How cold a sum of hot games can be

Temperatures do not add established the bound and the failure together. The mean of a sum is the sum of the means, exactly and always; the temperature of a sum is at most the largest temperature in it, and can be anything from that down to nothing at all.

A bound with an interval underneath it invites one question. Where in the interval does a sum actually land?

Eight hundred and sixty-four of the 1,035 sums are exactly as hot as their hottest part. One hundred and sixty have fallen to the floor of the scale: 124 of those are numbers outright, with nothing left to fight over, and 36 sit at temperature exactly nought, which is the bottom rung of the scale rather than a number. Eleven are anywhere in between.

The interval the bound allows is almost entirely empty.

The pool, and what it is made of #

Forty-five positions. Every value born by day two that is hot at all; every switch $\{a \mid b\}$ whose walls are halves between $-2$ and $2$; and five fights with follow-ups written out by hand, because a pool of plain switches would contain no bends and the bends turn out to be the whole of the middle of the answer.

The temperatures in the pool run from a quarter to two and a half, and several of them repeat — which is what makes the equal-temperature condition testable at all. A pool of forty-five positions with forty-five distinct temperatures would have produced 1,035 sums at the maximum and no information.

That the drops need equal temperatures is therefore a claim about a pool with plenty of ties in it, and the number to hold onto is that the ties are 171 of 1,035, which is one pair in six.

One tie in the pool is worth naming in advance, because the rest of the essay keeps returning to it. $\{0 \mid -1\}$ and $\{1 \mid 0\}$ are both worth a half to move in, and their diagrams are reflections of each other: the left wall of one is the right wall of the other turned over, their stops are 0 and −1 against 1 and 0, and their means are opposite. That is what being a negative looks like drawn.

What a drop requires #

The 171 sums that are colder than their hottest part have one thing in common, and it is a strict condition rather than a tendency: every one of them is a sum of two components with exactly equal temperatures.

Not approximately equal. Equal. If one component is hotter than the other by any amount at all, the sum comes out at the hotter one’s temperature, on every pair in the sweep.

That is not hard to see once it is stated. Cool both components together and the cooler one freezes first; above that height only one component still has anything worth doing in it, and a sum with one live component behaves like that component. The sum’s walls can only meet where both are still moving, which is at or below the second temperature — and above that height the sum inherits the hotter one’s diagram outright.

So the drop needs a coincidence, and coincidences are rare in a pool assembled from arbitrary positions. Of the 1,035 pairs here, 171 have matching temperatures.

And equal temperatures do not force one #

The converse is false, and it is worth exhibiting rather than asserting, because a rule that fires on every equal-temperature pair would be a much simpler and much less interesting rule.

$\{1 \mid -1\}$ and $\{5 \mid \{4 \mid 0\}\}$ both have temperature one. Their sum has temperature one. Whatever is happening in the 171 drops is not happening here.

$\{0 \mid -1\}$ and $\{1 \mid 0\}$ also have equal temperatures — a half each — and their sum is zero. Two fights, each worth a full point to take, and together worth nothing at all.

Why the second one collapses #

The mechanism is the mirroring strategy, and it is the cleanest thing in the subject.

$\{1 \mid 0\}$ is the negative of $\{0 \mid -1\}$: exchange the players and negate the values, and one becomes the other. So the sum is $G + (-G)$, which is zero — not “worth nothing on average”, but equal to zero, a second-player win, and a position either player may safely ignore for ever.

The strategy that proves it is to copy: whatever the opponent does in one component, answer in the other. The answer always exists because the components are mirror images, so the copier always has a move and the opponent runs out first.

That is the extreme case of the drop, and 160 of the 171 are of exactly this kind: the sum freezes completely. The fight in one component is answered by the fight in the other, and the two annihilate.

Nothing about either component’s own diagram anticipates that. $\{0 \mid -1\}$ has walls leaning towards each other from a mean of minus a half, meeting at a height of a half; it is a perfectly ordinary small fight, and everything that happens above is a fact about the pair.

The commoner collapse, which is not a mirror #

Only 22 of the 171 drops are mirror pairs, so the mirroring strategy explains an eighth of them and something else explains the rest.

The something else is visible in the smallest case there is. Two copies of $\{0 \mid -1\}$ add to $-1$ — a number, temperature below the floor, no fight at all. Neither copy is the other’s negative; they are identical.

What happens is the pairing off. Each component is worth $-\tfrac12$ on average and a full point to move in; a player who takes one is answered in the other, and after the exchange both fights are over and the score is the sum of the means. The copying strategy works here for the same reason it works for a mirror pair, except that the copier is answering a move with the same move rather than with its reflection, and what makes that legal is that the two components have the same shape.

That is the case the pile of copies is about, seen from the side of temperature. An even pile of a plain switch is worth exactly its mean and has nothing at stake; an odd pile is worth the mean plus one more fight. So identical components collapse in pairs, which is the commonest way a sum in this pool loses its heat.

The words plain switch in that sentence are doing work, and the next section is where they start to earn it.

The eleven in the middle #

Eleven sums land strictly between nought and the maximum, and they are the only inhabitants of the whole interval the bound leaves open. Every one of them has a component whose thermograph bends.

The first is $\{-\tfrac12 \mid -2\}$ added to $\{2 \mid \{1 \mid 0\}\}$. Both have temperature three quarters; the sum has temperature one half. The second component is not a plain switch — Right’s move in it leads to another fight rather than to a number — so its right wall has a corner in it, and the corner is what stops the two components cancelling cleanly.

The component responsible is worth seeing alone, because its shape is the only thing the eleven have in common.

Nine of the other ten are of the same kind: one plain switch and one fight with a follow-up, temperatures equal, sum somewhere between. The remaining two are the sharper case, and they are the reason the pairing-off account of the previous section needed the words plain switch in it.

So the answer to what decides where in the range the sum falls is not the temperatures, which are equal in every drop; it is the shape of the two walls, and specifically whether either of them has a corner. Identical components are not enough on their own: two copies of a plain switch freeze, and two copies of a bent one do not.

A reader who has met a thermograph with two bends will recognise the object: a bend is where a follow-up stops mattering, and two positions whose follow-ups stop mattering at different heights cannot cancel each other over the whole range in which they are both alive.

The reading that failed #

There is a second candidate explanation and it deserves reporting because the sweep refutes it.

A hot position leans. Its left stop sits above its mean and its right stop below, and the two need not be symmetric: $\{5 \mid \{4 \mid 0\}\}$ has a mean of four, a left stop of five and a right stop of four, so it is worth one point to Left to move and nothing to Right. The plausible idea is that two positions cancel when they lean in opposite directions — one a fight Left wants and one a fight Right wants.

Measured over the 171 drops, the number that lean in opposite directions is nought. Not a weak correlation: none of them. Every drop in this pool is between two components leaning the same way or not leaning at all, and the reading is simply wrong.

It is worth keeping the failed reading in view, because it is the reading a player’s intuition supplies and it is the sort of thing that survives indefinitely if nobody counts.

What the bound is good for #

The bound is loose as a statement — it permits an interval that is 99 per cent empty — and it is exactly right as a description of what happens.

Read as the sum is as hot as its hottest part, unless two parts are equally hot, in which case it may collapse, it is a rule a player can use. It says: to find where the action is on a board of independent fights, take the hottest region and expect the sum to be worth about what that region is worth. It says: when two regions are equally hot, look at whether they are answering each other, because if they are the pair is finished.

That reading is what licenses the endgame account to take a board’s regions in order of what is at stake and add the corrections alternately. The move order follows the temperatures, and it is legitimate precisely because the sum’s temperature tracks the hottest part — the situation this sweep says is the overwhelmingly common one.

That is why playing in a hottest component is a rule with a theorem behind it rather than a heuristic. The theorem needs the sum’s temperature not to exceed the maximum, and the sweep says the sum’s temperature usually equals the maximum, which is what makes the rule track the board rather than merely being safe.

The check that had to be made #

Zero of the 1,035 sums are hotter than the bound permits.

Reporting a zero is not padding. The bound is a theorem and the sweep is a computation, and a computation that agreed with a theorem in 1,034 cases would mean either the theorem or the implementation was wrong. Running the comparison and finding no violations is the only way this figure establishes that its own thermograph code is computing what the theorem is about.

The same figure carries a second check of the same kind. The mean of every sum equals the sum of the means, which is the half of the pair that does compose, and it holds on all 1,035 without exception. A pool in which means failed to add would be a pool in which something had gone wrong long before the temperatures were reached.

What a player takes from this #

Three sentences, and each is a measurement rather than a slogan.

Look at the hottest region and expect the board to be worth about that much to move in. It is right 864 times in 1,035, and where it is wrong it is wrong in the safe direction — the board is worth less than the estimate, never more.

When two regions are equally hot, check whether they answer each other. Equal temperatures are the only situation in which the estimate fails, and they are not rare: one pair in six here. The check is cheap, because it is the question of whether one region is the other’s reflection or its twin, and both are visible on the board rather than in an evaluation.

A pair that answers each other is finished. Not “worth little” — finished. Whichever player moves in one of them, the other answers, and the exchange returns the position to the same state minus two moves. A player who keeps treating such a pair as live is spending moves on a component that has already been settled, which is the practical cost of not knowing this.

The third sentence is the one with teeth, because the fights in question look valuable. $\{0 \mid -1\}$ and $\{1 \mid 0\}$ are each worth a whole point to whoever takes them, and together they are worth exactly nothing.

What the sweep cannot say #

The pool is a pool of values, not of positions from a game, so nothing here says how often two regions of a real Domineering or Go board have equal temperatures. That matters for the reading above. On a board with many small regions, the temperatures available are few and ties may be common, in which case drops are common and the rule of thumb is worse than one pair in six suggests. Measuring that needs a game rather than a day of the construction.

The second limitation is that everything above concerns two components. A sum of three can have its two hottest cancel and leave the third standing — a drop produced by a mechanism no pair can exhibit — and the arithmetic of that case does not follow from the pairwise one. The bound still holds, since the maximum over three is the maximum, but the characterisation of when it is met does not, and this sweep says nothing about it.

Why identical components are the extreme case #

The collapse happens when the two temperatures are exactly equal, and the commonest way for that to happen is for the components to be the same position. That is worth turning over, because it is a mechanism rather than a coincidence.

Two copies of one game are the one configuration where a mirroring strategy is available: whatever the opponent does in one copy, answer in the other, and the two moves cancel. That is the strategy that is a symmetry applied to a pair rather than to a board, and it is the only strategy in this subject that needs nothing computed. That strategy is what makes $G + (-G)$ worth nought, and on $G + G$ it does something weaker but related — it lets the second player keep the two copies in step, so neither copy can be turned into an advantage the other cannot answer.

A fight that can always be answered is a fight with nothing at stake, and a temperature is what is at stake. So the collapse is the mirror strategy showing up in the temperature scale, and the equality of the two temperatures is a symptom rather than the cause.

The qualification the eleven force is that always has to mean always. Copying answers the opening move in either copy of any position; it does not answer what that move leaves behind, and when the copies have follow-ups the answering runs out before the fighting does. Two copies of a plain switch freeze; two copies of $\{2 \mid \{1 \mid 0\}\}$ fall from $\tfrac34$ to $\tfrac12$ and stop there, and the same happens to two copies of $\{ \{4 \mid 2\} \mid 0\}$, which fall from $\tfrac32$ to 1. Those are two of the eleven, so the interval is not populated exclusively by mismatched pairs — it is populated by pairs with a bend anywhere in them.

That reading predicts the two things this sweep finds and neither is obvious from the numbers alone. Equal temperatures are necessary — if one part is hotter, the mirror is unavailable and a move in the hotter part cannot be answered in kind. And equal temperatures are not sufficient, because two different positions can share a temperature without either being an answer to the other, and then there is nothing to mirror.

So the right way to read the pairs that collapse is as pairs where a strategy exists, and the right way to read the pairs that do not is as pairs where two fights of the same size are still two different fights. The temperature is measuring the availability of an answer, which is what it was defined to do — and this is the case where the answer is the position itself.

What the pool leaves out #

A last thing, which is a limitation of the drawing rather than of the sweep. A thermograph shows a position’s walls and hides its options, and two positions with identical diagrams can behave differently in a sum — the diagram is a summary, and the sum is computed from the games. Every figure above draws diagrams and every number above was computed from positions, and the reader should not take the first as the evidence for the second.

Where the ladder goes next #

The thermograph anchor has six rungs to here, and the one above takes the obvious next step and finds it available only where it is useless.

When two thermographs can be added tests the natural repair for temperatures not adding: add the whole diagrams instead, wall by wall, since a thermograph is a pair of functions and functions add. Over the same 120 pairs, 92 come out exactly right and the 28 that do not are exactly the pairs with two hot components — which is to say, the construction works precisely when at most one of the parts has anything at stake, and a fight beside a cold thing is not a case anybody needed a construction for.

Two things do survive and both are worth carrying. The added walls always bound the true ones, so a solver holding two diagrams gets an interval for the sum’s stops and a ceiling on its temperature for nothing. And the mast is always exact, because the mean of a sum is the sum of the means — so the top of the diagram composes perfectly and the feet do not.

That split has a reason this page’s measurement makes visible in advance. A mean is a limit and limits survive bounded error; a temperature is a threshold and a threshold moves under any perturbation at all. The score composes and the schedule does not, which is the same division the endgame accounting has always made.