The operator chosen for one game

Cooling takes a position and a tax. The tax is a parameter, and everywhere else on this site it is treated as one: a thermograph is the family of cooled positions at every tax at once, and the temperature is the tax at which a particular position gives up.

Chilling fixes the tax at one and stops. It is not a special case with a theorem behind it; it is a choice, made because the game it was invented for mostly runs at that temperature, and the interesting question about a choice like that is how well it fits.

What chilling is for #

A Domineering board is a rectangle of squares. Left places a domino vertically, Right places one horizontally, and a player with nowhere to place has lost — the game whose whole small catalogue this site computes, and the one the operator here was aimed at. The values of every small board are computable and they are mostly hot: $2 \times 2$ is $\{1 \mid -1\}$, $2 \times 3$ is $\{2 \mid -\tfrac12\}$, and a board with both players wanting to move in it is the normal case.

Hot positions are expensive to work with. A sum of hot positions has to be played out to be evaluated, and the orthodox account that adds up the pieces is only exact when the pieces are simple. Cold positions are cheap: a sum of numbers is a number, computed by addition.

So the ambition is to make Domineering cold, and the tax is the instrument. Charge a point for every move, all the way down the tree, and the fights should disappear.

What the measurement says #

Thirteen of the fifteen boards come out cold. The $2 \times 2$, worth $\{1 \mid -1\}$, chills to $\ast$. The $2 \times 4$, whose value is $\{\{2 \mid 0\} \mid 0\}$ and takes four levels to write, chills to $0$. The boards that were already numbers — $1 \times 2$, $3 \times 4$, $4 \times 3$ — are untouched, since cooling has nothing to tax in a number.

The two failures are $2 \times 3$ and its transpose. Both have temperature $\tfrac54$, which is above the tax, so charging one point per move is not enough to close the fight: $\{2 \mid -\tfrac12\}$ chills to $\{1 \mid \tfrac12\}$, which is still a switch.

That is the whole of the exception, and the shape of it is worth stating. No board of temperature at or below one stayed hot, and both boards of temperature above one did. The operator does exactly what a fixed tax does — it settles every fight worth less than the tax and leaves the rest — and the fit is a fact about the distribution of Domineering temperatures rather than about the operator.

There is a second exception, in the other column, and it belongs here rather than in a footnote. Two of the thirteen cold boards cannot be warmed back: $2 \times 4$ and its transpose, whose temperature is zero. The tax exceeds what they had at stake, so they freeze to their plain means, and the round trip has nothing to rebuild them from. Meanwhile the two boards that stay hot come back exactly, because a tax that never reached their mast never threw anything away.

So the two tests fail on disjoint pairs and for opposite reasons: coldness fails where the temperature is above the tax and recovery fails where it is below. Eleven of the fifteen boards pass both, and the two thirteens in the table are a coincidence of counting rather than a set.

Why the answers have stars in them #

Thirteen boards come out cold, eleven of them as plain numbers and two as a number with $\ast$ attached, and the star is not an artefact of either board.

Cooling by exactly the temperature is the boundary case: at the tax the recursion is still the right description and the mean is not, and the recursion produces a position whose two options coincide. $\{2 \mid 0\}$ cooled by one is $\{1 \mid 1\}$, which is $1\ast$, not $1$.

So a board whose temperature is exactly one chills to its mean plus a star, and the catalogue’s two starry answers are exactly its two boards of temperature one — $2 \times 2$ and $3 \times 3$, both worth $\{1 \mid -1\}$, both chilling to $\ast$. Nine of the fifteen boards are numbers before the tax is charged and are untouched by it; of the six that are not, two sit at the tax, two below it and two above.

That is why the Go and Domineering literature’s statement of the theorem says number or number plus star rather than number. The star is what a tax charged at precisely the freezing point leaves behind, and a catalogue with any board at the tax will contain one.

The round trip, which works here and not elsewhere #

Heating by one recovers thirteen of the fifteen boards exactly — every board except $2 \times 4$ and its transpose. That is a striking contrast with what the same round trip does to positions in general.

Over the 1,474 values born by day three, cooling and heating back by one recovers 27, and 15 of those are numbers that never moved in the first place. Twelve positions of 1,459 survive the operation. Here, thirteen of thirteen do.

The reason is the same reason the fit is good. Cooling loses information when it freezes a position to its mean, which happens when the tax exceeds the temperature; the infinitesimal that was under the freezing point is discarded and heating has nothing to reconstruct it from. On a Domineering board of temperature exactly one the tax never exceeds the temperature — it meets it — so the recursion runs all the way down and nothing is thrown away.

And that mechanism predicts its own exceptions, which is the reason to trust it. The only boards in the catalogue whose temperature is strictly below the tax are $2 \times 4$ and $4 \times 2$, at zero, and they are exactly the two the round trip loses. Nothing else is below one; nothing else is lost.

What the arithmetic looks like once it works #

The payoff for making a board cold is that a sum of cold boards can be added rather than played, and it is worth seeing the difference in size.

Take three Domineering regions worth $\{1 \mid -1\}$, $\{\{2 \mid 0\} \mid 0\}$ and $-\tfrac32$. Evaluating their sum directly means building a game whose canonical form runs to several lines and whose construction touches every combination of the three components’ options. Chilled, the same three regions are $\ast$, $0$ and $-\tfrac32$, and the sum is $-\tfrac32 + \ast$ — an addition anybody can do in their head, and one whose answer is exact for the chilled board.

What that answer is about takes one more step, because the chilled sum is not the original sum. Recovering the original means warming the total back, and warming is not additive, so the step has to be taken carefully. That is precisely the caveat cooling adds and heating does not exists to record, and it is why chilling is presented in the literature with a theorem about which positions it may be applied to rather than as a general shortcut.

There is a second route to the same destination — the orthodox account, which adds up what each region is worth and then the largest stake, minus the next, and so on down — and the two agree on boards where every fight is worth at most the tax. On a board with a hotter region they come apart, and the account has no column for the difference.

An operator chosen rather than derived #

There is a general point here and it is worth separating from the measurement.

Most of the apparatus on this site is forced. The canonical form is forced by the requirement that equal positions have equal representatives. The simplicity rule is forced by the recursion. The temperature is forced by the thermograph. Nobody chose them, and asking whether they fit is not a meaningful question.

Chilling is chosen. Somebody looked at a game, noticed that most of its positions had temperature one, and fixed the tax there. That is a modelling decision of exactly the kind that appears everywhere outside this subject and almost nowhere inside it, and it comes with the obligation that modelling decisions come with: say what it was chosen for, and measure how well it does.

The measurement is fifteen boards, thirteen fits, two misses, and a rule that predicts which is which. The rule matters more than the count, because a rule can be carried to a board the sweep did not reach and a count cannot.

One thing the table deliberately does not report is who wins. Chilling is applied to reason about sums, not to decide single boards, so every claim on this page is about what a board is worth in company rather than about who takes the last domino from it alone — and the outcome of each board is settled by its own value long before the tax is charged.

What the two misses cost #

Two boards out of fifteen is a good fit, and it is worth asking what happens to the ones that do not fit, since a rule that is right most of the time is only useful if the exceptions announce themselves.

They do. A board that stays hot after chilling is visibly still hot — the chilled value is a switch, printed as one — so a player using the operator never has to wonder whether a result is trustworthy. The failure mode is loud.

What it costs is that the board cannot be added to the others. The whole purpose of chilling is that the chilled values live in a small set closed under addition, so an endgame becomes an arithmetic problem; a component that chills to a switch is a component the arithmetic cannot swallow, and it has to be played out.

Where the tax came from #

Berlekamp introduced chilling for Go, not for Domineering, and the Go version is the one with the theorem attached. A Go endgame region under chilling by one becomes an object in a small algebra of numbers and stars, and the whole endgame becomes an addition — which is what made Mathematical Go a book about a real game rather than about a toy.

Domineering acquired the operator afterwards, by analogy, and the analogy holds up because Domineering’s temperatures cluster the same way. The measurement above is the analogy checked rather than assumed, on the fifteen boards this site can evaluate.

What neither version has is a reason for the number one that comes from inside the theory. It is the temperature at which those games mostly run, and games that run at other temperatures need other taxes — which is to say the operator does not generalise, and does not claim to.

Would another number have been better? #

A chosen operator invites one question the sections above do not ask: what happens at a different tax? The catalogue is fifteen boards and the sweep costs nothing, so the choice can be tested rather than defended.

The first two columns move in opposite directions, and that is the whole finding. The third is the two tests applied to the same board, and it is the column the section below turns on.

A larger tax makes more boards cold. Charge five quarters and all fifteen freeze, including the two the operator misses — so if coldness were the only goal, one is the wrong number and the right one is a quarter higher.

A larger tax recovers fewer. At five quarters the round trip returns eleven of fifteen rather than thirteen. The boards lost are the ones whose temperature is exactly one — $2 \times 2$ and $3 \times 3$: taxed by more than they are worth, they freeze to their plain means, the star that a tax at precisely their own temperature would have left is discarded, and warming has nothing to reconstruct it from.

So the two boards gained in the first column are paid for with two lost in the second, and the third column says what that costs: eleven boards satisfy both tests at a tax of one, and eleven satisfy both at five quarters. The trade is exactly even on the criterion that requires both.

So the numbers do not pick a tax, and it is worth saying so plainly. One is the largest tax at which the recovery count is still at its maximum of thirteen; five quarters is the smallest at which every board comes out cold. Neither dominates the other, and the two tie at eleven on the conjunction. Choosing one is choosing to value the round trip over coldness — which is a defensible preference, and not a measurement.

What the corner is a fact about #

That is a much better defence of the number than “Domineering mostly runs at that temperature”, and it is worth being clear about what it does and does not establish.

It does not make the operator derived, and after the third column it does not even make it forced. Nothing in the theory of cooling says where an operator should sit; that is a judgement about which of the columns matters, and somebody who only ever wanted cold values and never wanted to warm anything back would rationally charge five quarters and take all fifteen.

What the sweep establishes is that the choice is defensible against the alternatives, on a stated criterion, over the whole catalogue rather than over the examples that motivated it — and that the criterion has to be stated, because on the conjunction of the two the choice is a tie. That is the standard a modelling decision should be held to, and it is a different standard from the one the operator has usually been offered under.

And the corner is a property of Domineering. It sits at one because this game’s temperatures are $0$, $1$ and $\tfrac54$ — a cluster at one with two outliers just above, which is exactly the distribution that puts a corner at the cluster. A game whose temperatures were spread evenly over a range would have no corner at all: every increase in the tax would trade coldness for reversibility at a steady rate, there would be no distinguished number, and an operator with a fixed tax would have nothing to recommend it over any other.

So the sweep answers the question the essay’s title raises. The operator was chosen for one game, and the reason it could be chosen at all is that this game’s temperatures have a shape — and the shape, not the theory, is what puts the number at one.

What a fixed tax is really assuming #

Underneath the choice of one there is an assumption worth naming, because it is the assumption every fixed-parameter approximation makes.

A tax of one is right for a board whose fights are worth about a point each. The Domineering catalogue has temperatures of $0$, $1$ and $\tfrac54$, so the assumption is nearly true and the two failures are the nearly. A game whose temperatures were spread over $0$ to $6$ would have no good fixed tax at all: charge one and most fights survive, charge six and everything freezes to its mean and every distinction is lost.

So the operator is not merely chosen — it is chosen on the strength of a distribution, and the distribution is a property of the game rather than of the theory. That is why the same construction transplanted to Amazons or to Clobber would need its own number, and why nobody quotes a chilling operator for a game whose temperatures nobody has surveyed.

The fifteen-board table above is that survey, for the one game this site can survey completely. Its shape — a cluster at one, two outliers a quarter above — is what makes the choice defensible, and it is the kind of evidence that has to be gathered before an operator like this is worth defining rather than after.

What the solver computed, and how #

The fifteen boards are every rectangle up to four squares on a side and twelve squares in area, evaluated by the ordinary Domineering recursion and reduced to canonical form.

Each value is cooled by one. The result is tested for being a number, and if it is not, the value with $\ast$ added is tested — which is the check for number plus star done by construction rather than by inspecting a printed form. Then the chilled value is heated by one and compared with the original for equality, using the ordinary comparison.

The temperature of each board is read off its thermograph, and the claim that ties the table together is checked rather than described: every board that stayed hot has temperature above one, and every board with temperature above one stayed hot. Both directions, because one of them alone would leave the correlation hanging.

The two thirteens are counted separately and the overlap is counted too, which is what the second line under the table reports. That line exists because the two counts being equal is the sort of coincidence a sentence quietly turns into an identity — and the four boards where the tests disagree are the whole content of the mechanism above.

Where the model stops #

Fifteen boards. Domineering positions are not only rectangles — a real game reaches boards with holes in them, and the regions a board breaks into are irregular by the time the endgame arrives, and finding the regions is itself work — none of those are in the sweep. A general theorem about chilled Domineering would have to cover them and this does not.

The theorem the literature states is also stronger than what is measured here. It is about Domineering positions in general, not about rectangles, and it carries a converse: the chilled values live in a small algebra whose members can be added directly. What is checked here is the shape of the answer on fifteen positions, which is the part an exact evaluator can reach.

And the two failures are failures of the tax, not of the idea. Cooling $2 \times 3$ by five quarters gives a number, and there is nothing wrong with doing so — except that the whole point of chilling is to use one tax for every board on the table, and a tax that varies by board is not an operator, it is a temperature measurement with extra steps.

Where the ladder goes next #

The rung below asks what cooling by one does to positions in general and finds that it destroys almost everything. This rung points the same operator at the game it was chosen for and finds it fits. The rung above is the converse: whether the chilled values really do form the small closed algebra the theorem claims, which needs a sum of chilled boards rather than a table of them.

Two neighbours sit alongside. Cooling adds and heating does not is why the round trip above is worth checking at all. And the endgame, accounted for is what the whole exercise is in aid of: a board of independent fights whose total can be added up rather than played out.