Colder exactly when the residues cancel

A hot position splits into three pieces. A number and a fight charged each position exactly its own temperature — cooled it by that amount — and found what was left to be its mean plus a residue, an infinitesimal that no number reports: a star on 942 of the 1,122 hot values born by day three, and never nothing. The hotter residue survives then added hot positions two at a time. When the two differ in temperature the answer is clean — the sum keeps the hotter part’s residue and nothing of the colder’s — because cooling is additive and a part cooled past its own temperature is a bare number. When the two are equally hot, three things happened: on 880 of 3,142 pairs the residues added and the sum stayed as hot; on 2,237 the sum grew colder; on 25 it cancelled to a number.

That essay closed on the case it could not explain. When two fights of one temperature cancel into something colder, what decides whether they do, and what is left?

The first question has a complete answer, and it is the residues.

The residues decide the direction #

Take two positions G and H at the same temperature t. Cooling is additive — the sum cooled by any tax is the parts cooled by that tax, added — so the sum cooled by t is G cooled by t plus H cooled by t. Each of those is its part’s mean plus its part’s residue. The sum cooled by t is therefore the two means added, plus the two residues added.

Now there are two cases. If the residues cancel — if one is the negative of the other — the sum cooled by t is a plain number. A position that becomes a number when cooled by t was already frozen below t, so the sum is colder than its parts, or is a number outright. If the residues do not cancel, the sum cooled by t is a number plus a non-zero infinitesimal, which is exactly what a position looks like when cooled by precisely its own temperature; the sum is as hot as its parts, and its residue is the two residues added.

So the direction of an equal-temperature sum is decided by an infinitesimal, and the figure checks it both ways. The step the argument takes on trust is the definition of temperature itself: a position cooled by exactly its own temperature is its mean plus something non-zero, and only above its temperature does it become a bare number. That is how cooling is defined — a position freezes at the first tax where it is infinitesimally close to a number, and the tax at which it freezes is its temperature — but a definition applied at a boundary is where a mistake would hide, so the counts are worth having. On the 3,142 pairs from days two and three, the 880 sums that stay as hot all have residues that do not cancel, and the 2,262 that grow colder or become numbers all have residues that do. On 3,000 equally hot pairs sampled from day three the split is 439 against 2,561, with the same clean division. The argument covers every pair of equally hot positions; the counts are there because the argument has one step — “cooled to exactly a number means colder” — that deserves to be tested against the recursion rather than trusted.

The reason the cancelling case is so common is that the residues are so few. Most hot positions born by day three have residue ∗, and ∗ is its own negative, so any two of them cancel. That is why nearly three quarters of the equal-temperature pairs grow colder: not because fights usually cancel, but because most fights end in the same infinitesimal, and that infinitesimal annihilates itself.

Doubling a fight #

The sharpest consequence is about copies. Two copies of one position are always equally hot, so the criterion applies to every one of them, and it reduces to a question about a single symbol.

A fight doubled grows colder exactly when its residue is its own negative. On day three that means ∗ and ∗2: 958 values, and every one of them cools when doubled. The other 164 — residues ↑, ↓, ↑∗ and ↓∗ — stay exactly as hot when doubled, and the doubled fight’s residue is the part’s residue added to itself.

That is a statement a player can use. Two identical fights on a board — two corners of a Go position with the same shape, two identical regions of Domineering — are worth playing out as a pair or leaving as a pair depending on something no count of points shows: whether the fight’s last infinitesimal is a star. If it is, the two fights together are colder than either, so neither is urgent; if it is an up, the pair is as hot as each, and the right to move first in them keeps its full value. The temperature of the part says nothing about which.

The residues as a sequence saw the simplest case of this at the first step of its piles, without needing a cause: copies of the plain switch $\{2 \mid 0\}$ alternate between a fight and nothing, and two of them are the number 2. The plain switch’s residue is ∗, and the pair cools because the star cancels itself.

Most cancellations go all the way down #

When the sum does grow colder, how much colder?

On the pool from days two and three the answer is: all the way. Every one of the 2,237 colder sums has temperature nought — both of its stops equal, a number plus an infinitesimal — and the other 25 are numbers outright. The two fights cancel entirely and something infinitesimal is left over. In the day-three sample the same is true for most pairs, but at parts of temperature three quarters and five quarters, 394 and 349 sums stop at one half: a smaller fight inside the two larger ones survives their cancellation.

Those intermediate cases are what the survivor question is really about, and the rest of this essay asks what the survivor is made of.

What a player can read before adding #

The criterion is short enough to use at a board. Two fights at one temperature, each summarised by its mean, its temperature and its residue, can be sorted before anything is added. If the residues do not cancel, the pair is exactly as hot as each fight and the pair’s residue is their sum: an up beside a star gives up-star, and the right to move first in the pair is worth the full temperature. If they do cancel — two stars, an up against a down — the pair is colder, possibly much colder, and neither fight is as urgent as its own temperature says, because each is the other’s answer.

That second case is the one ordinary play gets wrong. A player who reads each fight’s temperature sees two equally urgent fights and plays in one; the criterion says the two are, together, a smaller fight, and the tempo might be better spent elsewhere on the board. Temperatures do not add measured how much colder sums can be than their parts; the criterion names the pairs for which that happens, and names them by their residues alone.

One thermograph, four survivors #

The natural hope is that the survivor can be read off the diagrams. A thermograph records a position’s two stops at every tax, and every quantity the hotter residue survives needed for the unequal case — which part is hotter, its mean, its residue — can be read from one. If the survivor of an equal-temperature cancellation were a function of the two parts’ thermographs, a player could carry the diagrams instead of the games.

It cannot be. $\{1 \mid 0\}$ and $\{1 \mid 0, \ast\}$ differ only in an extra Right option, $\ast$, whose stops are both nought — the stop the option $0$ already supplies — so the extra option changes no wall and the two thermographs coincide. The same is true of $\{2 \mid 1\}$ and $\{2 \mid 1, 1\ast\}$. All four parts are switches of temperature one half with mean a half or three halves and residue ∗, so by thermograph and by residue they are two parts drawn twice.

Their four sums are not two sums. $\{1 \mid 0\} + \{2 \mid 1\}$ is the number 2: the fights cancel with nothing left. Put the extra option in either part and the sum is 2 plus the infinitesimal $\{\{1 \mid 0\} \mid \{\ast \mid -1\ast\}\}$. Put it in both and the survivor is different again, $\{\{1 \mid 0, \ast\} \mid 0\}$. The thermographs of the parts are identical across all four rows and the sums do three different things.

The survivor is not in the thermographs #

One example makes the point possible; the census makes it the rule.

Group the 2,262 colder pairs from days two and three by their two thermographs. There are 124 groups, 106 of which hold more than one pair, and in every one of those 106 the survivor differs between the pairs in the group. The sum’s own thermograph differs in 83. Only the sum’s temperature is stable: it varies in fifteen groups, and in every one of them the variation is between a number and temperature nought — between nothing left over and an infinitesimal left over, which is the survivor’s content again rather than its temperature.

The day-three sample says the same thing at lower density: 311 groups with more than one pair, the survivor varying in 192 and the temperature in 13. And the figure’s first footer rules out the obvious rescue. On both pools a part’s residue turns out to be fixed by its thermograph — two parts with the same diagram have the same residue — so adding the residues to the thermographs adds no information, and the survivor is still undetermined.

The comparison with when two thermographs can be added is exact. That essay found the diagram of a sum computable from the diagrams of its parts only when at most one part is hot. The criterion here is the other side of that coin: two equally hot parts are the case where the diagrams stop being enough, and the residues — which the diagrams do determine — decide only the direction of the sum, not its content.

Built below the parts’ temperature #

What does determine the survivor is also what the criterion’s proof uses, pointed one step lower.

$\{2 \mid \tfrac12\}$ and $\{1\ast \mid \{\ast \mid -1\}\}$ are both hot to three quarters, both with residue ∗, so their sum is colder: it has temperature one half. Cooling is additive at every tax, so the sum cooled by one half is the two parts cooled by one half, added. Cooled by one half, the first part is $\{\tfrac32 \mid 1\}$ and the second $\{\tfrac12 \mid \ast\}$ — both still games, because one half is below their temperature. Their sum is the sum cooled by one half, and the survivor is read from it.

That is the general answer, and it explains the census. The survivor of an equal-temperature cancellation is the parts cooled by the sum’s temperature, added — and the sum’s temperature is below the parts’ own, so the parts enter the survivor as games, with every option they have at that tax. A thermograph keeps, of each part at each tax, only its two stops. When the sum falls all the way to nought, as it does on nearly the whole pool from days two and three, the parts enter uncooled, and every option either part has — including an extra ∗ that no wall can see — can reach the survivor.

The diagrams fix the temperature and nothing else #

The survivor’s content is out of the diagrams’ reach. Its temperature is not.

A bend is a height below a position’s temperature at which one of its thermograph’s walls changes slope — where, as the tax rises, a different option starts to govern a stop. The bend is in the stops found that the bends are where a diagram stops being predictable from the stops alone; here they are where a cancellation can stop.

The count is exact on both pools. Of the 2,500 colder sums in the day-three sample, 878 are of parts whose diagrams have no bend at all, and every one of those falls to temperature nought. The other 1,622 have a bend somewhere in one part or the other; 879 of them fall to nought anyway, and the remaining 743 stop at a temperature that is, every time, the lowest bend of either part. Never above it, never between bends. On the pool from days two and three only 72 colder pairs have a bend at all, and all of them fall to nought, which is why that pool’s survivors were all at nought.

And whether a bent pair stops at its bend or falls through it is decided by the diagrams: across the 502 groups of the sample, no two pairs with the same two diagrams disagree. So the thermographs carry exactly three facts about an equal-temperature sum. They carry the direction, through the residues they fix on these pools. They carry the survivor’s temperature — nought or the lowest bend. And they carry nothing of what the survivor is.

That division has a plausible reading, offered as a reading and not a proof. A thermograph is built from stops, and the lowest bend is the lowest height at which a stop changes which option governs it. Below that height each part’s walls are straight lines running from its two stops, which is the shape of a plain switch, and two switches at one temperature with cancelling residues cancel completely — the first worked figure’s $\{1 \mid 0\} + \{2 \mid 1\}$ is the model. So a survivor can be hot only if something in one of the parts is still fighting below its temperature, and the diagram records the height at which that something starts. What the diagram cannot record is what the something is.

The convention under the counts #

Residues, temperatures and means are those of a number and a fight: cooling by the tax t charges each move t, a position’s temperature is the least tax at which it cools to a number plus an infinitesimal, and its residue is the position cooled by its own temperature less its mean. A number has no temperature in this sense and no residue; “colder” includes becoming a number. Two games are counted equal when their difference is a second-player win, and a residue “cancels” another when their sum is equal to nought in that sense. Thermographs are compared wall by wall, to a precision of 1/1024 at every breakpoint, which is finer than any value in these pools.

What the pools cannot show #

The criterion is proved; the pools only illustrate it. Its one step beyond additivity — a position cooled by exactly its temperature is never a plain number — is what the counts check, and they check it on 6,142 sums.

The census is of two pools, not of all positions. On them a part’s residue is fixed by its thermograph, which is why the residues add nothing to the diagrams; that need not hold in general, and a pool where it fails would give the residues a chance to carry information the diagrams lack. The survivor statistics are about groups within these pools. And the sample is seeded and small: 3,000 of the 186,162 equally hot day-three pairs.

Still open: how far apart the survivors of one diagram can be #

The diagrams fix the survivor’s temperature and leave its content free, and the natural next measurement is the size of that freedom. Within one group of pairs — the same two diagrams — the survivors are different infinitesimals when they fall to nought and different fights when they stop at a bend. Whether they nevertheless lie within a bounded band of each other — within an up of each other, say, or between two multiples of the same infinitesimal — would say how much a player loses by carrying diagrams instead of games in exactly the case where diagrams fail. Atomic weight is the instrument for weighing infinitesimals of this kind where it applies — it is defined on the all-small games, and whether the survivors at nought are all-small is itself part of the measurement. The measurement is the spread of atomic weights within each group, and a spread of nought would mean the diagrams lose the survivor’s name and keep its weight.