Cooling by exactly one

Cooling arrives on this site as a way of reading a diagram. Charge a tax of $t$ on every move, ask what each player gets, and the two answers are the two walls of the thermograph at height $t$. The tax is a dial; the diagram is what the dial produces.

But cooling is an operator, not a reading. $G_t$ is a game — the position with every move taxed, all the way down the tree — and holding on to that game rather than to its two scores is what the Go and Domineering literature does under the name chilling, which is cooling by exactly one.

The definition, and the case at the boundary #

Three clauses, and the third is the one worth watching.

$$G_t = \begin{cases} G & \text{if } G \text{ is a number} \\ \text{mean}(G) & \text{if } t > \text{temperature}(G) \\ \{\, G^L_t - t \mid G^R_t + t \,\} & \text{otherwise} \end{cases}$$

A number has nothing to tax: whoever moves in it is already committed, and the arithmetic does not change. Above the temperature the position has frozen — both players have stopped wanting to move — and it is worth its mean value exactly.

Between those two, each option is cooled and then charged the tax. And at exactly the temperature the recursion still applies, which is where the star comes from: $\{2 \mid 0\}$ has temperature 1, and cooling it by 1 gives $\{2 - 1 \mid 0 + 1\} = \{1 \mid 1\}$, which is $1\ast$ and not $1$. The mean with an infinitesimal hung on it.

That star is not a rounding artefact. It is the residue of a fight that has just stopped being worth having, and it is precisely why the operator used in Go endgame analysis carries an explicit convention about stars rather than being plain division by a temperature.

Three regimes, and which positions are in which #

A fixed tax sorts the positions into three groups, and the sorting is what makes the operator useful and what makes it lossy.

Numbers pass through unchanged. There is nothing in a number to tax: whoever moves in it is already committed to an amount, and taxing a move that gains nothing changes nothing. Fifteen of the 1,474 day-three values are numbers.

Positions colder than the tax freeze. Their temperature is below 1, so a tax of 1 is past the height at which they stopped being worth moving in, and they collapse to their mean values. That is 1,238 of the 1,474 — five sixths of the day. A single tax of one is a blunt instrument at day three because day three is mostly cold.

Positions hotter than the tax shrink. $\{4 \mid 0\}$ becomes $\{3 \mid 1\}$: the same fight with a point taken off each side, still a fight, still worth moving in. Sixty-one values are strictly hotter than 1 and 160 sit exactly at it.

The three regimes are the reason a fixed tax is a choice. Cooling by 1 is right for a game whose typical fight is worth about a move, and every Domineering and Go endgame this site has drawn is such a game — which is why the operator exists at that setting and not at another.

What it does to a catalogue #

Domineering is the natural place to point the operator, because its small boards carry values of every kind: integers, fractions, switches and forms with no name.

The $\pm 1$ boards are the case the operator was built for. A position worth $\pm 1$ is a fight over one point: whoever moves takes it. Taxing every move by one removes exactly that fight and leaves $\ast$ — a position where the points are settled and the move still matters. Everything about who is ahead has been divided out, and what remains is the question of who runs out first, which is what atomic weight measures.

That is the whole idea behind chilling as a tool: a Go or Domineering endgame is a sum of local fights, most of them worth about the same, and taxing away the common part turns a question about points into a question about tempo.

The star at the boundary #

The third clause of the definition is the one that produces the operator’s signature, and it deserves its own look.

At $t$ strictly below the temperature the recursion produces a fight. At $t$ strictly above it the position is the mean. At $t$ exactly the temperature the recursion still applies and it produces the mean with a star: $\{2 \mid 0\}$ cooled by 1 is $\{1 \mid 1\}$, which is $1\ast$.

That is not an artefact of writing $\{1 \mid 1\}$ rather than $1$. The two are different games: $1\ast$ is confused with $1$, a player moving in it goes to $1$ and the opponent then has nothing, and no amount of rearrangement makes it a number. The recursion produced it and the naming routine reports it.

The Domineering catalogue shows the same thing from the board. The $2\times2$ and $3\times3$ boards are both worth $\pm 1$, both have temperature exactly 1, and both cool to exactly $\ast$ — the mean is zero and the star is what survives. Two boards of different sizes, taxed into the same infinitesimal, which is the operator doing precisely what it was built for: dividing out the part of a fight that is about points and leaving the part that is about tempo.

Heating does not undo it #

Cooling has an obvious candidate for an inverse. Heating by $t$ hands the tax back:

$$G^t = \{\, G^{L,t} + t \mid G^{R,t} - t \,\}, \qquad \text{with a number left alone.}$$

For $\{2 \mid 0\}$ the round trip works: cooling gives $1\ast$, and since $1\ast$ is not a number, heating it applies the recursion and returns $\{2 \mid 0\}$ exactly. For $\ast$ it does not: cooling freezes $\ast$ into $0$, heating leaves $0$ alone, and the star is gone for ever.

The question is which of those two is typical, and it is not the first.

Two separate losses are visible in that table.

Below the tax, everything is lost, and the reason is stated in the definition: those positions freeze into their means, a mean is a number, and heating a number does nothing. 1,238 of the 1,474 values are in this band. Cooling by one is a very blunt instrument at day three, because most values born by day three are cold.

Above the tax, most positions are still lost, and this one is less obvious. Of the 61 values with temperature above 1, five come back. $\{1\ast \mid -2\}$ cools to $\{0 \mid -1\}$ and heats back to $\{1 \mid -2\}$ — the star inside the Left option was frozen away at a level of the tree where the tax had already exceeded the local temperature, and the refund cannot know it was ever there.

How much is thrown away, counted directly #

The round trip measures the loss from one side — how many positions can be recovered — and there is a more direct measurement available, which is how many distinct things are left after the tax.

Cool every one of the 1,474 day-three values by one and count the answers. There are 29.

That is the whole of the operator’s output on a day of the construction: 1,474 values in, 29 out. Nothing about the round trip is needed to see the size of the collapse, and 29 is a much starker number than 27 survivors, because it does not depend on any claim about inverses.

The distribution is as lopsided as the count. 433 of the values cool to zero — that is the single largest class, nearly a third of the day, every one of them a position whose whole content was a fight smaller than the tax. Four more classes hold 177 apiece, at $\pm\tfrac14$ and $\pm\tfrac12$: those are the cold positions whose means happen to be quarters and halves, arriving from very different trees and leaving indistinguishable. Star has 62 members and $\pm 1\ast$ have 29 each.

Only three of the 29 classes have a single member, so all but three of the answers are shared. Two of them are the extreme integers $\pm 3$, which are numbers the operator never touched, and the third is $\{1 \mid -1\}$ — the switch whose temperature is exactly the tax, which is the boundary case the star comes from.

A class with 433 members in it is what “the operator is not injective” looks like at scale. The $2\times2$ and $3\times3$ Domineering boards landing on the same star, two sections above, is not a curiosity about two boards; it is one instance of a collapse that is happening everywhere, and the sizes of these classes are the honest measure of what a fixed tax costs.

Why the collapse is this severe here #

Two things make the number as small as 29, and only one of them is about the operator.

Day three is mostly cold. Five sixths of its values have a temperature below one, so five sixths of the input freezes to a mean before the tax has done any work at all. The image is then dominated by whatever means those positions have — and day-three means are drawn from a very short list of small dyadic rationals, which is why four classes of 177 appear at $\pm\tfrac14$ and $\pm\tfrac12$. Most of the collapse is a collapse of cold positions onto a handful of numbers.

And the hot part is small. Sixty-one values are strictly hotter than the tax and 160 sit exactly at it; between them they supply the star, the two $1\ast$ classes and the handful of remaining fights. There is simply not much material at day three for a tax of one to leave standing.

So the 29 is a joint fact about the operator and about where it is being pointed, which is the same caution the round-trip census carries and is worth repeating in this form. Point the same tax at a pool of Go endgame regions, where every fight is worth about a move by construction, and the cold band that dominates here is largely absent — the operator would be working on its intended input rather than on a day of the universal construction.

That is the reason to read both measurements together. The round trip says the operator has no inverse worth the name; the fibres say how much it merges, and where; and the temperature bands say that most of the merging on this particular input was decided before the operator was applied.

What is lost, and where it goes #

The two losses in the census have different causes and it is worth separating them, because only one of them is about the tax being too large.

Freezing is the obvious one. A position below the tax becomes a number, a number has no structure left, and heating a number does nothing. Everything about the position other than its mean is gone, and no operator applied afterwards can recover it — the information is not hidden, it is absent.

Depth is the subtle one, and it accounts for the failures above the tax. Cooling is recursive: the options are cooled before the tax is applied, and an option whose own temperature was below the tax froze at that level. So a position can be hot at the top, survive the tax there, and still lose an infinitesimal three levels down where the local fight had already cooled out.

$\{1\ast \mid -2\}$ is the example. Its temperature is $\tfrac32$, comfortably above the tax; cooling gives $\{0 \mid -1\}$ and heating that back gives $\{1 \mid -2\}$. The star inside the Left option is what went missing, at a level of the tree where the tax exceeded what was going on locally, and a refund applied at the top has no way of knowing it was ever there.

That is why the Go literature’s warming operator is not simply heating. It carries a rule about stars precisely to put back what the recursion’s lower levels drop.

What the solver computed, and how #

cool(G, t) and heat(G, t) are the two definitions above, written as recursions over the canonical form and memoised nowhere — they are cheap because the trees are small.

Both are new machinery here, and the reason is worth stating: the cooling this collection has had all along returns the two numbers at height $t$ and nothing else. A pair of numbers cannot be added to anything, compared with anything, or heated back. Every question in this essay needs the cooled game, and none of them could be asked before.

The census runs over the 1,474 values born by day three, obtained by the antichain enumeration described elsewhere. For each, the temperature comes from the thermograph, the round trip is heat(cool(G, 1), 1), and the comparison with $G$ is a difference game solved by the recursion — not a string comparison of names, which would have counted two spellings of one value as a failure.

Where the model stops #

This is cooling, and chilling is cooling plus a convention. In the Go endgame literature the operator applied to a position is cooling by one with a rule about how stars are treated, so that its inverse — Norton’s warming operator, written $\int$ — recovers the position exactly. This site computes plain cooling and plain heating, and the measurement above is the reason the literature needs the extra clause: without it, the inverse recovers 12 positions in 1,459.

A tax of one is a choice, and the last figure is the evidence: the same six positions at a tax of two sort themselves differently, and one that survived a tax of one does not survive a tax of two. Nothing distinguishes 1 as a temperature except that most small Domineering and Go fights are worth about one move, which is a fact about those games and not about the theory. The last figure shows the same operator at 2 behaving differently on the same positions.

The operator is applied to values, not to boards. Cooling a Domineering position here means cooling the game its board evaluates to, and a reader should not take the figure as showing a taxed game of Domineering — there is no such game with a rulebook, and the tax is an operation on the value.

And cooling does not commute with addition. $(G + H)_t$ is not generally $G_t + H_t$, for the same reason temperatures do not add: the sum can freeze at a temperature neither part froze at. Every claim here is about a single position; nothing above licenses cooling a board one region at a time.

An operator with a purpose, not a trick #

It would be easy to read a census reporting 27 survivors out of 1,474 as a verdict against the operator. It is not, and the reason is what the operator is for.

Chilling is not used to store a position and get it back. It is used to change the question: an endgame of a dozen local fights, each worth about a point, becomes — after a uniform tax of one — a sum of infinitesimals, where the points have cancelled and what remains is a question about tempo that atomic weight answers directly.

Losing information is the point of the operation. What matters is that the right information is lost, uniformly, across every region of the board, so the comparison between regions survives. The census measures how much is thrown away, which is the honest number to have beside a tool whose job is throwing things away.

The one place it becomes a problem is the inverse. A player who chills a board, computes an answer and wants to translate it back into points needs an operator that undoes the tax exactly — and plain heating is not it, on 1,447 of 1,474 values.

Where the ladder goes next #

The operator opens two rungs. One is the inverse this essay could not build: the warming operator, its star convention, and the theorem that makes the pair exact rather than merely nearly exact. The other is the use the operator was invented for — taking a whole endgame, chilling every region, and deciding the sum by atomic weight instead of by points, which is a claim about sums that no single position can test.