Where a slot loses a wall

A growing organ sits in a slot between two neighbours, and the two neighbours are its two walls — the two contact families the counter returns. Removing one is one experiment; removing both is the fourth cell of a two-by-two, and it is the cell the design exists for.

On most lattices the pair removal costs far more than either wall alone — that is the interaction the design was built to find. On two rungs it costs nothing over the larger wall alone, to within half a degree, and those rows were reported at 84 and 85 per cent of their own rungs.

What three samples a rung can say #

The slot design was run on twenty-four lattices, three to a rung at 15, 50 and 85 per cent of it. Position inside the rung was the candidate the design was built to test and it decided nothing about the interaction’s sign.

The one thing it did decide was this: both rungs that go free go free at their 85 per cent position and not at their 50 or 15 per cent ones. Three samples say where a change is between, and nothing about what kind of change it is.

What the position did not decide was the sign, which is settled by the larger counted number instead on 22 of the 24 rows. Position moves the interaction by three to fourteen degrees across a whole rung and decides nothing.

Why the kind matters #

If the fourth cell’s cost declines smoothly to nothing, then free is a word doing too much work — a threshold at half a degree cutting a continuum, and the two free rows are the two that happened to fall below it.

If it falls in one step, there is a rise at which a slot loses a wall, and that is a thing to name. The two readings have the same three samples behind them and different consequences for everything the word is used in.

The rung swept #

The golden 5/8, because it is one of the two rungs with a free row and because it is the one whose free row is not the whole rung. The Lucas 1/3 rung is free at all three of its positions, which is the arithmetic case that file separates out — a transition cannot be located on a rung with no other side.

The golden 5/8 runs from a rise of 0.01791 down to 0.00689 and the slot table holds six rows on it, at 15, 34, 50, 61, 84 and 85 per cent. Two of those six came from the six lattices the earlier design ran on and were kept rather than re-measured, which is why the sweep’s coarse end is already populated.

The first ten positions #

Ten rises between 70 and 97 per cent of the rung, at three per cent apart. That is where 84 and 85 sit, so it is where a transition near them would be.

Every one of the ten is free. So the change is not in that stretch at all: it is higher up the rung than the design that found it, and the two rows reported at 84 and 85 per cent were not the start of anything.

Which is worth pausing on as a lesson about searching rather than about slots, because the bracket was already in hand. The three-position design has the 50 per cent row with two walls and the 85 per cent row free — so it had already established that the change sits between 50 and 85 per cent, and the first sweep went to 70–97, which is nine tenths inside the free half of that bracket.

Ten runs were spent confirming something one row already said. A bisection of the bracket the design supplied would have taken 35 percentage points to one in about six steps, and to the five-decimal grid in a few more — call it ten runs against the twenty-nine this search used, with the answer in the same place.

The reason it went to 70–97 is not carelessness either, and naming it is the useful part. The two free rows were reported at 84 and 85 per cent, and a number a table reports is where attention goes; the row at 50 per cent that bracketed them from the other side is a row about something else and was not being read as a bound. A search aimed at where a phenomenon was seen rather than at where it is bracketed will start inside the phenomenon.

That is a cheap habit to fix and it costs nothing to state: before sweeping for a change, write down the two nearest rows on either side of it, and start between them rather than beside the one that showed it.

The next ten #

Ten rises between 61 and 70 per cent, at one per cent apart. The row at 61 per cent — the lattice at a rise of 0.01, which the slot table already held — has two walls. The row at 62 per cent does not.

So the change is between 61 and 62 per cent of the rung, which is a step of about nine parts in ten thousand in the rise.

And nine more #

Nine rises at the five-decimal grid the ladder is named on, between 0.00999 and 0.00991. The change sits between 0.00998 and 0.00997.

That is one step of that grid — one part in a thousand of the rise, since the grid is absolute and the rise there is about a hundredth. The transition is located to the resolution the ladder is described at and no further.

The size of the step #

Above the transition the pair removal throws the next organ 163.59 degrees round. Below it, 9.14 degrees — which is what removing the larger wall alone does, to the last digit of the azimuth grid.

A fall of a hundred and fifty-four degrees, across a change of one part in a thousand in the rise. That is not a slope.

Both values sit exactly on the grid #

The azimuths are chosen from 1,536 samples of the circle, a quarter of a degree a step. 163.59 degrees is 698 of those steps and 9.14 is 39, both exactly.

So neither number is a fit or an average. They are grid positions, and the same two grid positions come back at every rise on their own side of the transition.

The sign changes too #

The signed displacement is +163.59 above the transition and −9.14 below it. That is the check that matters most, because a displacement is folded to a half turn and a value near 163 degrees is close enough to 180 for a fold to be the whole story.

It is not. An angle creeping past a half turn and coming back would give −16 and then −9, not +164 and then −9. The two placements are 172.7 degrees apart and on opposite sides.

Folding is not optional here — a lattice and its mirror are one lattice under this rule, so a displacement has to be reported in a form that does not distinguish them — and the price of folding is that a check like this one has to be made explicitly.

Nothing else moves across it #

The obvious objection to a jump like that is that the lattice changed. Five quantities say it did not.

The two walls are 5 and 8 on both sides. The settled divergence is 137.438 degrees on both sides. The block the cut opens is 2 on both. All three removals wreck the stem on both. And the lags the doubled cut leaves rigid are 5, 10 and 15 on both.

So what changes is where one organ goes #

That is the whole of it, stated plainly. Same lattice, same walls, same block, same settled divergence, same surviving lags — and the first organ placed after the doubled cut lands in a different place.

Which makes the transition a fact about a placement rather than about a lattice, and puts a boundary on how much the word free is entitled to carry.

Why it happens where it does #

No account is offered. Two observations are on record and neither is an explanation.

The first is that the transition is nowhere near the rung’s handover, which sits at 0.01558 — about 34 per cent of the rung — while the transition is at 61 per cent. The second is that the fourth cell’s cost is a flat 163.59 degrees at every position above it, from 15 per cent down, rather than climbing towards a boundary.

A cost that is constant and then abruptly a different constant is what a placement that picks one of two candidate azimuths would look like. That is a guess and it is written down as one.

It would be checkable. The organ after a cut is placed at the minimum of a sum over its neighbours across 1,536 candidate azimuths, so a run whose sum has two nearly equal minima would flip between them as the rise moved a hair. Nobody has looked at the sum itself.

What the free rows were, then #

Not a discovery about the fine end of a rung. The two rows reported free at 84 and 85 per cent are free because everything below 62 per cent of that rung is free.

The design placed three samples per rung and two of them — at 15 and 50 per cent — fell above the transition, so it reported the change as being between 50 and 85 per cent. That is correct and it is a bracket sixteen times wider than the thing it brackets.

The other rung that goes free #

The Lucas 3/4 rung also has a free row, at its 85 per cent position. It has not been swept finely and the same question applies to it.

Whether its transition is also a single step, and whether it sits at a similar fraction of the rung, is about fifteen minutes of stems. It would say whether the shape found here is a property of slots or of this rung, which is the same generalisation question two bands cut whole had to ask about a different quantity — and where the answer turned out to be that one of the two cases was the unusual one.

And the rung that is free throughout #

The Lucas 1/3 rung is free at all three of its positions, and it is free for an arithmetic reason rather than a measured one: on it, removing both walls costs exactly what removing the larger costs, so the interaction is minus the smaller wall’s own cost by cancellation.

That is not the same phenomenon. It is a rung whose smaller wall costs nothing to remove alongside the larger, and it is separated out of every claim the slot table makes for exactly that reason.

What twenty-nine positions cost #

Each is a two-by-two — an intact stem, the smaller wall removed, the larger removed, and both — so twenty-nine positions is a hundred and sixteen runs and about three minutes.

That is cheap for a result that turns a bracket of thirty-five per cent of a rung into a bracket of one grid step. The reason it is cheap is that the underlying design was already written and takes a rise; nothing new had to be built.

The design’s own economy #

That is worth generalising. The expensive part of an experiment is usually the apparatus, and the apparatus here is a function that takes a rise and a seed angle and returns four numbers.

Once such a function exists, locating a transition is a matter of calling it thirty times. The slot table’s own three-per-rung design was chosen when the apparatus was new and the question was whether the interaction had a sign at all.

What the transition does not say #

It does not say the interaction changes sign at that rise. The interaction is the fourth cell less the sum of the two singles, and it goes from +129.4 to −25.3 across the step — but that is arithmetic on the fourth cell’s fall rather than an independent quantity.

Below the transition the interaction is minus the smaller wall’s own cost, by the same cancellation the Lucas 1/3 rung shows. So on the fine two-fifths of this rung the interaction is not evidence about an interaction.

Which shrinks the table a little #

The slot table has twenty-four rows and calls four of them free, and every claim it makes is quantified over the other twenty. This sweep says the four are the visible part of a larger free stretch.

It does not add rows to the table — the twenty-nine positions are one rung — but it does mean that a table sampling a rung at 15, 50 and 85 per cent will land in a free stretch about a third of the time on a rung like this one, and will report that as a property of the position.

That is the same caution a single rise per rung earned for a different quantity, and the same one which offsets wreck earned for a third.

What the sweep did not have to assume #

One property of the design is worth crediting. The lattice at each position is not asserted to have walls of 5 and 8; the walls are counted from the points of the very stem the cuts are made in.

That matters here because a sweep of twenty-nine rises is a sweep across a rung, and a rung is defined by its counted pair holding. Reading the pair off each stem rather than assuming it means the sweep would have reported a change of pair if there were one, and there is not.

And what a rung’s boundary would have looked like #

The rung ends at 0.00689 and the sweep stops at 0.00709, which is 97 per cent of it. That is deliberate: a rise at the very edge of a rung is a rise whose counted pair is in doubt, and a doubtful pair would put doubt into every cell of its two-by-two.

Nothing near the transition is anywhere near that boundary. The transition is at 61 per cent and the nearest edge is a third of a rung away.

What a reader should carry #

That the second wall of a slot stops costing anything at a single rise, located to one part in a thousand, with the lattice unchanged on both sides of it.

And that three samples a rung reported the change as being between 50 and 85 per cent of a rung when it is at 61 — which is not an error, but is a bracket sixteen times wider than the thing inside it.

What the picture at the top shows #

Three curves against the rise, coarse on the left: what the smaller wall’s removal costs, what the larger one’s costs, and what removing both costs.

The first two run gently across the whole stretch. The third sits flat at 163.59 degrees, falls to 9.14 in one step, and then runs along beside the larger single removal for the rest of the rung. The vertical rule is the step it falls at.

The one line #

The second wall of a slot goes free between a rise of 0.00998 and 0.00997 on the golden 5/8 rung — one step of the grid the ladder is named on — with the pair removal’s displacement falling from +163.59 to −9.14 degrees while the walls, the block, the settled divergence and the surviving lags are all unchanged.

So the word free names a transition rather than a threshold on a slope, and the two rows reported at 84 and 85 per cent were the visible end of a stretch beginning at 61.