A number and a fight

A hot position is usually described in two numbers. Its mean is what it settles at; its temperature is how much is at stake in it. Both are read off a thermograph, and the pair is the closest thing this subject has to a summary: worth about this much, worth about this much to move in.

The two numbers are not the position, and the question is what the gap between them consists of. It has an exact answer, and the answer is obtained by charging the position precisely what it is worth.

Cooling by the right amount #

Cooling by a fixed tax is a blunt operation: too small a tax leaves the fight, too large a tax freezes it into a number. Cooling a position by its own temperature is neither, because the temperature is the exact height at which the fight stops being worth having.

What comes out is not the mean. It is the mean with something attached, and the something is the whole subject of this essay:

$$G_{t(G)} \;=\; \text{mean}(G) \;+\; \varepsilon, \qquad \varepsilon \text{ infinitesimal}.$$

For a plain switch the arithmetic is visible without any machinery. $\{a \mid b\}$ has mean $\tfrac{a+b}{2}$ and temperature $\tfrac{a-b}{2}$; cooling by that temperature gives $\{a - t \mid b + t\}$, and $a - t = b + t = \tfrac{a+b}{2}$. So the cooled game is $\{m \mid m\}$, which is $m\ast$ — the mean with a star.

What “cool by its own temperature” means as a game #

The operation deserves a sentence in playing terms, because “charge a tax” is a metaphor until it is a rule.

Cooling by $t$ builds a new game in which every move costs the mover $t$ points. A player who moves in the cooled game gains whatever the move gains and pays the tax, so as $t$ rises the incentive to move falls, and at some height nobody wants to move at all. That height is the temperature and the value at it is the mean.

Cooling by the position’s own temperature is therefore the last moment at which the position is still worth moving in. Any more tax and the fight is over; any less and there is still something to fight about. It is the boundary case, and boundary cases are where the interesting residues live.

For a plain switch the picture is arithmetic. $\{a \mid b\}$ taxed by $\tfrac{a-b}{2}$ becomes $\{m \mid m\}$ where $m$ is the mean: both players’ options have been dragged to the same number, and a game whose two options are the same number is that number with a star on it. The star is what is left of a fight after the points have been taxed away — a position where nobody gains anything by moving and moving still matters.

Every hot value born by day three #

A switch is the easy case, and the interesting question is whether the shape survives away from it. Every value born by day three was cooled by its own temperature and had its mean subtracted.

Three things are worth pulling out of that count.

The residue is never zero. Not once in 1,122. A position that cooled exactly to its mean would be one where the value really is the number — and the theory says that a game equal to a number is a number, which a hot game is not. The count is the theorem made visible: something always survives.

The residue is almost always a star. Eighty-four per cent. That is the switch case, and it says most hot values behave like switches once the points are taxed away, whatever their trees look like.

The rest are ups and downs. A residue of $\uparrow$ rather than $\ast$ means the position has, buried under its fight, a slight advantage to Left that no number reports and that survives cooling. 122 values of the 1,122 carry one.

Where the residue is not a star #

Nine hundred and forty-two of the 1,122 leave a star, and the other 180 are the interesting minority.

Up and down account for 122. A residue of $\uparrow$ means the position, once its points are taxed away, is still slightly better for Left — not by any number, but by an amount that decides games where the numbers cancel. Those are positions with an asymmetry the mean cannot see: Left’s options and Right’s are not mirror images, and the difference survives cooling.

Up-star and down-star account for 42, and $\ast 2$ for 16. A residue of $\ast 2$ means the cooled position behaves like a Nim heap of two — it is confused with zero and with $\ast$, and it needs two different moves to cancel.

None of that is visible in the pair of numbers a thermograph reports.

The two numbers and the third thing #

So the summary “mean and temperature” is exactly two-thirds of a position, and the missing third is infinitesimal.

That sounds dismissible, and it is why this site keeps returning to it. An infinitesimal is smaller than every positive number, so it cannot change what a position is worth by any amount a scorer would record. It decides who wins anyway, whenever the numbers happen to cancel — which in a real endgame is exactly when the game is close.

Why the mean is the right number to have in the first place is the mean-value theorem: the value of $n$ copies of a position stays within a bounded distance of $n$ times the mean, and the bound does not grow with $n$. It is that theorem which licenses treating a hot position as its mean plus an error, and which fixes the error at the temperature. The residue is what remains inside that error once the fighting is over.

Why the residue can never be nothing #

The census reports zero exceptions in 1,122, and the reason is worth having as an argument rather than as a count.

Suppose a hot position cooled by its own temperature came out exactly equal to a number. Cooling never turns a game into something it was not: the cooled game’s options are the cooled options, taxed. If the result is a number, then above that temperature the position is that number too — which is the definition of the temperature being lower than it was.

Put the other way: the temperature is the least tax at which the fight stops, and at exactly that tax the position is still, in an infinitesimal sense, a fight. The residue is what “still, in an infinitesimal sense” means. The count is the theorem made visible, and the theorem is that the temperature is a minimum rather than a threshold with slack in it.

That also explains why the residue is so often a star. A star is the smallest thing a position can be while still being worth moving in — it is confused with zero and with nothing else nearby — and a fight taxed to its last penny is exactly a position worth moving in and worth nothing.

The residue appears all at once #

A reader meeting “cool until the fight stops” naturally imagines the fight shrinking — the position getting less hot, the two options converging, the infinitesimal emerging as what is left when the shrinking finishes. That is not what happens, and the difference is worth being exact about because it says what kind of quantity the residue is.

Tax a hot position by anything less than its own temperature and what comes out is another hot position. This is immediate rather than empirical: cooling by $t'$ reduces the temperature by exactly $t'$, so a position of temperature $t$ taxed by $t' < t$ has temperature $t - t'$, which is positive. A position with positive temperature has its two stops apart — that is what positive temperature means — and a position whose stops are apart is not within an infinitesimal of any number at all.

So along the whole way up there is nothing infinitesimal to see. At $t' = 0.9\,t$ the position is still a fight worth a tenth of what it was; at $0.99\,t$ it is still a fight; and at $t$ exactly it is a number with a star or an up hanging off it. The residue does not fade in. It is absent, absent, absent, and then it is the whole of what is left.

The switch shows the mechanism in one line. $\{a \mid b\}$ taxed by $t'$ is $\{a - t' \mid b + t'\}$, whose options stay distinct until $t'$ reaches $\tfrac{a-b}{2}$ and collide exactly there. Every intermediate tax leaves two different numbers to fight over; the last one leaves the same number twice, and a game whose two options are the same number is that number with a star.

That is why the temperature is the right place to look and the only place. It is not a scale on which the position is measured but a single height at which the position changes kind, and this measurement is a census of what is standing at that height.

The residues do not add #

The decomposition tempts a further step that the earlier sections warn off in general terms, and it is worth naming the specific reason, because a reader who has followed the endgame account will want to take it.

The step is: a board is a sum of components, each component is its mean plus a residue, so the board is the sum of the means plus the sum of the residues. The first clause is true and the third does not follow — and the reason is not subtle. Each component here was cooled by its own temperature, and the components have different temperatures, so the sum of the dissociations is not the dissociation of anything. There is no single tax that was charged.

Cooling the sum by the sum’s own temperature is a different operation with a different answer. Temperatures do not add: a sum’s temperature is bounded by its hottest part and is often lower, so the tax the sum would be charged is smaller than most of its components paid individually. Charge it, and the cooler components have not been taxed to their own boundary — they are still fights, in the small — while the hottest one is taxed to exactly its own. What comes out is not a sum of residues and generally not a number plus an infinitesimal.

That is the precise content of “not a factorisation”. The three quantities are three answers about one position, computed from that position, and they belong to it alone. A board has its own three, and they have to be computed from the board.

Where the model stops #

A day is not the subject. Day three has 1,474 values and the site cannot enumerate day four, so every count above is a count over one day of the construction. Every count above is over the values born by day three, and the residues there are all-small — the strong form of infinitesimal, where every follower gives both players a move or neither. That is not general. $\{10 \mid \{9 \mid 1\}\}$, whose follow-up is hotter than the position itself, leaves a residue that is smaller than every positive number and is not all-small: it is a tiny, and tinies live outside day three.

The decomposition is not a factorisation. “Mean plus a fight plus a residue” describes what cooling produces; it does not let the position be reassembled. Heating the cooled game recovers the original in a small minority of cases, and where the residue was frozen away it cannot be recovered at all.

The residue is not an error term. The mean-value theorem bounds how far $n$ copies of a position stray from $n$ times the mean, and that bound is a number — the temperature. The residue is not inside that bound in the sense of being small compared with it; it is a different kind of object, smaller than every positive number, and the two should not be added up. The sharpest way to see that they are unrelated is to raise the stake and watch the residue not move.

And the temperature has to exist. A number has no temperature to cool by — the generator refuses a number rather than dividing by nothing — so the statement is about hot positions only, and 352 of the 1,474 day-three values are not hot.

What the solver computed, and how #

The residue is three operations: thermograph for the mean and the temperature, cool(G, t(G)) for the cooled game, and a subtraction whose result is reduced to canonical form and named. The infinitesimal test is not allSmall, which is a property of the form; it is both stops equal to zero, which is exact and which is what “smaller than every positive number” means.

One thing had to be fixed before any of it could be trusted, and it is worth recording.

The thermograph was wrong on four of the 1,474 values. Its two walls are built by taking a pointwise maximum and minimum over the options’ walls, and the routine that did that evaluated both functions at the union of their corner points and joined the results with straight lines. That is exact only when the two never cross between two corners. On $\{-\tfrac12, \{1 \mid -1\} \mid -1\}$ they do: Left’s two option walls are $-\tfrac12 - t$ and the constant $-1$, which cross at $t = \tfrac12$, a point neither of them has a corner at. The chord ran through the crossing instead of round it, the walls met at $t = \tfrac13$ rather than $\tfrac14$, and the position was reported with a mean of $-\tfrac23$ and a temperature of $\tfrac13$.

Both are impossible, and that is how it was caught: a mean of two-thirds is not a value any of these games can settle at. Six copies of the position have both stops at $-4\tfrac12$, so the mean is $-\tfrac34$, which is what the corrected routine returns. The repair inserts the crossing point, and across the 1,474 values it changes four answers and leaves 1,470 alone.

Two things about the repair are worth recording. It changes four of the 1,474 day-three answers and leaves 1,470 alone, so nothing in the site’s earlier temperature essays moves. And after it, no day-three mean or temperature is a non-dyadic number at all — which is the property that made the fault visible in the first place, now holding everywhere instead of failing in four places.

The lesson is the one this site keeps relearning: the symptom was an impossible number, and the check that caught it was a fact about the subject rather than a fact about the code. Nothing in the figure was misdrawn, no assertion fired, and 24 gates were green.

Reading a hot game as three things #

Putting the pieces together gives a way of holding a hot position in mind, and it is worth writing down even though it is not a theorem.

A hot game is a number — its mean, which is what it settles at and what adds across a board. It is a fight — its temperature, which is what moving first in it is worth and what decides where a player moves. And it is a residue — an infinitesimal, which decides who wins when the numbers have cancelled.

The three are not independent parts a position can be reassembled from; heating does not undo cooling, and nothing here licenses treating the decomposition as a factorisation. What they are is three questions with computed answers, and a reader who has all three knows more about a position than one who has the first two.

The endgame accounting this site draws elsewhere uses the first two and drops the third. A board of independent fights is accounted for by adding the means and then taking the stakes largest first, alternating, and that account is exact while every region is a plain switch — because a plain switch’s residue is a star, and a sum of stars is either nothing or a star, which decides the last move and nothing before it. Put a region with a fight inside it on the board and the residues stop being stars, they stop cancelling, and the account is out by a point. The residues are where that point goes.

Where the ladder goes next #

The residue is the part of a hot position that survives the fight, so the natural next rung is what happens when several of them are added: the means add, the temperatures do not, and the residues add as infinitesimals — which is the situation where atomic weight stops being a curiosity and starts deciding endgames. The other direction is the operator that would make this decomposition reversible, which cooling by itself is not.