Phyllotaxis
the spirals are counted, not admired
Stems and cones

One step of the grid, again

A search of the fine end stepped from one rise to another a hundred grid steps away, and the family a cut keeps changed somewhere between. Cutting the rises in between puts the change inside one step, with the lattice identical on both sides.
14 min read 20 figures State the claim, state the testCounted, not admiredOne parameter, one family

Worth reading first: The organ that was taken away · Where a handover sits.

A search of the fine end of both branches went looking for a lattice whose wrecked cuts keep a lag outside the four the census holds. It found one: the Lucas branch at a rise of 0.006, on the 7/11 rung, three of whose offsets keep a lag of 11.

The search stepped from 0.0070 to 0.0060 in one move. At the coarser rise every cut that wrecks keeps 7; at the finer, three keep 11. So the change is somewhere inside a hundred steps of the grid the ladder names its rises on, and nothing said where.

The Lucas 7/11 rung between 0.007 and 0.006, cut at every rise. One row per offset, one column per rise of the search, coarse on the left. A pale cell is a rise at which that offset's cut recovers; a dark cell is a cut that wrecks and keeps a lag of 7; a warm cell is one that keeps 11. The nine intermediate rises of the ten-thousandth grid were cut first and the nine of the hundred-thousandth grid inside the one bracket they opened. The change sits between 0.0064 and 0.00639, one step of the grid the ladder names its rises on, and above it no cut keeps 11 anywhere.
Fig. 1 The rung between the two rises the fine-end search stepped over, cut at every rise of two grids.

The design

Two grids, the second only where the first says something happened.

The coarse pass cuts the nine intermediate rises of the ten-thousandth grid — 0.0069 down to 0.0061 — plus the two ends the search already has, so this table contains the rows it is a refinement of rather than being a rival reading of them.

The fine pass takes every adjacent pair whose answer differs and cuts all nine interior rises of the hundred-thousandth grid between them. That is the grid the ladder itself is named on, and it is where a slot was found to lose its second wall inside one step.

Twenty rises at the fine end of both branches, cut at every offset. One row per rise searched, coarse at the top of each block. The bar names the counted pair the stem shows and the numbers on the right are the lags its wrecking cuts leave standing. A pale row is a rise whose settled divergence has left the branch it was started from by more than 20 degrees, which is what happens below the ladder's finest rung — the pairs there are 2 and 4, 8 and 16, 11 and 22, which are not two consecutive terms of any additive sequence. One rise on the Lucas branch keeps a lag of 11 while still on it.
Fig. 2 The fine-end search this refines, whose Lucas list stepped from 0.0070 to 0.0060.

Why refine every bracket rather than bisect

A bisection assumes the answer changes once and monotonically, and neither is known in advance. The golden 8/13 band changes back and forth nineteen times, so an assumption of monotonicity on a quantity of this kind has been refuted on the object next door.

Refining every bracket costs more when there are several and the same when there is one. A budget caps the total, and a bracket left unrefined is reported as unrefined rather than dropped.

In the event there was one bracket, so the two designs would have agreed — which is not a reason to have chosen the weaker one.

The Lucas 7/11 rung between 0.007 and 0.006, cut at every rise. One row per offset, one column per rise of the search, coarse on the left. A pale cell is a rise at which that offset's cut recovers; a dark cell is a cut that wrecks and keeps a lag of 7; a warm cell is one that keeps 11. The nine intermediate rises of the ten-thousandth grid were cut first and the nine of the hundred-thousandth grid inside the one bracket they opened. The change sits between 0.0064 and 0.00639, one step of the grid the ladder names its rises on, and above it no cut keeps 11 anywhere.
Fig. 3 The twenty rises the search cut, of which nine are the coarse grid and nine the fine one inside a single bracket.

What it cost

Twenty rises, eighty-nine cut stems, two hundred and eighty stems in all counting the controls, and about seven minutes.

That is cheap because a rise here is thirteen offsets tried and about four wrecking, rather than a band’s worth. The expense in this thread is bands: 1,120 to 1,890 cut stems each.

The comparison is worth making because the two designs answer different questions. A band holds the lattice still and moves the rise a little; this moves the rise across nearly a sixth of a rung and lets everything move with it.

The Lucas ladder, rung by rung. Each bar is one rung — a run of rises over which a counter returns one pair — drawn from its coarse end on the left to its fine end on the right, with the whole bar scaled to the same width so that positions inside different rungs can be compared. The mark on each bar is the rise at which the two contact steps change places, and it falls between 6 and 40 per cent of the way down on every rung that has one. The dots are the lattices this collection's ablation census was grown at, dropped onto the rungs they belong to. They are not spread across the bars; three of them sit past the mark.
Fig. 4 The Lucas branch’s rungs, with the stretch of the 7/11 rung this search cuts.

The answer

Between h = 0.00640 and h = 0.00639. One step of the grid.

Above it, at every one of the eight rises from 0.0070 down, every wrecking cut keeps 7 and none keeps 11. Below it, at every one of the eleven rises down to 0.0060, some cut keeps 11.

So it is a transition and not a stretch where both occur. There is no rise in the search at which the answer is ambiguous, none at which it changes back, and no interval over which the two families alternate.

The Lucas 7/11 rung between 0.007 and 0.006, cut at every rise. One row per offset, one column per rise of the search, coarse on the left. A pale cell is a rise at which that offset's cut recovers; a dark cell is a cut that wrecks and keeps a lag of 7; a warm cell is one that keeps 11. The nine intermediate rises of the ten-thousandth grid were cut first and the nine of the hundred-thousandth grid inside the one bracket they opened. The change sits between 0.0064 and 0.00639, one step of the grid the ladder names its rises on, and above it no cut keeps 11 anywhere.
Fig. 5 The located transition: no rise above it keeps eleven and every rise below it does.

Which is the second such transition

The first is the slot that loses its second wall, on the golden 5/8 rung, located between 0.00998 and 0.00997 — also one step of the same grid, also with nothing continuous changing across it.

Two transitions on two rungs of two branches, both one part in a thousand of the rise wide, both in quantities that are integers or discrete states read off a run. That is a pattern worth naming even at two cases: the discrete readings in this thread change sharply, and the sweeps that have looked have found no gradual case.

What would refute it is a quantity of the same kind that changes over ten grid steps rather than one. Nothing has produced one.

The golden 5/8 rung swept at 27 rises, with each removal's cost. How far the first organ placed after a cut moves, at every rise the sweep visits, coarse on the left. Removing both walls costs far more than removing the larger one alone above a rise of 0.00998, and exactly what the larger one costs below it. The change happens in one step of the grid the ladder is named on: 163.59 degrees at 0.00998 and 9.14 degrees at 0.00997, which is a fall of 154.5 degrees for a change of one part in a thousand in the rise.
Fig. 6 The other transition located to one grid step, on a different rung and a different quantity.

And nothing about the lattice moves across it

This is the check that makes a transition a transition rather than a lattice change.

The settled divergence is 99.2578 degrees on both sides, to every digit the azimuth grid carries — the grid is 1,536 samples of the circle, a quarter of a degree a step, and the two readings are identical rather than close. The counted pair is 7/11 on both sides.

The ordering of the two contact hops is the same on both sides: the 11-hop is shorter at both rises, as it is at every rise from 0.008 down. Their lengths change by about a part in a thousand across the step, in the direction they were already moving.

Two rises on the Lucas 7/11 rung, each located to one step of the grid. The whole of the Lucas 7/11 rung, coarse on the left, with the two rises this round located. The two contact steps change places between 0.00804 and 0.00803, found by walking all 378 rises of the rung and evaluating hop lengths, with no cut stems at all. The family a cut leaves standing changes between 0.0064 and 0.00639, found by cutting 89 stems. They are 164 steps of the grid apart, which is 45 per cent of the rung's whole span, so the geometry crossing is not what moves the survivor.
Fig. 7 The two located rises on this rung, of which the geometry crossing is far above the transition.

So what changes is which cuts wreck

At 0.00640 the offsets that wreck are 5, 6 and 7, and all three keep 7. At 0.00639 they are 5, 6, 7 and 9, and the new one keeps 11.

Nothing changes its mind. An offset that was wrecking and keeping 7 goes on wrecking and keeping 7; an offset that was recovering starts wrecking, and what it keeps is the family the search was looking for.

That is a different shape from the one the question assumed. The rise this search locates is the rise at which the lattice first keeps 11 anywhere, and it is located by an arrival rather than by a change of answer.

What each offset keeps, above and below 0.00639. Every offset that wrecks anywhere in the search, with the family it keeps above the transition and below it. Offsets 5, 6, 7, 9 keep the same family at every rise they wreck at, on both sides. What changes at 0.00639 is that offset 9 begins to wreck at all, and what it keeps is the larger of the counted pair. One offset does change its own answer, and it cannot be located, because it does not wreck at the rises in between.
Fig. 8 What each offset keeps above and below the transition, and the one that arrives with the new family.

Which offsets wreck, again

The wrecking set moves from rise to rise across the whole search, exactly as it does across every band cut whole.

At 0.0070 it is {4, 5, 7}. At 0.0069 it is {5, 6, 7}. At 0.0068 it is {5, 7, 8}. At 0.0067 it is {5, 7}. At 0.0066 it is {5, 6, 7} again. At 0.0065 it is {5, 6, 7, 8}.

Six consecutive rises, six membership changes, one part in a hundred and forty apart in the rise. The band sweeps found the same thing over stretches of rise; here it is visible between adjacent readings.

Which offsets wreck across the Lucas 7/11 band. One row per offset and one column per rise, with a mark where a single removal at that offset wrecks the stem. The set is not the same at every rise: on this band one offset wrecks at only 27 of its 124 rises, in several separate stretches, while others wreck at all of them. So a census taken at one rise of a band and a census taken at another are censuses of different sizes, and every claim of the form "at every offset that wrecks" is quantified over a set the rise decides.
Fig. 9 The wrecking set moving across the same rung’s band, which is the coarser view of the same instability.

What that does to a census

A census cuts every offset at a chosen rise and reports what it finds. The offsets that wreck at that rise are its rows.

On this stretch, a census taken at 0.0067 has two rows and one taken at 0.0060 has six. Neither is wrong. They are censuses of different sizes because the rise decides membership, and the rise was chosen for coverage rather than for membership.

That is the same boundary the third band put round the same claims, and it is worth repeating because the ablation census’s ten lattices are the base of most of this thread.

Every stem that never repaired, and the lag it kept. The 19 offsets across six lattices at which a single removal leaves a stem that never returns to its divergence. For each one: which organ was removed, the period of the block of angles the stem settles into, the lag whose hop survived the cut unchanged, and how many whole turns the stem gains over one period of that lag. The block and the surviving lag are the same number in every row. The marked row is the one whose survivor is not a parastichy number of the lattice that was cut — a hop 6.8 times the length of a contact hop, which no census would report and which the rule held rigid all the same.
Fig. 10 The census this stretch sits below, whose membership depends on which rises were chosen.

The eleven-keeping set fills in

Below the transition the offsets keeping 11 grow: {9} at 0.00639 and 0.00638, {8, 9} at 0.00637, {9, 10} at 0.00636, {8, 9} again for four rises, then {8, 9, 10} from 0.00631 down to 0.0060.

So the arrival is not a single event but a stretch over which more offsets join. The located rise is where the first one does, and that is the quantity the fine-end search was asking about.

By 0.0060 — the rise the search actually cut and the one the exchange’s fifth hop cluster comes from — three offsets keep 11 and three keep 7, which is the row the census gained.

The 20 rows of the exchange table, gathered by the lag they kept. One bar per surviving lag, its length the number of rows the table holds at that lag, with the hop that lag's stems keep written beside it. The hop is nearly constant inside a lag, so a correction fitted over 20 rows is fitted over five hops — which is the denominator that matters and is much smaller than the row count suggests. Adding a lag to the table is worth more than adding rows at a lag already in it.
Fig. 11 The extended census, whose eleven-keeping rows come from the rise at the bottom of this search.

Offset 8 is the one that changes its mind

And it cannot be located. It wrecks at 0.0068 keeping 7, at 0.0065 keeping 7, and at 0.00637 keeping 11 — and it does not wreck at 0.0064, 0.00639 or 0.00638.

So its own change is bracketed across thirteen steps of the grid, and the bracket contains the transition without that meaning anything. It is the same limit the third band ran into: an offset’s answer exists only where its cut wrecks, and the wrecking set moves faster than the answer does.

Two sweeps, two questions, one limit. Naming it in both places is the point, because it is structural rather than a matter of resolution.

What each offset keeps, above and below 0.00639. Every offset that wrecks anywhere in the search, with the family it keeps above the transition and below it. Offsets 5, 6, 7, 9 keep the same family at every rise they wreck at, on both sides. What changes at 0.00639 is that offset 9 begins to wreck at all, and what it keeps is the larger of the counted pair. One offset does change its own answer, and it cannot be located, because it does not wreck at the rises in between.
Fig. 12 The one offset whose own answer changes, whose change is bracketed across thirteen grid steps.

What a rung buys that a band does not

A band holds the settled divergence still to within a twentieth of a degree, which is what makes it the right object for asking whether the survivor changes with nothing else moving.

A rung is looser. Across this stretch the divergence moves from 99.141 to 99.344 degrees — two tenths, four times a band’s tolerance — so a change anywhere in it could in principle be a change in the lattice.

What makes the located transition safe is that the divergence is held across the step rather than across the stretch. The two rises either side read identically, so whatever the divergence is doing over the search as a whole, it is doing nothing there.

The two contact steps change places inside the 5/8 rung. Measured at every rise on one rung, where a counter returns 5 and 8 spirals throughout. The settled divergence slides from 136.6406 to 137.8672 degrees. The ratio of the two contact steps falls to 1.0131 at a rise of 0.016 and the ordering changes hands at 0.015: above it the shorter step belongs to the 5-family and below it to the 8-family. Neither quantity is available to a counter, which is shown positions and reports a pair, and both of them move while that pair does not.
Fig. 13 A rung’s geometry, which moves across its span where a band’s is held still.

The refusal

A search whose two ends agree has nothing to bracket, and the reading must say so rather than reporting a transition between the only two rises it looked at.

That is checked by asking for a stretch at the coarse end of the same rung, one grid step wide, where every cut keeps 7 on both sides. It opens no bracket and spends no fine cuts, which is the answer it must give.

The check costs two rises rather than a rung, which is the same economy a refusal on a band was rewritten for after one provoked its refusal by cutting seventy rises to learn one boolean.

The Lucas 7/11 rung between 0.007 and 0.006, cut at every rise. One row per offset, one column per rise of the search, coarse on the left. A pale cell is a rise at which that offset's cut recovers; a dark cell is a cut that wrecks and keeps a lag of 7; a warm cell is one that keeps 11. The nine intermediate rises of the ten-thousandth grid were cut first and the nine of the hundred-thousandth grid inside the one bracket they opened. The change sits between 0.0064 and 0.00639, one step of the grid the ladder names its rises on, and above it no cut keeps 11 anywhere.
Fig. 14 The search’s own ends, both of them rises the fine-end search had already cut and agreed with.

Why this rung and not another

Because it is the only rung on the ladder whose cuts have ever kept a lag of 11, and the reason is arithmetic rather than luck. A surviving family is a contact family of the lattice, so a lag of 11 needs a counted pair with an 11 in it, and on the two branches those are the golden 8/13 rung and the Lucas 7/11.

The golden half has no object at all. Every golden rise inside the 8/13 rung keeps 8 or 4, and at not one of them does the cut leave the longer of the two counted steps standing — the 13 is there and is never what survives.

So the whole question lives on one rung of one branch, and this search is the whole of what can be asked about it without leaving the ladder.

The 17 rows of the exchange table, gathered by the lag they kept. One bar per surviving lag, its length the number of rows the table holds at that lag, with the hop that lag's stems keep written beside it. The hop is nearly constant inside a lag, so a correction fitted over 17 rows is fitted over four hops — which is the denominator that matters and is much smaller than the row count suggests. Adding a lag to the table is worth more than adding rows at a lag already in it.
Fig. 15 Every family a wrecking cut has been seen to keep, of which eleven appears on one rung only.

What the transition is not

It is not the two contact steps changing places. That happens on this rung too — it is what makes it a rung with a handover — and it happens at 0.00804, a hundred and sixty-four grid steps higher.

It is not the counted pair changing. The pair is 7/11 from 0.00947 down to 0.00570, which is the whole rung, and the transition sits comfortably inside.

And it is not the stem leaving its branch. The settled divergence is within four tenths of a degree of the Lucas angle at every rise of the search, where the rises below the ladder’s finest rung sit thirty to eighty-eight degrees off it.

The two shortest hops across the Lucas 7/11 rung, and where they cross. The lengths of the two shortest hops at every one of the 378 rises of the rung, fine on the left, computed from the settled divergence rather than measured off a cut. They cross exactly once, between 0.00804 and 0.00803, which is this rung's handover. The rise at which a cut first keeps the longer family is 164 steps of the grid further down and is marked separately; nothing happens to either length there.
Fig. 16 The two contact hops across the whole rung, crossing once and far above the transition.

The window the answer is read through

A surviving family is the smallest lag whose hop the wrecked run holds rigid, read over a window at the top of the run. That window is a hundred and twenty organs, and it is one of the instrument settings this collection has learned to distrust.

It matters less here than elsewhere. The reading is a boolean per lag — held or not held — and across the census the rigid lags hold their angle to within a fraction of a degree while the others miss by tens, so the verdict is a separation rather than a threshold.

Where the window does bite is on a survivor large enough that a hundred and twenty organs gives few repeats of it. At a lag of 11 that is eleven samples a class, which is comfortable; at a lag of 20 it is six and the reading has to be declined.

A period of 8, and the two classes that are not with the rest. The same wrecked stem, folded on the lag it kept: one row per residue class, each drawn at the mean displacement of its own organs against the level the rest of them share. The bar through each row is the spread inside that class, and the widest of them is 0.87° — so within a class the displacement is a constant. six classes sit at the common level. The two that do not sit at 134.3° and -134.2°, equal and opposite to within 0.0 per cent, and they are neighbouring residues. The stem's own divergence is 137.44°, so an exception is one organ's step.
Fig. 17 The profile a surviving lag is read from, folded onto the period it repeats at.

What a hundred grid steps hid

The fine-end search that produced this question is a search rather than a sweep, and it says so: ten rises a branch, chosen to cover the two rungs where an 11 could appear and to run down past the ladder’s finest rise.

Its step is not a resolution. Between 0.0070 and 0.0060 it takes one step, which is a hundred rises of the grid the ladder is named on, and between 0.0040 and 0.0035 it takes one step of fifty. A search designed to answer does this exist anywhere is under no obligation to be uniform.

What it cannot do is locate anything, and this round is the price of that: the search answered its own question in eight minutes and left a hundred-step bracket that costs seven more to close.

Twenty rises at the fine end of both branches, cut at every offset. One row per rise searched, coarse at the top of each block. The bar names the counted pair the stem shows and the numbers on the right are the lags its wrecking cuts leave standing. A pale row is a rise whose settled divergence has left the branch it was started from by more than 20 degrees, which is what happens below the ladder's finest rung — the pairs there are 2 and 4, 8 and 16, 11 and 22, which are not two consecutive terms of any additive sequence. One rise on the Lucas branch keeps a lag of 11 while still on it.
Fig. 18 The search’s own rises on the Lucas branch, whose steps are chosen for coverage rather than for resolution.

What would be different one rung up

Nothing that can be tested, because the 4/7 rung has no 11 in its pair and no cut on it can keep one. The question does not exist there.

That is worth saying because it bounds the generality of everything here. This is one transition, on one rung, in one quantity, and there is no second instance of it available on this ladder — not because nobody has looked but because the arithmetic forbids one.

A second instance would need a different ladder: a different falloff exponent, or a different geometry, either of which changes the rungs and the pairs they carry. That is a different question rather than a replication of this one.

The golden ladder, rung by rung. Each bar is one rung — a run of rises over which a counter returns one pair — drawn from its coarse end on the left to its fine end on the right, with the whole bar scaled to the same width so that positions inside different rungs can be compared. The mark on each bar is the rise at which the two contact steps change places, and it falls between 12 and 35 per cent of the way down on every rung that has one. The dots are the lattices this collection's ablation census was grown at, dropped onto the rungs they belong to. They are not spread across the bars; seven of them sit past the mark.
Fig. 19 The golden branch’s rungs, of which one carries a thirteen that no cut has ever kept.

What the seven minutes bought

A rise located to one part in a hundred thousand, with the lattice held across it, on a quantity that had been known to a hundred grid steps.

That is a good return for eighty-nine cut stems, and it is worth comparing against the two alternatives. A blind sweep of the whole rung at the same grid is 378 rises of cutting, about two hours, and would locate the same rise no better. A finer version of the fine-end search — twenty rises a branch instead of ten — would have halved the bracket and cost as much as this did, without closing it.

The design that works is: search coarsely for existence, then refine inside whatever bracket the search leaves. It is the same shape as the three sweeps that located a slot losing a wall, and it works for the same reason — the quantity changes once, so the bracket is real.

What would have made it fail

A quantity that changes back. Nothing in advance ruled that out: the golden 8/13 band changes nineteen times across 126 rises, so a survivor that alternates was the shape most recently observed on a neighbouring object.

The design handles it — every bracket is refined, not just the first — and it cost nothing here because there was one bracket. Had there been three, the fine pass would have spent twenty-seven rises instead of nine and reported three located changes.

That is the difference between a design that assumes an answer and one that admits it. The budget is what stops the second from being unbounded, and a bracket left unrefined is reported as unrefined rather than dropped.

What the twenty rises hold besides the answer

A record of the wrecking set across a sixth of a rung at the finest resolution anybody has taken it. Seven consecutive rises give seven different sets of wrecking offsets, and that is the same instability the band sweeps found over stretches of rise — here between neighbours.

It also holds the survivors at every one of those rises, which is what makes the located transition a statement about the lattice rather than about one cut: the answer is read from four to six offsets at each rise, not from one.

Neither was the point of the search, and both come out of it for nothing, because the sweep returns everything it measured rather than the summary the question needed.

What is claimed

That on the Lucas 7/11 rung the family a wrecking cut leaves standing first includes 11 between two rises one hundred-thousandth apart; that no rise above keeps it and every rise below does; and that the counted pair, the settled divergence and the ordering of the two contact hops are identical on both sides.

That the change is an offset arriving rather than an offset changing its answer, and that the one offset which does change its answer is not locatable at this resolution.

And that this is the second discrete transition in the thread located to one step of the grid the ladder is named on, with nothing continuous moving across either.

The Lucas 7/11 rung between 0.007 and 0.006, cut at every rise. One row per offset, one column per rise of the search, coarse on the left. A pale cell is a rise at which that offset's cut recovers; a dark cell is a cut that wrecks and keeps a lag of 7; a warm cell is one that keeps 11. The nine intermediate rises of the ten-thousandth grid were cut first and the nine of the hundred-thousandth grid inside the one bracket they opened. The change sits between 0.0064 and 0.00639, one step of the grid the ladder names its rises on, and above it no cut keeps 11 anywhere.
Fig. 20 The whole search: twenty rises, eighty-nine cut stems, one transition and one grid step.

Where this goes next

Twenty rises at the fine end of both branches, cut at every offset. One row per rise searched, coarse at the top of each block. The bar names the counted pair the stem shows and the numbers on the right are the lags its wrecking cuts leave standing. A pale row is a rise whose settled divergence has left the branch it was started from by more than 20 degrees, which is what happens below the ladder's finest rung — the pairs there are 2 and 4, 8 and 16, 11 and 22, which are not two consecutive terms of any additive sequence. One rise on the Lucas branch keeps a lag of 11 while still on it. The lag that never survives A correction to the exchange's size rests on four hop clusters, and a fifth would be the first real test of it. The prediction was written for a golden lattice at a lag of eleven. No golden lattice on this ladder reaches one, and the reason is a fact about the rule rather than about the search. The golden 5/8 rung swept at 27 rises, with each removal's cost. How far the first organ placed after a cut moves, at every rise the sweep visits, coarse on the left. Removing both walls costs far more than removing the larger one alone above a rise of 0.00998, and exactly what the larger one costs below it. The change happens in one step of the grid the ladder is named on: 163.59 degrees at 0.00998 and 9.14 degrees at 0.00997, which is a fall of 154.5 degrees for a change of one part in a thousand in the rise. Where a slot loses a wall Two rungs were reported to go free at 84 and 85 per cent of themselves — the second removal stops costing anything over the larger one alone. Three samples a rung cannot say whether that is a transition or a slope, and twenty-nine more say it is a transition one grid step wide. Two rises on the Lucas 7/11 rung, each located to one step of the grid. The whole of the Lucas 7/11 rung, coarse on the left, with the two rises this round located. The two contact steps change places between 0.00804 and 0.00803, found by walking all 378 rises of the rung and evaluating hop lengths, with no cut stems at all. The family a cut leaves standing changes between 0.0064 and 0.00639, found by cutting 89 stems. They are 164 steps of the grid apart, which is 45 per cent of the rung's whole span, so the geometry crossing is not what moves the survivor. Two rises far apart The whole handover thread rests on the claim that where the two contact steps change places is not where the survivor does. On one rung both rises are now located to a single step of the grid, and they sit forty-five per cent of the rung apart.

What links here

Computed from the collection, not written here: the essays that point at this one.

Shares its objects with

Essays that name at least two of the same things, and that neither author linked.

Named objects

A flat tag is an object no other essay names yet.

AblationCensus designClaim testingContact familyDiscretisationHonest limitsHop lengthResolutionRigid hopRungSearchTransition