a count is worth from the other end. The placed mark’s error was fixed by the head and no amount of care removed it; the pin’s error is all care. A counter who knows their own slip rate can choose between them. One who does not should count at the band’s outer edge with a pin and a third of the way in with a place, and compare — a disagreement is a slip or a closing, and an agreement between two methods with different errors is worth more than either alone.

What a survey of many heads would inherit #

A survey of many heads inherits whichever error its counters carry, and the two leave different fingerprints in a table of reported pairs. Placed marks put their misreadings on both families equally and almost never on both at once, so a census of misread 34/55 heads would show 33/55, 35/55, 34/54 and 34/56 in about equal numbers. Pinned counts put them mostly on the 55-family — 34/54 and 34/56 four times as often as 33/55 and 35/55 — and add a thin tail two out. A table of reported pairs can therefore say something about how its counts were made, which the window nobody varied found a survey rarely records.

What neither method touches is what the head itself did before anyone counted it. A head whose organs sit off the ideal lattice moves the crossings themselves, and a head displaced before it is counted priced that separately; the slips here are added on top of whatever the head already did.

What changes for each pair #

The two Fibonacci pairs whose single miscounts all announce themselves, 21/34 and 34/55, lose most from a pin: their silent rate under a place was nought or nearly, and a pin’s double slips add one that grows with the square of the slip chance. At 1 in 500 on the outer edge, 21/34 is read silently wrong on 0.016 per cent of readings — small, and still more than a place ever did.

The pairs with unprotected single miscounts, 13/21 and 55/89, are where the pin’s family asymmetry can help, since it decides which family’s errors dominate and only one family’s errors are silent — as the 55/89 reading below shows in full.

55/89, where the slips land on the announcing family #

The pair above 34/55 shows the family asymmetry doing something useful. On 55/89 the single miscounts of the 89-family, 55/88 and 55/90, share factors with 55 and announce themselves; the single miscounts of the 55-family, 54/89 and 56/89, are coprime and silent, because 89 is prime. A placed mark, erring alike in both families, read 55/89 silently wrong on 4.7 per cent of readings across its band.

A pin puts most of its errors on the 89-family — at 1 in 500 on the outer edge, its count is wrong on 6.8 per cent of readings against 1.7 for the 55-count — and those are the announcing ones. The silent rate is then almost exactly the 55-family’s wrong rate: 1.67 per cent, a third of the placed mark’s, while the pin is wrong overall on 8.3 per cent against the placed mark’s 15.0 on the same circle. A third of the way in the pin is worse on both counts, 14.4 per cent wrong and 3.0 silent, against a place that is wrong on only 1.1.

So which pair is being read decides whether a pin’s family asymmetry is a cost or a gift. On 34/55 every single miscount announces itself and a pin’s extra errors are pure cost. On 55/89 the pin shifts the errors toward the family whose mistakes announce themselves, and on the outer edge it cuts the silent rate even where it does not cut the overall one.

What the law assumes #

The slip law is a choice. It makes a slip four times as likely when the room between spirals halves, and a different power would move every break-even: a law that ignored the spacing altogether would make the two families slip alike per crossing, and the larger family’s count wrong only by its extra crossings, 1.6 times as often rather than 4.1. The direction of every comparison here — the larger family worse, the outer edge better for a pin, the pin needing a steadier eye against a place a third of the way in — holds under any law in which crowding helps a slip.

Slips are taken as independent of one another. A tired eye’s are not, and a counter who slips once is likelier to slip again on the same trace, which would raise the two-out rate above what independence gives. The head is an ideal Vogel head; on a real one the crossings are less even round the circle, and the closest of them would set the slip rate.

Readings that would undo it #

A family’s crossings on any circle varying round it by more than four per cent, or the larger family’s spacing departing from the smaller’s times the ratio of the counts by more than two per cent. Under the law, at 1 in 500 on the outer edge of 34/55, the 55-count wrong less than three times as often as the 34-count, or both counts wrong other than as often as the two chances multiply. A pinned 34/55 never two out and never silent at 1 in 100 a third of the way in. On any pair, a pin at the outer edge beating a place a third of the way in at a slip chance no smaller than it needs to beat a place on its own circle. Each is checked whenever the counts are read.

Still open: a slip that remembers #

Every slip here is independent of the last, so a count’s error is a sum of separate accidents and two slips the same way are rare. An eye that has lost its place once is likelier to lose it again nearby — a stretch of the head where the florets are damaged, a moment of distraction — and slips that come in runs make two-out errors, the silent ones on 34/55, far commoner than independence allows. The next measurement lets each slip raise the chance of the next over a stated number of crossings and asks how much clustering it takes for a pin’s silent rate on 34/55 to reach a place’s error rate, and whether counting each family twice, from two pins, catches what clustering adds.

Measured, a slip that remembers is caught by a second pass: runs of slips raise the pin’s silent share about fivefold and lower its total error, and at no reasonable eye does the silent share reach a place’s error. Two passes that must agree leave a tenth of a per cent wrong, with runs or without, so long as the passes are independent.