Why cooling has to add

Cooling and heating are introduced as a pair: one charges a tax on every move, the other pays a bonus, and a position cooled and heated by the same amount looks as though it ought to come back. The sweep that set them against a sum found that they are not a pair. Heating a sum differed from heating its parts and adding on 263 of the pairs it tried. Cooling never differed, on any of 1,768.

That sweep was careful about what its cooling result meant. It reported a search rather than a theorem, and it went further: it said that cooling is not additive in general, that freezing a position to its mean must on some pair throw away an infinitesimal the sum still needs, and that the pools — day two, and twenty hot positions — were simply too shallow to show it. It named where to look: positions carrying an infinitesimal underneath a low temperature, at fractional taxes.

That search has now been run where it pointed, and the answer is not that the pools were too small. Cooling is additive on every pair of short games. It is a theorem of the subject — Conway proved it in On Numbers and Games — and it is load-bearing: the method that values a Go endgame by chilling each region on its own and adding the results is sound because of it. A search could never have found a counterexample, and what is worth measuring instead is why not, because cooling’s clause fires inconsistently just as heating’s does.

The search, where it was told to look #

The weak point the earlier account identified is real in the sense that it is where cooling does its most drastic thing. Cooling a position by a tax t charges t for every move all the way down the tree, until the tax exceeds the position’s temperature; past that point the position is frozen to its mean, a plain number, and whatever infinitesimal it carried — a star, an up — is dropped. If the parts of a sum froze at different taxes from the sum, an infinitesimal could be dropped on one side of the equation and kept on the other, and the two sides would differ by it.

So the targeted pool was built to have as many dropped infinitesimals as possible. It takes nine bases — switches of temperature a quarter, a half, one and so on, and two numbers — and adds to each of them one of nine infinitesimals: nought, ∗, ↑, ↓, ↑∗, ⇑, ∗2, and two further ones, {0 | ↑} and {↓ | 0}. After duplicates by value are removed there are 81 positions, nearly every one hot with something infinitesimal beneath it. Every ordered pair was cooled whole and in parts at six taxes: a quarter, a half, three quarters, one, one and a half, and two. Beside it, every eighth value born by day three — 185 values, from the census of 1,474 — was paired with every other at taxes of a half and one.

That is 39,366 pairs in the targeted pool and 68,450 in the day-three sample, 107,816 in all, sixty times the earlier sweep. Cooling distributes over every one of them. If cooling were additive only on small pools, this is the pool that would show it.

Why a bigger search was the wrong instrument #

It is worth being exact about what changed between the two accounts, because the earlier one did nothing wrong by the standard it set itself. It found no counterexample, said so with the size of its search, and declined to call the absence a theorem. That is the right habit. What it got wrong was the other half of the sentence — the claim that a counterexample exists and the pools were too small to reach it — and that half was not a measurement at all. It was a guess about where the theory’s edge lies, written in the same voice as the counts beside it.

The guess had a plausible mechanism behind it, which is why it was easy to make. Freezing does discard an infinitesimal; the star on 1∗ really is gone once the sum is frozen. What the mechanism left out is that discarding something on one side of an equation is harmless whenever the other side discards it too, or carries it in a form that cancels. A search over pairs can show that this happens on every pair it tries. It cannot show that it happens on every pair there is — that is a claim over infinitely many games, and the only thing that discharges a claim of that shape is an argument, the same lesson equal in every company drew about equality itself, where a finite test settles a quantifier over every context and a search over contexts does not. So the new search is not here to prove cooling additive. It is here because the earlier account named a place where a counterexample should be, and a named place can be checked.

A frozen sum beside hot parts #

The column that matters is the fifth. It counts the pairs on which the freezing clause is consulted inconsistently at the top level: the sum is colder than the tax, so it freezes to its mean, while at least one part is hotter, so it is cooled by the recursion instead. That is precisely the situation the earlier account was worried about, and it is common: 1,377 of the targeted pairs at a tax of a quarter, 2,583 of the day-three pairs at a half, 6,032 across the search.

The smallest such pair in the targeted pool shows how it comes out. G = {1 | 0} is a switch of mean one half and temperature one half. H = {1∗ | ∗} is another, the same switch with a star riding on both options, and it has the same mean and temperature. Their sum is 1∗ — the fights cancel, leaving the number one with a star still to play — and 1∗ has temperature nought.

Cool everything by one half. The sum is past its temperature, so it freezes to its mean: 1, with the star dropped, exactly the event the earlier account expected to cause trouble. The parts are not past theirs — a tax equal to a temperature is still cooled by the recursion, which at that height gives the mean with a star — so each part cools to ½∗. And ½∗ + ½∗ is 1, because ∗ + ∗ is nought. The star the sum discarded is not missing from the other side; the other side has two of them, and two stars are nothing.

The pair is not special at one tax. At every tax from an eighth to a half the sum is frozen and the parts are hot: at a quarter they are both {3/4 | 1/4}, a switch of temperature a quarter about a mean of a half, and two copies of one switch cancel, because a switch about its mean is its own negative about that mean. The parts’ fights are not cooled away; they are cooled to exact opposites of each other, and they annihilate on addition, so the sum of the parts is the very number the frozen sum was replaced by. The inconsistency is real at every one of those taxes, and it never shows in the answer.

Two facts the clause leans on #

Why does it always come out like that? The published proof is an induction over the two games, and the case that needs care is the one the worked pair illustrates. Two facts carry it, and both can be counted directly.

A sum is never hotter than its hotter part. If G and H are both colder than the tax, so is G + H. Over every pair of the targeted pool the sum’s temperature sits on or below the larger of the parts’, never above, and over all 107,816 pairs of the search not one sum is hotter. It is often colder — 1,401 of the targeted pairs lie strictly below the diagonal — because opposing fights cancel, as the worked pair’s did. So the freezing clause can fire inconsistently in only one direction. Both parts frozen and the sum not is impossible, and the search confirms it: it never happens. What does happen is the sum frozen and a part not, which is the case the worked pair shows.

The mean of a sum is the sum of the means. This is the mean value theorem — many copies of one game settle on the mean, and a sum’s copies are the parts’ copies side by side — and it holds on every pair of the search. It is what makes the frozen side of the equation the right number: when the sum freezes, it freezes to its mean, which is exactly the mean of G plus the mean of H. The other side, cooled by the recursion, has to arrive at that same number — and it does, because each part cooled by t stays within its own fight of its own mean, and when the sum is colder than t the two fights left on the parts are fights that cancel.

When two thermographs can be added met the first fact from the other side. A sum’s temperature is not the sum of its parts’ temperatures, and the added walls of two thermographs bound the true walls without matching them; but the masts always come out right, and the mast is the mean. Cooling only ever reads two things from a thermograph — whether the tax is above the temperature, and if so where the mast stands — and those are exactly the two things about a sum that adding its parts gets right.

Heating’s clause leaks too, and the leak costs #

The comparison with heating is the surprising connection in the measurement, because the earlier account explained heating’s failure by a property of its clause that cooling’s clause turns out to share.

The earlier account located heating’s failure in what an option of a sum is: one part’s option plus the whole of the other part. Heating leaves numbers alone, and a number inside one part becomes, inside the sum, a number plus something that is not a number — which heating does not leave alone. The exempt set, it said, has to be closed under adding the other part, and the numbers are not.

Cooling’s freezing clause is consulted inconsistently in the same way and for the same reason. An option of a sum is a part’s option plus the other part, and a frozen option inside one part is, inside the sum, a frozen thing plus something hot — which is not frozen. On the earlier sweep’s own pools that happens at the top level on 42 pairs, and inside the trees, where nothing here counts it, it can happen again at every level. The difference is not in whether the clause fires consistently. It is in what the clause hands back. Heating’s clause hands back the number unheated, and nothing on the other side of the equation is obliged to match it. Cooling’s hands back the mean, and the mean adds, and the side that did not freeze is forced by the arithmetic of the fight to arrive at the same number by cancelling. One clause leaks and pays for it; the other leaks into a quantity that was conserved all along.

That is a sharper form of the rule the earlier essay ended on. An operator defined by a recursion with an exception distributes over a sum when the exception’s set is closed under the sum — or when, closed or not, the exception returns something the recursion would have reached anyway. Cooling’s exception is of the second kind: a frozen position is replaced by the number its fight would have converged to, so firing the clause early or late changes the route and not the destination.

What cooling forgets, and why the forgetting is orderly #

Additivity has a consequence that is easy to measure and changes how the operator should be read. A map that preserves sums is a homomorphism, and a homomorphism that is many-to-one forgets a whole subgroup at once.

Cooled by a half, the 1,474 values born by day three land on 57 distinct values; by one, on 29; by two, on 18. And the most common image every time is nought: 292 values cool to nought at a half, 433 at one, 495 at two. Because cooling adds, the values that cool to nought are closed under adding — the sum of two of them cools to nought too — and any two values with the same image differ by one of them. So the loss is not scattered; it is a single, structured forgetting of everything in that set, which at a tax of one includes every infinitesimal and every fight colder than one whose mean is nought.

That is exactly the property the operator chosen for one game relies on without saying so. Chilling a Domineering board is cooling by one, and a board that chills to a number or a number plus a star has had its small fights forgotten in a way that respects addition. That is why chilling the regions of a board separately and adding is the same as chilling the board — the property that made the operator usable on the first position the theory told somebody something about. The round trip, which cooling by exactly one found losing nearly everything, loses it in this orderly way: what heating cannot restore is the forgotten subgroup, and heating is not obliged to respect it.

The conventions the counts rest on #

Cooling by t follows the standard definition: a number is unchanged; a position whose temperature is below t is replaced by its mean; otherwise every Left option is cooled and has t subtracted and every Right option is cooled and has t added, and the result is reduced. A tax exactly equal to the temperature is cooled by the recursion, not frozen, which is why {1 | 0} cooled by a half is ½∗ and not ½. Temperatures and means are read from each position’s thermograph, a number’s temperature taken as −1. Equality is the ordinary comparison of values, so a failure would be a genuine inequality, not a difference of printed forms. “The clause fires on the sum alone” means the sum’s temperature is below the tax while at least one part’s is not; deeper firings, inside the trees, are not counted. The targeted pool’s bases are {1/2 | −1/2}, {1 | 0}, {1/4 | −1/4}, {3/4 | 1/4}, {1 | −1}, {3/2 | 1/2}, {2 | 0}, 0 and 1/2; heating was not run on it, because at a tax of a quarter heating’s forms grow past what a run can hold, and its failures were already counted on the earlier pools.

What a search of this size still does not do #

The additivity of cooling is not established here. It is a published theorem, and 107,816 agreements are evidence that the implementation agrees with the theorem rather than evidence for the theorem itself — the count is consistent with it, aimed at the place where a mistake in either would show, and that is all a count can be. The account of why the inconsistent firings cost nothing is an illustration of the proof’s structure, with both of its supporting facts counted, and not a proof; the induction that turns them into one is in the literature and is not reproduced. Nothing here says how often the clause fires inconsistently deep inside the trees rather than at the top. And the pools are short games of small birthday: the theorem covers every short game, and the measurement covers the ones counted.

Still open: whether heating can be repaired by changing what its clause returns #

The analysis here suggests a test of its own reading. If the difference between the two operators is what their exceptions hand back, then heating with an exception that returned what the recursion would have reached — a number heated into the switch of that mean, rather than left alone — should distribute over sums where heating does not. That operator exists: it is the other thing the word heating names, building {x + t | x − t} from a number x, and the earlier essay kept it separate precisely because it behaves differently on numbers. Whether the recursion built on it, applied to every option, adds on the 263 pairs ordinary heating fails — and whether it then fails to terminate, as the earlier essay predicted any heating without a number clause would — is a finite question on the same pools, and it would say whether the asymmetry between the operators is in their exceptions or in their directions.