Where the value stops mattering

The same number in two currencies gave the bent class of translated sums a closed form — a stop falls short of its bound by exactly the value’s own hottest follow-up temperature — and closed by naming the class that still had nothing of the kind:

The rung below found 358 of the 372 straight pairs attaining the bound and 14 falling short … 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.

It is not the follow-up, the formula is shorter than the bent class’s, and it does not mention the value at all.

The fourteen #

The population is the ladder’s own: 120 day-three values against twelve hot addends, 1,440 pairs, with the error on each being how far the sum’s stops sit from the sum of the two positions’ stops. A bound with one number too many established that the error never exceeds twice the smaller of the two temperatures; which end a sum lands at established that a value with straight walls attains that bound exactly when the addend is at least half again as hot as it.

Fourteen do not. They are seven values against two addends — every one of the seven falls short behind both of the two coldest addends and behind none of the other ten — and their temperature ratios are $1$ and $4/3$, both under the threshold of $3/2$.

That much was known. What was not known is where their stops actually land, and it is here.

It is worth restating the quantity, because everything on this page is a number about it and it is easy to lose.

A position’s two stops are what it is worth to Left moving first and to Right moving first, with the fighting played out and nothing else on the board. Add two positions and the sum has two stops of its own, and the natural guess is that they are the sums of the parts’ stops. They are not: a player choosing between two components has an option a player facing one component does not, and the sum’s stops move. The error is how far. It is nought when the stops add, and it is bounded above by twice the smaller of the two temperatures.

So the whole anchor is one question asked four ways: given two positions and their thermographs, how far does adding them move the stops? The answer has been a bound, then a bound with a description of who attains it, then a formula on one class, and now a formula on all three.

The seven values #

The fourteen are seven values against two addends, and the seven are worth looking at before the formula, because they are not a random seven.

They are $\{ {\ast}2 \mid -2\}$, $\{ {\uparrow}{\ast} \mid -2\}$, $\{ {\downarrow} \mid -2\}$, $\{\{0, {\ast} \mid -1\}, {\downarrow} \mid -2\}$, $\{-1/2 \mid -2\}$, $\{-1/2, \{0, {\ast} \mid -1\} \mid -2\}$ and $\{-1/2, \{ {\ast} \mid -1\} \mid -2\}$. Every one has $-2$ as its only Right option, and their temperatures are $1$ and $3/4$ — the two largest in the straight class.

That is the threshold showing through rather than a coincidence. A pair falls short when the addend is not half again as hot as the value, and the addends here top out at temperature $1$, so only the hottest values can fail the test. The seven are exactly the straight values whose temperature is at least two thirds of the hottest addend’s, and there is nothing else about them.

Each falls short behind exactly the same two addends — $\{3 \mid \{2 \mid 0\}\}$ and $\{2 \mid \{1 \mid -1\}\}$, the two whose temperature is $1$ — and attains behind the other ten. So the fourteen are a two-by-seven block of the census rather than a scatter, which is what a threshold produces and a residue does not.

The addend’s temperature, and nothing else #

On all fourteen the stop moves by exactly $t_H$ — the addend’s temperature.

Not by the bound, which is $2\min(t_G, t_H)$ and is $2$ or $3/2$ on these pairs. Not by anything involving the value’s temperature, its stops, its follow-ups, or its bend, of which it has none. By the addend’s temperature, to the last bit of a dyadic rational, on every one of the fourteen.

That is a shorter formula than the bent class’s and a stranger one. The bent shortfall is a property of the value — the value’s own hottest follow-up — and the addend enters only by deciding which of two cases applies. Here the value has vanished from the answer entirely.

Two lines, and the class is finished #

With the fourteen accounted for, the straight class has a rule in two lines:

$$\mathrm{err} = 2\min(t_G, t_H) \quad \text{when } t_H \geq \tfrac32 t_G, \qquad \mathrm{err} = t_H \quad \text{otherwise}$$

exact on all 372 straight pairs.

What that changes is not the count — 358 and 14 were already known — but what the threshold is. The rung below found a condition separating a clean case from a residue, and reported it that way; it is the boundary between two expressions, each exact on its side. A threshold with a formula on both sides of it is a piecewise rule, and a piecewise rule is a closed form.

Where the value stops mattering #

The interesting half of the two-line rule is what the second line says.

Take the addend $\{3 \mid \{2 \mid 0\}\}$, whose temperature is $1$, and four straight values behind it in order of their own temperature: $1/4$, $1/2$, $3/4$ and $1$. The first two are above the threshold and their stops move by $1/2$ and $1$ respectively — twice their own temperature, which rises with the value.

The last two are below the threshold and their stops both move by $1$. A three-quarters-hot value and a one-hot value, behind the same addend, move their stops by the same amount.

So below the threshold the value’s own temperature drops out of the answer. That is not what a bound stated in both temperatures suggests, and it is why the fourteen looked like a residue: they are the cases where the quantity everybody was watching stops appearing.

The mechanism is the ceiling. A translated stop can move at most as far as the addend can push it, and the addend’s own capacity to push is its temperature. Above the threshold the value is the binding constraint and the answer is about the value; below it the addend is, and the answer is about the addend. The bound $2\min(t_G, t_H)$ is a bound in both temperatures precisely because it is trying to describe two regimes with one expression, and it can only do that by being loose in one of them.

Why the ceiling is the addend’s temperature and not twice it #

The second line has a specific number in it and the obvious candidate would have been twice the addend’s temperature rather than once, since the bound itself is twice a temperature. It is once, and the arithmetic of the two regimes explains which.

Above the threshold the error is $2 t_G$ — twice the value’s temperature, since the value is the colder of the two there. That is the value being moved as far as it can be moved: both its stops swing by its own temperature, and the two swings are in opposite directions, so the total displacement is twice it.

Below the threshold the value is the hotter one, and the movement is limited by what the addend can supply. An addend of temperature $t_H$ has a mast at height $t_H$ and its own stops are $t_H$ apart from its mean in each direction — so a single stop can be pushed by at most $t_H$, not by twice it, because there is only one addend to spend and it is being spent on one side.

That is a sentence rather than a proof, and it is the kind of sentence a derivation would begin from. It also predicts what the second line does when the addend is colder still, which is the test the current pool cannot run.

The census closes #

Putting the three classes together, every pair in the sweep now has an expression.

• The value is cold — a number or an infinitesimal, temperature nought. The stops add exactly: error nought, on all 660 such pairs.
• The value is hot with straight walls. The two lines above, on all 372.
• The value is hot with a bent wall. $\min(2t_G - f,\ 2t_H - 2f)$, where $f$ is the value’s own hottest follow-up temperature, on all 408.

Exact on 1,440 of 1,440, with no tolerance beyond the last bit of a dyadic rational. Four rungs of this anchor have been narrowing a bound; the bound is now a computation.

The control matters, because three expressions fitted to three classes can always be made to work if the classes are chosen after the fact. They were not: the classification is the bend, which arrived on a different ladder answering a different question, and the straight class’s rule applied to the bent class is right on 48 of 408 pairs — wrong on nine in ten. The expressions are not flexible; the classification is doing the work.

Three constants, and what they have in common #

This anchor and the switches ladder have now produced three constants between them, and it is worth putting them side by side because two of them are the same number.

A second level of stops found a bent value’s temperature under-read by half its follow-up’s temperature. The same number in two currencies found a bent value’s stop falling short of its bound by the whole of that follow-up’s temperature, and identified the two as one measurement in different units — a temperature is half a stop gap, so a half in one currency is a one in the other.

This page’s constant is not in that family at all. The second line is $\mathrm{err} = t_H$: a coefficient of one on a quantity belonging to the other position. There is no follow-up in it and no factor of two, and the value it is about does not appear.

That is a genuine difference and it is worth naming, because the temptation on a ladder that has found one constant twice is to expect it everywhere. The bent constant is about a value under-reading itself; this one is about a ceiling on how much one position can move another. Different mechanisms, and the arithmetic says so.

What the tables cannot show #

Every figure here is a table of temperatures and errors, and the thing they cannot show is the geometry that makes the second line true.

What a picture would have to draw is two thermographs and their sum’s: three pairs of walls, with the sum’s stops marked at the ground and the two parts’ stops marked beside them, and the gap between where the stops land and where they would land if they added. Below the threshold that gap is capped by the addend’s mast, and seeing it capped is the argument this page has as prose and not as a figure.

That drawing is possible — this site draws thermographs constantly — and it is a drawing of one pair. The finding is that fourteen pairs behave one way and 358 the other, and a picture of one is a picture of one. Reading a thermograph is where the walls get drawn and read directly, and it is the right page to look at the geometry this one is counting.

What this does not settle #

It is a census and not a theorem. Three expressions, exact on 1,440 pairs, with no derivation of any of them from the definition of a sum. The same number in two currencies says the same of its own half, and the position is unchanged: this is a complete description of a population rather than a proof about translations.

The population is one pool of addends. Twelve hot addends with follow-ups, chosen on an earlier rung to span two families, against the first 120 day-three values. The threshold at $3/2$ and the second line’s $t_H$ are both statements about that pool; the twelve addends have eight distinct temperatures between them, and the fourteen occur behind the two coldest.

Fourteen is a small sample for a formula. Seven values and two addends, and the formula fits all fourteen exactly — but a formula with one term that happens to equal one of the numbers in front of it is the kind of fit that a wider pool refutes. What would test it is a colder addend still: the second line predicts the stop moves by that addend’s temperature, whatever it is, and the pool contains nothing colder than $1$.

And the classification is not free. Deciding whether a value’s wall bends means computing its thermograph, which is the object the whole ladder is trying to avoid. So the closed form is a description rather than a shortcut, and this page says so in the same terms the two rungs below it did.

The threshold is measured, not derived. Three halves is where the two lines meet on this pool, and nothing here says the ratio is exactly three halves rather than something near it — the pool’s ratios jump from $4/3$ to $2$ with nothing between, so any threshold in that interval fits the data equally well. The rung below reported $1{\cdot}5$ and this page inherits it, and both should be read as somewhere between four thirds and two.

Normal play, short games, and stops read from the ordinary thermograph.

What a closed census is worth #

Four rungs of narrowing produce a table of three expressions, and it is fair to ask what a reader gains from that over the bound they started with.

The bound answers how bad can it be. The closed form answers what will it be, which is a different question and the one a player has. Two components, two thermographs, one classification and one expression: the stops of the sum, without building the sum. On this pool that is a saving of the whole canonicalisation — the sum of two day-three values is a day-four value and the recursion that reduces it is the expensive part of everything on this site.

What it is not is a general method, and the difference is the classification. Deciding which of the three classes a value is in means asking whether its walls bend, which means computing its thermograph, which is affordable and is not free. So the closed form replaces a large computation with a small one rather than with a look, and the honest description is that it makes the sum cheap once the parts are understood.

Where the ladder goes next #

The translation anchor has seven 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, how far inside the bound the bent ones land, and now the last class without a formula.

The rung above is the derivation. Every one of the three expressions is a measured identity over a census, and two of them are short enough to be provable: the stops add when the value is cold is very nearly a definition, and the stop moves by the addend’s temperature when the value is too hot for the bound is a statement about a ceiling that an argument about walls ought to reach. The bent case’s $\min(2t_G - f,\ 2t_H - 2f)$ is the hard one and would come last. A ladder whose whole census is closed and none of it proved is in exactly the position the cutcake ladder was in before its proof arrived, and the way out was the same: write the statement so that the variable it does not depend on is visibly absent.

Two neighbours are worth the trip. A bound with one number too many is where the bound was confirmed and where a third temperature was found not to sharpen it, and it is the page whose looseness this one explains. And the bend is the condition is where the classification comes from, on a different ladder entirely, and it is worth reading beside a census it partitions exactly.