How hot a background has to be

What is left when the small change is thrown away built the reduced canonical form — the ordinary reduction with one clause added, throwing away everything below every number — and found the classes it produces. Two values in one class differ by an infinitesimal and are supposed to be interchangeable in any position hot enough not to notice.

It closed by asking for the quantity that would make hot enough mean something:

“Infinitesimally close” is exact and “safe in practice” is not, and the quantity that would join them is a bound: how hot must a background be, in terms of its temperature and its stops, for a class’s members to be interchangeable in it?

A background at a tax of an eighth is safe on every pair. One at a quarter is not. One at a half is not, three quarters is not, one is not — and two is safe again. Reading down the column of failures gives nineteen, nought, twelve, fourteen, one, two, nought, which is not a curve of any shape.

There is no threshold. The question was asked in the wrong currency.

What the sweep does #

Take the five classes of day-two values that have more than one member. The largest holds seven — nought and every infinitesimal the day produces — and the other four hold two, two, three and three. Add each pair’s two members to the same background and compare the outcomes. If they differ, the reduction discarded something that background could see, and the substitution the reduced form licenses is unsafe there.

The backgrounds are switches: $\{x + t \mid x - t\}$, a fight over $2t$ points with nothing underneath it. A switch is the position whose temperature is exactly $t$ and which carries nothing else, so a result found against switches is a result about temperature rather than about whatever else a richer background might be carrying.

Seven taxes, three means, twenty-nine pairs: 609 comparisons, forty-eight of which come out differently.

Forty-eight of 609 is a rate of about one in thirteen, and the rate is the least informative number on the page. What matters is which thirteen, and the whole essay is the answer to that.

Why there is no threshold #

The non-monotonicity is not noise and it is worth reading off the table directly.

At $t = \tfrac18$ every swap is safe. At $t = \tfrac14$ twelve fail. That is a background getting hotter and the reduction getting less safe, which is the opposite of the story the bound was supposed to tell.

The reason is that a switch’s temperature says how far its two walls are apart and says nothing about where they are. $\{\tfrac18 \mid -\tfrac18\}$ and $\{\tfrac12 \mid 0\}$ have temperatures of an eighth and a quarter — and the second has a stop exactly at nought, which the first does not.

An infinitesimal added to a position with a stop at nought can move the outcome. Added to a position whose stops are both strictly positive or both strictly negative it cannot, because the position is already decided by a margin no infinitesimal reaches.

That mechanism is measured next door over ten thousand shifts, and the same three-way split can be made over this page’s own sweep, where the question is not “what does adding an up do” but “does a swap change anything”.

Note which of the two empty rows is the surprising one. That a total whose stops are clear of nought — a comfortable win — should be immune is obvious. That a total whose stops straddle nought strictly should be equally immune is not, because straddling nought is exactly the configuration a reader calls “close”: 324 of the 609 comparisons are in that row, more than half the sweep, and not one of them moves. A fight that has room on both sides of nought is decided by an amount with a size, and a size is what an infinitesimal does not have.

So the quantity that decides safety is where the stops are, and temperature is a statement about the distance between them. A bound in temperature cannot express a condition about position, and that is the whole of why the requested bound does not exist.

The criterion that does hold #

Stated in the currency the mechanism is in, the condition is short and is checkable by looking:

If the reduced form plus the background has neither stop at nought, the substitution is safe.

Over the 609 comparisons the criterion fires eighty-eight times and every one of the forty-eight outcome changes is among them. That soundness is asserted rather than reported: a swap that changed an outcome away from a stop of nought would mean the mechanism above is wrong, and the figures below refuse to draw at all rather than print it.

It is sufficient and not necessary. Eighty-eight firings against forty-eight changes means the criterion warns forty times where nothing goes wrong, and that is the right shape for a safety condition: it is a licence to substitute, not a prediction that substituting will fail.

What separates a warning that matters from one that does not is the thing the criterion never looks at — the infinitesimal being swapped.

The reason for the split is worth stating exactly, because the obvious version of it is too strong. A star is confused with nought, so it can push a total either way and it usually does. An up is strictly positive, so it can only ever move the total in Left’s favour — and on a total already balanced in Left’s favour it moves nothing. It is tempting to conclude that a positive difference is therefore always harmless, and the table says otherwise: it is harmless about two thirds of the time. The remaining third is the case where the total is balanced against Left, and an up is exactly enough to tip it.

So the sharper condition would have to look at the total’s stops and at the sign of the discarded infinitesimal, and it would stop being readable off an endgame count. A criterion cheap enough to apply at the board is a criterion that warns about a third more positions than it needs to.

Why the question was asked the way it was #

The reason a temperature bound looked like the right thing to want is worth spelling out, because the intuition behind it is good and only the currency is wrong.

The rule of thumb is the reduced form is safe against a hot enough background, and the picture behind it is a fight so large that a fraction of a move cannot decide it. That picture is correct as far as it goes. A switch of temperature two, with a mean of nought, has stops at $2$ and $-2$ and is nowhere near a decision; adding an up to it changes nothing, and the sweep agrees on every pair.

What the picture leaves out is that a hot background can still be balanced — its stops can straddle nought with one of them exactly on it — and a balanced position is precisely where a fraction of a move decides everything. $\{1 \mid -1\}$ has a temperature of one and its right stop is $-1$; $\{\tfrac12 \mid 0\}$ has a temperature of a quarter and its right stop is nought. The second is the dangerous one and it is the colder one.

Which of the sweep’s twenty-one backgrounds are in that condition is a short list, and the list is the reason the temperature column looks so disorderly: a temperature appears three times in the table, once for each mean, and the three cells behave differently.

Heat is not the same as decisiveness, and the whole of the false expectation is the two being conflated. Big is not the same as hot makes the same separation for move choice, and this is the third quantity in the same family: hot, big, and balanced are three different things a fight can be.

The three come apart in every combination. A switch of temperature two with a mean of ten is hot, big and not balanced. One of temperature a quarter with a mean of an eighth is cold, small and balanced. And the position a player has to be most careful about is the third: small enough to ignore, decided by nothing, and exactly where an infinitesimal is the whole of the argument.

The pairs, and which ones are fragile #

Five classes, twenty-nine pairs, and the failures are not spread evenly across them.

That top class is the one to look at, and it is not the pair a reader would expect. It is not $\{0, \ast\}$: it is $0$, $\ast$, $\uparrow$, $\downarrow$, $\uparrow\!\ast$, $\downarrow\!\ast$ and $\ast 2$ together, because every one of them is infinitesimally close to nought and the reduction is willing to identify all seven. Nought and star are the sharpest of its twenty-one pairs — one is a second-player win and the other a first-player win — but $\uparrow$ against $\downarrow$ is a worse swap still, since those two are on opposite sides of nought.

The class $\{-1, -1\ast\}$ is that pair translated one unit down, and it produces three changes where its mirror image $\{1, 1\ast\}$ produces one. The asymmetry is in the sweep and not in the mathematics. The backgrounds are drawn from three means — nought, a quarter and a half — all of them non-negative, so a reduced form of $-1$ has more ways to land a total’s stop on nought than $+1$ does. Adding a number moves both stops by that number and changes nothing else, so the criterion is genuinely translation-invariant; a grid of means symmetric about nought would have produced two equal counts, and the one drawn here is not.

The classes with a hot member fail least. A pair whose members are both switches differs by an infinitesimal that is buried under a fight, and burying an infinitesimal under a fight is what makes it invisible — which is the intuition the temperature bound was reaching for, correct as far as it goes and about the members rather than about the background.

What this means for using the reduced form #

The practical reading is a rule a player or a program can apply.

A reduced form may be substituted whenever the position it sits in, plus the reduced form itself, is decided by a margin — that is, whenever neither stop of the total is nought. That is a condition on the total, not on the background alone, which is the second thing the original question had slightly wrong: it asked how hot the background must be, and the criterion is about what the background and the reduced form come to together.

In an endgame that is a strong condition and it is usually met. A board with several fights on it is rarely balanced to the last infinitesimal, and a position whose total has a stop exactly on nought is a position the players are about to argue about — which is exactly when a careful player stops rounding.

So the rule of thumb survives, with its justification changed: the reduced form is safe unless the game is on a knife edge, rather than unless the game is cold.

That replacement is worth more than the original would have been. A temperature threshold, if one had existed, would have been a number to remember and to check a board against. A condition about a stop being nought is something a player already computes: the stops are what an endgame count is, and a total that comes to exactly nought is the position both players are already staring at.

A third reading of the same table #

There is one more thing in the table and it is easy to miss: the row at temperature nought.

A background that is a plain number is the coldest thing there is, and nineteen swaps fail against it — more than at any other row. That is the case the rule of thumb is clearest about and it is the case where the reduction is least safe, which is the story running the right way round for once.

A number has both its stops at itself. So a number background of nought has both stops on nought, and every pair is exposed; a number background of a half has both stops at a half, and only a reduced form of $-\tfrac12$ brings the total back to a stop of nought. The criterion covers the cold case and the hot case with the same sentence, which is the mark of it being about the right thing — and it explains why the cold row is the worst without needing coldness to be the reason.

So the table has three regimes and only two of them were expected. Cold backgrounds are dangerous, as everybody says. Very hot ones are safe, as everybody says. And in between, safety depends on where the walls sit rather than on how far apart they are — which is the regime a real endgame spends all its time in.

What the sweep does not settle #

The pool is the twenty-two values born by day two, which gives five classes and twenty-nine pairs. That is a small pool and it is small in a specific way: the infinitesimals separating class members are all stars and ups, and none of them is a tiny — a value below every multiple of up. Whether the criterion still holds when the discarded part is much smaller is a question a wider pool would answer.

The backgrounds are switches. A background with a follow-up — a bent wall — has stops that come from deeper in its tree, and nothing here says the criterion reads them the same way. The mechanism suggests it does, since the criterion is stated in terms of the stops of the total and a bent-walled position has stops like any other, but suggesting is not measuring.

And the criterion is sufficient and unquantified. It says safe where it holds and says nothing where it does not, and the forty of eighty-eight firings that turn out harmless are not characterised. A sharper condition would separate them, and it would have to look at the infinitesimal rather than at the stops.

The convention, named #

Normal play, disjunctive sum. Every outcome above was obtained by adding the two games and evaluating the total, rather than by any rule about how outcomes combine — outcomes do not add, so there is no shortcut and none was taken.

The reduced canonical form is the ordinary reduction with one extra clause: a position infinitesimally close to a number is replaced by that number. That clause is what creates the classes, and it is why the members of a class differ by an infinitesimal exactly. The stops are read off the thermograph at zero tax, which is the same recursion the temperature comes from and one implementation rather than two.

Where the ladder goes next #

reduced-form has two rungs to here: the reduction and its classes, and now the condition under which its licence to substitute is good.

The rung above answers the other question the rung below named — the sum’s reduced form computed from the parts’ — and the answer is one word longer than the obvious guess. Add then reduce again finds the operation to be addition followed by a second reduction, which is needed on 431 of 3,600 pairs of day-three values and on not one of the 1,751 pairs with a cold part. So the homomorphism is real and it is not free: putting two reduced forms side by side creates an option that only becomes dominated once both fights are on the board, and nothing in either part announced it.

Two more rungs go after that option. The option nothing names scores eight rules for predicting it from the parts; the best reaches four in five, and on a pool closed under negation it falls under half — which says the near-miss was a property of the population rather than a rule almost working. And not a domination in that order overturns the assumption underneath all of it: on none of the 525 deletions is a surviving option greater than or equal to the deleted one, so the second pass is not an ordinary domination at all. Read in the order the reduced form actually works in — both stops at least as good — every deletion with a survivor is a domination, the dominator is unique on all but twelve, and it always comes from the other part.

That last finding closes a loop with this page. The criterion here is stated in stops because temperature could not express it, and three rungs up the arithmetic of reduced forms turns out to be a domination in the order on stops rather than the order on values. Stops are the currency this anchor works in, and the temperature bound the rung below asked for was never going to be denominated in the right thing.

Two neighbours are worth the trip. What an infinitesimal does to a fight is where the mechanism behind the criterion is measured over ten thousand shifts, and where the stop-at-nought condition is established. And where the fight stops is where the stops are defined and first shown not to determine a position — which is the same fact this page turns into a licence.