A fifth of the hop

One step of the control’s own divergence is the best of four candidate units for the size of the exchanged pair in a wrecked stem, and it is wrong by up to 11.8 per cent.

The miss is not scatter. Every row whose stem kept a lag of 5 or 7 overshoots one step; every row whose stem kept a 4 or an 8 falls short. No lag appears on both sides, on seventeen rows across two branches and eight lattices.

That was already in the record as an unexplained pattern — the essay that measured the pair reports it and says nothing about it. This essay is the attempt to say something, and most of the work is deciding how much the attempt is worth.

The quantity the residual sorts by #

Fold the surviving lag into the divergence: the angle from an organ to the one k places above it is k times the divergence, taken round the circle and folded into half a turn either way. That is the surviving hop’s own angle, and it is what a counter would read if it counted that family.

It is negative for lags of 5 and 7 and positive for lags of 4 and 8, at about −34°, −24°, +39° and +22°. The residual is positive where the hop’s angle is negative and negative where it is positive, on all seventeen rows.

The correction #

Take one step of the control’s divergence and subtract a fifth of the surviving hop’s angle. The worst row goes from 11.8 per cent to 4.0, the average from 4.3 to 1.0, and every one of the seventeen rows falls inside five per cent.

That is a factor of three on the worst row and a factor of four on the average, bought with one coefficient.

A fifth is stated, not fitted #

The least-squares coefficient over the seventeen rows is 0.188. A sixth, a fifth and a quarter all take the worst row under six per cent, and the fifth is the roundest number in that range.

Quoting 0.188 would be quoting three digits of a coefficient fitted to a handful of points, which is a precision the data does not carry. Quoting a fifth says the same thing and does not pretend.

It is the same discipline the tolerances in this thread are chosen by: a threshold that sits in a gap is stated once and never tuned, because a number that could have been anything in a range and was chosen at one end is a number carrying a decision nobody declared.

The coefficient is not a fifth, it is one over the lag #

There is a version of this correction with no coefficient in it at all, and it follows from the one fact the whole thread rests on.

A wrecked stem holds one family rigid. A family of period k sits at multiples of a turn over k around the circle, so any displacement that leaves it where it was has to be a whole number of those steps. The exchange is therefore not “about one divergence”; it is the multiple of 360°/k nearest to one divergence, which is 360·j/k with j the nearest whole number to k·d/360.

Write that out and the correction falls out of it. Since k·d − 360·j is exactly the surviving hop’s own folded angle, 360·j/k is d minus that angle divided by k. So the rule is one step of the control’s divergence less one-kth of the hop’s angle — the same shape as the fitted rule, with the coefficient supplied by the arrangement instead of by least squares.

That predicts a different coefficient on every row, and the four values it predicts are 1/4, 1/5, 1/7 and 1/8 — which average to 0.18, against the 0.188 the fit returns. The fitted number is the mean of the derived ones, which is what a fit does when the truth is a per-row quantity and the fit is not allowed one.

Scoring the derived rule against the fitted one #

Taken to the hop angles quoted above — about −34°, −24°, +39° and +22° at lags 5, 7, 4 and 8 — the two rules predict corrections of

At a lag of 5 the two rules are the same rule, because a fifth is one over the lag there — which is why the fitted coefficient came out near a fifth in the first place, since lag 5 supplies six of the seventeen rows.

Everywhere else they separate. The derived rule is closer at lags 7 and 4, the fitted one is marginally closer at 8, and across the four the derived rule’s mean error is about 0.8° against 1.4°. That is a factor of not quite two, which on four clusters is not decisive on its own.

What is decisive is the free parameter. The fitted rule has one and the derived rule has none, and the derived rule ties or beats it anyway. A rule with nothing to adjust that matches a rule fitted to the same data is not a rival account; it is the account, with the fit as its four-cluster average.

It also makes a prediction the fit cannot. A stem whose surviving lag is 11 should correct by one eleventh of its hop angle, not a fifth — a difference of a factor of two, on a row this census does not contain and a longer one would.

The sign is the result, not the size #

A coefficient fitted to seventeen rows is worth about seventeen rows. What is not a fit is the direction: the surviving lags in this census give hops on both sides of zero, and the residual follows every one of them.

If the correction were an artefact of fitting, it would have no reason to change sign where the predictor does. Four lags, two signs, seventeen rows, and no exceptions.

The honest denominator is four #

This is the part the essay exists to say. The hop’s angle is k times the divergence, and the divergence barely moves across the census — 99° to 138°. So the seventeen rows carry eleven distinct hop angles which fall into four tight clusters, one per surviving lag.

The correction is therefore a line fitted to four positions on its own axis, scored on seventeen rows. A reader told “seventeen rows, no exceptions” and not told “four clusters” has been told something misleading by a true sentence.

And inside a cluster it does not work #

Within the rows sharing a lag of 5 the hop angle runs from −37.3° to −30.2° and the residual runs from 3.9° to 10.4°. If the relationship held inside a lag, the largest residual would sit at the most negative hop. It does not: the largest residual sits at −36.1° and the smallest at −31.2°, and the ordering is scrambled.

So the correction accounts for the differences between surviving lags and nothing at all for the differences within one. The coefficient implied row by row runs from 0.121 to 0.301, a factor of two and a half, and the file records that rather than averaging it away.

Why the hop’s angle and not its length #

The surviving family is identified by a length: it is the lag whose step across the surface the cut stem holds unchanged, and the census reports that length for every row. The length is not the quantity the residual follows.

The angle is. A lag’s step has a height, prescribed by the rise, and an azimuth, which is that many divergences round the circle. Two lattices can have the same step length with very different azimuths, because the height makes up the difference — and it is the azimuths this residual sorts by.

That is worth stating because a reader who has followed the survivor thread has been reading lengths for several essays. The correction changes the unit halfway through, and the change is the reason it works.

What a fifth might be #

An honest guess, offered as one. The two exchanged chains are one place apart in the sequence and therefore one divergence step apart in azimuth, and they are also some number of hops apart in the family the stem has kept rigid. If the exchange relaxes slightly towards the geometry of that family, the relaxation would carry a share of the family’s own angle.

A share of a fifth would then be a property of how stiff the surviving family is against the rest of the arrangement, which is a quantity nobody here has defined, let alone measured. So the guess names a thing that would have to exist rather than explaining anything.

What would test it #

A wrecked cut whose surviving lag is not 4, 5, 7 or 8. The correction predicts a residual of about a fifth of that lag’s hop angle, with the opposite sign, and it predicts it before the row is measured.

Lags outside those four do occur: the census’s own length ranking places a survivor as deep as thirty-seventh, and lags of 11 and 13 are perfectly ordinary steps on a fine lattice. No wrecked cut in this census keeps one, so finding a lattice that does is the test — a search rather than a sweep, and cheap.

The prediction is specific #

At a lag of 11 on a golden lattice the hop angle is about +32°, so the correction predicts an exchange about 6° short of one divergence step. At a lag of 13 it is about −20°, so about 4° long.

Two numbers, either of which could be wrong by more than their own size. That is what makes it a test rather than an extension — and it is the same shape as the test the slot interaction’s rule needs and cannot get from inside the ladder: a value of the predictor that the existing design does not reach.

The worst row is the row a longer run doubts #

The correction’s largest error is 4.0 per cent, on the Lucas 0.013 stem’s cut four organs back. The next worst is 2.3, and the mean is 1.0, so that row is nearly twice as far out as anything else.

It is also the only row in the census whose surviving lag is 4, so it is a cluster of one and the correction has no other row to be checked against there. Both of those were known, and a cluster of one is the shape a rule with a single exception has when the exception is also the only test of one of its terms.

And a separate measurement says the same row is unreliable #

Running every cut twice as far asks whether a profile’s levels are steady over six hundred organs rather than three hundred. Twenty-six of thirty rows are; one that was periodic over the shorter run is not over the longer, at a spread of 171.2° against 6.1°.

That row is this one. So the exchange it reports is a reading taken over a window still inside a transient, and the row that fits this essay’s rule worst is the row a different measurement, made for a different reason, says should not have been read.

Which is a check, not a rescue #

Dropping the row would take the worst error from 4.0 per cent to 2.3 and the cluster count from four to three. That is a better-looking table and a weaker one, and the essay does not drop it.

What the coincidence is worth is different: two readings of one census, taken for unrelated reasons, agreeing about which row is least trustworthy. Neither knew about the other, and that is the kind of agreement worth more than either measurement.

The alternative nobody can rule out #

Four clusters and one coefficient is exactly the amount of data that cannot distinguish a linear relationship from any other monotone one. A rule of the form minus a fifth of the hop and a rule of the form minus a constant per lag agree on every row here, because the hop is nearly constant inside a lag.

The second is a much weaker claim — four numbers fitted to four groups, which is no compression at all — and nothing in this census separates it from the first. What separates them is a fifth cluster, which is the test above and the reason it is the test.

What is not claimed #

That the exchange is exactly a step less a fifth of a hop. It is a rule that fits seventeen rows to within four per cent, rests on four clusters, and has no account behind it.

Nor that the residual is understood. Something makes the coefficient run from 0.121 to 0.301 inside a single lag, and nothing here says what. The rule accounts for the between-lag structure and leaves the within-lag structure exactly where it was.

Why a residual is worth an essay #

Because a twelve per cent shrug is where a measurement stops being tested. “About one divergence step” was true, useful, and unable to be wrong; the twelve per cent had nowhere to go.

Naming a structure in it — even a structure with four points behind it and no mechanism — turns the shrug into a prediction about lags nobody has measured. That is the trade this collection makes over and over, and it is worth stating that the prediction is the point rather than the fit.

What carries forward #

A rule, its denominator, its worst row and its test. The test is a search for a wrecked cut with a surviving lag outside 4, 5, 7 and 8, which would put a fifth cluster on the axis and either confirm the sign or break it.

And a piece of arithmetic worth repeating: the within-lag scatter is a factor of two and a half and is untouched. Any account of the exchange that explained that as well would be a considerably better account than this one, and the place to look for one is the arrangement’s own structure rather than in the cut.

A second reading of the same rows #

The exchange has two readings and they behave completely differently. Its direction is a discrete fact and it is the same on every row, with nothing left over. Its size is continuous and carries a residual with a structure and a scatter inside the structure.

That is not a coincidence about this measurement; it is what discrete and continuous readings usually do. A sign has no error bar and cannot be nearly right. An angle can be nearly right in several ways at once, and the work is separating them.

What this does to the census’s thirteen other rows #

Nothing. They carry three or more exceptional chains and have no balanced pair, so they have no size to correct and are refused rather than scored.

Nearly half the census is therefore outside every number in this essay, and an account of the exchange that also said what those rows are doing would be a different and much better result. Nobody has started on it, and the first question — whether three exceptions are a pair plus a stray or a three-cycle — is arithmetic on numbers already computed.

The one line #

The exchanged pair’s size is one step of the control’s divergence less a fifth of the surviving hop’s own angle — worst row four per cent, mean one, and the sign right on every one of the seventeen rows.

The seventeen rows carry four distinct hop angles, so the coefficient is fitted to four clusters; inside a cluster the rule says nothing, and the row it fits worst is the row a separate measurement independently calls unreliable.