The same number in two currencies

Which end a sum lands at sorted 1,440 translated sums into three classes and found the bent-walled ones falling strictly inside the bound, every time, with nothing said about how far. It closed on that, and named a candidate:

The rung above is the bent case, made quantitative … the switches ladder already has the number that ought to govern it: a bent wall is under-read by exactly half its follow-up’s temperature, so the natural conjecture is that the stop moves by the bound less something proportional to the same quantity. Both numbers are in hand for all 408 pairs, and fitting one against the other is an afternoon.

The conjecture is right, the constant is one rather than a half, and the two constants are the same number.

The shortfall is the follow-up’s temperature #

Write $f$ for the temperature of the value’s hottest follow-up — its hottest option that is not a number — and the bound for a stop’s movement is $2\min(t_G, t_H)$, twice the smaller of the two temperatures.

On 400 of the 408 bent pairs the stop falls short of that bound by exactly $f$. Not approximately: to the last bit of a dyadic rational, on four hundred pairs, with no residue to describe.

The rung below’s conjecture asked for something proportional to $f$ and the constant is one. The switches ladder’s constant is a half, and taking it literally into this measurement gives the shortfall on none of the 408 — which is the control this page needs, because a conjecture whose constant is anything at all fits somewhere.

It is worth saying that the rung below could not have found this. It had the class and the fact that the class never attains; what it did not have was the follow-up temperature attached to each value, which is a quantity the switches ladder computes and this anchor had never needed. Half a follow-up out is where that quantity was first isolated, and importing it is the whole of this page’s method.

The eight exceptions are one condition. Every one of them has the addend colder than the value, so the bound $2\min(t_G, t_H)$ is set by the addend rather than by the value — and there the shortfall is $2f$ instead of $f$.

One expression #

The two cases collapse. The error — how far the sum’s stop moves from the sum of the two values’ stops — is

$$\min\left(2t_G - f,\ \ 2t_H - 2f\right)$$

on all 408 bent pairs, exactly. The first term is the value’s own bound less one follow-up; the second is the addend’s bound less two. Whichever is smaller is what happens.

That is a closed form for a quantity that was a residue two rungs ago. A bound with one number too many established that $2\min(t_G, t_H)$ bounds the movement and that adding the addend’s follow-up as a third number does not sharpen it; what sharpens it is the value’s follow-up, and it sharpens it to an equality on the class where the bound was never attained.

It is worth being explicit that this is a statement about bends. The expression is exact on 408 of 408 bent pairs, on 276 of 660 cold ones and on 238 of 372 straight ones — so it is not a formula that happens to fit everything, and the class the rung below identified is the class it belongs to.

The shortfall belongs to the value #

Thirty of the 34 distinct bent values in the census produce the same shortfall against every one of the twelve addends. The addend does not enter it at all.

The four that do not are the four hottest values in the census, at temperature $\tfrac54$, and they are the only ones some addend in the pool is colder than. So the addend’s whole contribution is to decide which of the two cases applies, and the size of the shortfall is a fact about the value being translated.

That extends the rung below’s finding rather than complicating it. That page found which end a sum lands at to be almost entirely a property of the value; this finds where between the ends to be entirely one.

Why the constants differ by two #

The two ladders’ constants disagree and neither is wrong, and the reconciliation is a unit conversion.

A second level of stops measures a bent value’s temperature: the stop reading says half the gap between the two stops, the true temperature is larger, and the difference is $f/2$.

This page measures the same value’s stop: the translation bound says the stop moves by $2\min(t_G,t_H)$, it moves less, and the difference is $f$.

A temperature is half a stop gap. So an error of $f/2$ in a temperature is an error of $f$ in the gap the temperature is half of, and the two ladders have found one quantity in the two currencies each of them happens to work in. Checked on the same 34 values, the ratio is exactly two on every one.

That is worth a paragraph beyond the arithmetic, because it is the kind of coincidence that is usually not one. Two ladders — one about a single value’s thermograph, one about what happens when two values are added — arrived at the same correction, and the correction is the hottest follow-up’s temperature. What the follow-up’s temperature is measuring in both cases is the part of the position the stops cannot see: a bent wall means an option takes over the wall part way up, and a stop is a reading at the bottom of the wall. Anything the wall does above the bottom is invisible to a stop, and the size of what it does is $f$.

There is a third place the same number could be looked for and has not been. A bent wall is a fact about a thermograph, and a thermograph is a fact about cooling; the natural third currency is the cooled game — how far a value has to be cooled before its bend disappears, which is $f$ again by construction. Three ladders would then be quoting one quantity, and the third has not been measured because nothing has asked it to.

What the anchor now has #

Six rungs in, the translation anchor has a complete account of one operation, and it is worth assembling because the pieces were written separately.

Add a number to a fight and everything is predictable: what a number does to a fight — both stops move by the number, the temperature does not move, and the position is a first-player win exactly while the number lies strictly between the negated stops.

Add an infinitesimal and neither stop moves and the outcome changes anyway, which is what an infinitesimal does to a fight and is the sharpest statement of what a pair of stops leaves out.

Add a fight and the stops stop adding, by an amount bounded by twice the smaller temperature — what a fight does to a fight and the bound two rungs below. The bound is attained when both walls are straight and the addend is hot enough; it is missed by nought when the value is cold; and it is missed by the value’s follow-up temperature when a wall bends.

So the three cases of what can be added now have three answers, and the third has three sub-cases with a formula for each. That is as complete as this site’s anchors usually get, and the thing that made it complete was a quantity — the bend — imported from a different ladder entirely.

It is worth noticing which direction the import went. The bend was established on the switches ladder as a condition for the stop reading of a single value to be exact, and it turns up here deciding how a sum behaves. Nothing about the first use suggested the second. A property that decides whether a reading of one position is exact turns out to decide how that position combines with others, which is a stronger statement than either ladder set out to make.

What a proof would look like #

The expression is short enough to be worth saying what an argument for it would need, because for once the shape is visible.

$2t_G$ is the whole width of a value’s confusion interval, which is what a stop could move by if the addend were hot enough to expose all of it. A bend means the wall is governed by a different option above the height $f$ — so the top $f$ of the wall belongs to the follow-up rather than to the value, and a translation that exposes the wall exposes only the part below the bend. Subtracting $f$ once is subtracting the part of the wall the value does not own.

The second term is the same argument from the addend’s side, and the factor of two in $2f$ is where the argument is least clear. When the addend is the colder, the bound is the addend’s, and the value’s bend costs twice — which reads as the follow-up is charged on both walls and is the piece that would have to be checked rather than asserted.

Eight pairs is a thin base for the second half. The first half rests on four hundred.

What would settle it is a pool with more addends colder than the values in it. The census’s twelve addends were chosen for a different question — a bound with one number too many wanted addends whose own follow-ups varied — and their temperatures run from one to three and a half, against values that top out at five quarters. So the second case is rare in this census by construction rather than by nature, and a pool built the other way round would populate it.

Reading the formula as advice #

A closed form on this site is usually worth restating as something a player could act on, and this one is, with one qualification.

The expression says: when two hot positions are added, the sum’s stops fall short of the sum of the stops by nearly the whole of twice the smaller temperature — and the only thing that claws any of it back is a bend in the value’s wall, worth exactly its follow-up’s temperature.

For a player that reads as a warning about a particular kind of position. A fight whose best answer is itself a fight — the shape a second level of stops is entirely about — behaves better in a sum than its stops suggest. Its stops are a pessimistic reading of it precisely because the follow-up is invisible to them, and the pessimism is worth $f$.

The qualification is that a stop is not a score. Everything here is about how the two players’ first-move results combine, and the game is decided by the whole sum played out rather than by a stop. What a fight does to a fight is where that gap is drawn, and it is the reason this anchor keeps producing bounds rather than values.

Why a conversion factor is evidence #

Finding the same constant at twice the size on two ladders is the sort of thing that could be a coincidence, and it is worth saying why it is not — because the check is arithmetic rather than statistical.

A temperature is half the gap between a position’s two stops. That is a definition, not a measurement, so any quantity expressible in stops has a value in temperatures exactly half as large, and any law stated in one currency has a twin in the other with a factor of two between them.

So a half there and a whole here is not two constants that happen to be related; it is one constant, reported in two units. The check that they really are the same is to convert one and see whether it lands on the other, which it does — and the check would have failed if the two laws were about different quantities that happened to have similar numbers.

That is the difference between an analogy and an identification. An analogy notices that two results look alike; an identification supplies the conversion and verifies it, at which point the two results are one result and any evidence for either is evidence for both.

It also says what to do when a constant looks familiar. Convert before comparing. Two ladders on this site measure things in stops and two measure them in temperatures, and a constant carried between them without the factor of two is a constant that will be wrong by a factor of two — which is exactly the size of error a reader is least likely to notice, because it looks like a different but plausible law.

What this does not say #

The census is 120 values and twelve addends. Those are the rung below’s population unchanged, which is what makes the two pages comparable, and it is 34 distinct bent values rather than a survey of bent values in general. Every one of them has a bent left wall and a straight right one, which is a fact about the day-three values the census draws from and not something the expression assumes.

The eight exceptions are four values. The doubling is observed on eight pairs made from four values and two addends, so the addend being colder doubles the shortfall is a rule fitted to four values. It is stated as the second case of the expression because the expression is exact, not because the case is well populated.

Nothing here is proved. The expression is an identity checked on 408 pairs. The argument in the section above is a reading of why it might hold and is explicitly not an induction; the site’s habit is to say which is which, and this is the first.

And the picture cannot show a shortfall. A stop is a number and a bound is a number and the difference between them is a third number; there is no drawing of that, and the figures are tables. What could be drawn is the thermograph whose bend causes it, and the bend is the condition draws those.

The convention, named #

Normal play throughout, and every value computed by the recursion.

A position’s stops are what each player gets moving first and playing on until somebody faces a number, with no tax charged. Its temperature is the height at which its thermograph’s two walls meet, and for a straight-walled position it is half the gap between the stops.

A wall bends when its slope is not one throughout, below the meeting point — which happens when the option holding it up stops holding it up, either because another option takes over or because that option has cooled out.

The follow-up temperature $f$ of a value is the largest temperature among its options that are not numbers. A value with only numbers as options has $f = 0$ and a straight wall.

The error of a pair is how far the sum’s stop is from the sum of the two stops, taken as the larger of the two sides’ discrepancies. The bound is $2\min(t_G, t_H)$, established two rungs below, and the shortfall is the bound less the error.

The census is every pair of a day-three value from the first 120 and an addend from a list of twelve — 1,440 pairs, of which 408 have a bent-walled value.

Where the ladder goes next #

The translation anchor has six rungs: what a number does to a fight, what an infinitesimal does, what a fight does, the bound with one number too many, which end a sum lands at, and now how far inside the bound the bent ones land.

The rung above is the straight class’s fourteen. The rung below found 358 of the 372 straight pairs attaining the bound and 14 falling short, with the boundary at an addend at least half again as hot as the value — and this page has now given the bent class a closed form while those fourteen still have only a threshold. The obvious question is whether they have a shortfall with a formula too, and the obvious candidate is the same one: a straight-walled value has $f = 0$ by definition, so whatever the fourteen are missing the bound by is not the follow-up, and finding what it is would complete the census.

Two neighbours are worth the trip. A second level of stops is where the other half of this quantity lives, and reading the two together is the clearest example this site has of one number wearing two constants. And the bend is the condition is where the bend became a condition rather than a symptom, and it is what makes the bent class a class worth giving a formula to.