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Chain counting — the series

3 essays on one idea, from the one that introduces it to the one that assumes the rest.
  1. Tracing one family: 4 chains. Every node is joined to the node one repeated displacement away, and the chains that result are drawn separately. There are 4 of them, which is the parastichy number of this family. No index of arrival was used anywhere, which is what lets the same count be made on a pattern where 2 primordia appear at once.

    Counting without an index

    A person counting spirals on a cone puts a finger on one scale, follows a family round, and counts how many distinct chains there are. That needs no order of arrival — and building it turns out to be strictly more general than the counter that reads the order of arrival, and to find a bug in the counting of a bijugate stem that nothing had caught.

    part 3 · lattices
  2. How often three counters read a band's own pair, against the width of the band, on an ordinary stem. Bands of 40 to 320 organs slid up an ordinary stem grown at T = 300 over a rise from 0.05 to 0.0005, each read three ways. The counter without an index reads the pair at 65%, 66%, 87%, 88%, 88%, 88%, 89%, 100%, 100%, 92%, 54%, 23%, 8%, 2%, 1%, 0%, 0%, 0%; the counter families required to cross reads the pair at 65%, 66%, 87%, 88%, 88%, 88%, 89%, 100%, 100%, 92%, 54%, 22%, 8%, 2%, 1%, 0%, 0%, 0%; the counter with the index reads the pair at 100%, 100%, 100%, 100%, 100%, 100%, 100%, 100%, 100%, 100%, 100%, 100%, 100%, 100%, 100%, 100%, 100%, 100%. The index-free counter reads nine bands in ten or more only from 80 to 100 organs.

    A counter that cannot be slid

    The counter that needs no order of arrival follows each family into chains and counts them, and on an ideal lattice it agrees with the counter that does. On a stem whose rise falls it works only inside a band of widths, and outside the band it returns a pair rather than refusing: the rung below when the band is too narrow for the larger count, a pair on no rung when the band spans more than about a third of a rung of rise. The upper edge moves with the rate, so a width that is right on one stem is wrong on another, and on the fastest bijugate stem measured no width works at all.

    part 4 · lattices
  3. The band the counter chooses for itself, against how slowly the stem grows. The counter is given positions and no width. It fits the decay of the rise through the spacings of its own band's whorls and takes the band that spans a third of a rung of rise: 49 organs on the stem that falls over 150; 97 organs on the stem that falls over 300; 193 organs on the stem that falls over 600. A stem grown four times as slowly chooses a band 3.94 times as wide, because a third of a rung is a statement about rise and a width is a number of organs. That is exactly why no one width fits the seven stems, and why the previous reading found a stem for which none worked.

    A band that follows the rise

    The index-free counter reads a growing stem only inside a band of widths, and the upper edge is a third of a rung of rise rather than a number of organs — so a width right on one stem is wrong on another. A counter that fits the decay of the rise through the spacings of its own band's whorls, and takes the band spanning a third of a rung, chooses 49, 97 and 193 organs on stems falling over 150, 300 and 600. Given no width at all it reads within eight points of the best of eighteen fixed widths on six of the seven stems, matching it exactly on one and beating it on two. Refusing any band too narrow to have shown the rung above the pair it counted removes every reading of the rung below, on every stem — and costs between five and fifty-one points of correct reading to do it.

    part 5 · lattices

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