What is left when the copies pair off

The same fight, eight times over put a pile of copies on the table and measured how far it strays. The answer is that $n$ copies of a position stay within its own temperature of $n$ times its mean, however large $n$ gets, and that the distance stops growing after the second copy.

It closed by naming what it had not asked. The difference is a game, not a number. What that game is has a far more interesting answer than how large it is.

The object #

Write $\mu$ for the mean of $G$ and define

$$r(n) \;=\; n \cdot G \;-\; n\mu$$

reduced to canonical form. The subtraction is exact: $n\mu$ is a dyadic number and the negative of a number is a number, so nothing is approximated at any point. The sum is reduced at every step, because an unreduced pile of eight copies is the product of eight game trees and has nothing to do with the size of its value.

$r(n)$ is what the pile has left over after the paired-off fighting is accounted for. If the position were a number, $r(n)$ would be zero for every $n$; the whole content of temperature is that it is not.

Two things make the sequence worth computing rather than describing. The first is that nothing forces it to have a pattern — it is a sequence of games, and games have no ordering to be monotone in and no size to converge to. The second is that its terms are the objects a player would meet: a board carrying six equal fights, five of which have been taken, is exactly $r$ at some $n$ plus a number, and knowing which game it is is knowing what to do.

The plain switch, which is the case a reader can hold #

$\{2 \mid 0\}$ has mean one and temperature one. Its residues are

$$\{1 \mid -1\}, \quad 0, \quad \{1 \mid -1\}, \quad 0, \quad \dots$$

alternating for as long as the computation is run.

That is exactly right and it is worth saying in words. An even pile of a plain switch is worth precisely its mean: the two players take the fights in pairs, each takes half of them, and the score is the average. There is nothing left over — the residue is zero, and the pile is a number.

An odd pile is worth the mean plus one more fight. And the fight left over is not the original fight: it is $\{1 \mid -1\}$, the switch about nothing, worth a point to whoever takes it and zero on average. The mean has been extracted and what remains is pure tempo.

The alternation has a one-line reason, and it is the reason the rest of this essay is about. $\{1 \mid -1\}$ is its own negative: put two copies of it side by side and the second answers every move in the first, so the pair is worth exactly nothing. An odd pile carries one of them and an even pile carries none, and that is the whole of the sequence.

Two more positions in the sweep do the same thing. $\{1 \mid -1\}$ is already the switch about nothing, so its residues are itself and zero. And $\{\tfrac32 \mid -\tfrac12\}$ — a switch with a non-integer mean — produces $\{1 \mid -1\}$ and zero as well, because subtracting the mean turns any plain switch into the same object.

That is the first finding, and it is a small theorem hiding in a table: every plain switch of temperature one has the same residue sequence, whatever its mean.

Where it stops being simple #

$\{5 \mid \{4 \mid 0\}\}$ has mean four and temperature one, so the size measurement says its pile never strays more than a point from $4n$ — the same bound as the plain switch above.

Its residues do not settle. Eight copies produce eight residues, all different, and each is a nested expression larger than the last:

$$\{1 \mid \{0 \mid -4\}\}, \quad \{1 \mid \{0, \{1 \mid -3\} \mid -4\}\}, \quad \{1 \mid \{0 \mid \{-3 \mid \{-4 \mid -8\}\}\}\}, \; \dots$$

The first is a fight with one follow-up. The third has a follow-up three deep. The eighth does not fit on a line.

So the pile is bounded in size and unbounded in complexity, and those are different things. A reader told that a pile of copies stays within a point of its mean will picture something settling down; what is actually happening is that the position gets steadily more complicated while staying the same distance from a number.

Put beside the plain switch, the difference is visible without any residue arithmetic at all. Two copies of $\{2 \mid 0\}$ make a number; two copies of this one do not, and the reason is that the first exchange does not finish either copy.

What separates the two behaviours #

Four of the eight positions in the sweep settle with period two, one with period four, and three do not settle inside eight copies. The split is very nearly the presence of a follow-up.

A plain switch $\{a \mid b\}$ has numbers on both sides. Taking the fight ends it, so two copies pair off completely and the pile has no memory. A fight with a follow-up does not end when it is taken: the player who takes it leaves a position that is still hot, so a pile of them is a pile of fights in different states, and the state of the pile is not determined by its parity.

$\{\,\{6 \mid 2\} \mid \{1 \mid -3\}\}$ is the exception that shows the rule is about the shape rather than about follow-ups as such. It has a follow-up on both sides and it settles with period two — because its two follow-ups are symmetric about its mean, so taking one and answering in another copy returns the pile to a state it has been in before.

The one with period four #

$\{1 \mid \ast\}$ has a wall that is not a number, which is the case when a switch is not a switch is about. Its mean is a half and its temperature is a half, and its residues run

$$\{\tfrac12 \mid -\tfrac12 \ast\}, \quad \ast, \quad \{\tfrac12 \ast \mid -\tfrac12\}, \quad 0, \quad \{\tfrac12 \mid -\tfrac12 \ast\}, \; \dots$$

with period four rather than two. The star is what doubles the period: two copies of the position leave a star behind, and the star only cancels after four.

That is a nice small demonstration that the parity argument for switches is a statement about two things at once — the fight pairing off, and the infinitesimal pairing off — and that they need not have the same period.

The doubling is visible in the diagram of two copies, and what makes it worth drawing is that the pair looks almost settled and is not.

The sequence is a pile of copies of its own first term #

There is a rule producing $r(n+1)$ from $r(n)$, and it is one line. The mean is a number, so it distributes across the pile:

$$r(n) \;=\; n \cdot G - n\mu \;=\; n \cdot (G - \mu) \;=\; n \cdot r(1).$$

The residue sequence is a pile of copies of the residue of one copy. Checked on all four positions above and every $n$ the sweep reaches, with both sides reduced to canonical form independently: they agree everywhere.

That is a better statement of the object than the definition gives, and it changes what the table is a table of. $r(1)$ is the position with its mean subtracted — a game of mean nought, of the same temperature as $G$, and the only thing about $G$ the sequence depends on. Every position sharing an $r(1)$ has the same residue sequence, which is why the three plain switches in the sweep produce the same two entries: subtracting the mean from a plain switch of temperature one leaves $\{1 \mid -1\}$ whatever the mean was, and after that there is nothing left to distinguish them.

It also explains the two behaviours without appealing to follow-ups. A sequence settles with period two exactly when $r(1)$ is its own negative — because then two copies cancel and the pile has no memory — and $\{1 \mid -1\}$ and $\{\{9/2 \mid 1/2\} \mid \{-1/2 \mid -9/2\}\}$ both are. The period-four case is $r(1) = \{\tfrac12 \mid -\tfrac12\ast\}$, which is not its own negative but whose double is $\ast$, so it takes four copies rather than two. And $\{5 \mid \{4 \mid 0\}\}$ has $r(1) = \{1 \mid \{0 \mid -4\}\}$, which has no finite order at all.

So the question the table poses is not what happens to a pile but the older question what is the order of this element, asked in a group where most elements have none. The follow-up reading and this one agree on all eight positions, and this one says why: a follow-up is what stops $G - \mu$ being its own negative, and being its own negative is what pairing off actually requires.

Every residue is hot #

Not one of the sixty-four residues computed here is an infinitesimal. Every one of them has a positive temperature or is zero, and the ones that are not zero are fights.

That is worth dwelling on, because the phrase the fighting is over is how the mean is usually explained, and it is misleading in exactly this respect. Extracting the mean from a pile does not leave a small correction. It leaves another fight, of a temperature no greater than the original’s, which somebody still has to win.

The bound is checked rather than assumed: across all sixty-four, no residue has a temperature exceeding the temperature of the position it came from. The worst is two and a half, from the position whose own temperature is two and a half, and there are no exceptions. A residue hotter than its source would contradict the rung below, so the check is where that contradiction would appear.

The extreme case is the position whose two copies were drawn earlier. $\{\,\{6 \mid 2\} \mid \{1 \mid -3\}\}$ has a temperature of two and a half, and the leftover of an odd pile of it — $\{\,\{9/2 \mid 1/2\} \mid \{-1/2 \mid -9/2\}\}$ — has a temperature of two and a half as well. Subtracting the mean has removed the position’s worth and left every point of its heat behind. That is the bound being attained rather than broken, and it is the row a check of this kind exists to find.

The residue read as a strategy #

The table is arithmetic, and there is a strategy sitting underneath it that makes the arithmetic legible.

A pile of $n$ equal fights is played by copying. Whatever the opponent does in one copy, answer in another — which is available exactly while an unplayed copy remains. With an even pile the copier answers every move and the fights come off in pairs, which is why the residue is zero and the pile is worth its mean. With an odd pile one fight is left over at the start, and whoever takes it has the advantage the residue describes.

So $\{1 \mid -1\}$ appearing as the residue of an odd pile is not a coincidence of the algebra. It is the leftover fight, with its mean removed because the mean has already been counted into $n\mu$. A position worth a point to whoever moves is what one unpaired fight is worth once the average has been taken out of it, and the table is saying so in the only vocabulary the theory has.

The same reading explains the failure. A fight with a follow-up does not come off in one exchange: the copier’s answer leaves a position that is still hot, so the pile is never a set of untouched copies plus a leftover — it is a set of copies in several different states, and no pairing argument reaches it.

Two piles with the same bound and nothing else in common #

$\{2 \mid 0\}$ and $\{5 \mid \{4 \mid 0\}\}$ both have temperature one. Both piles stay within one of $n$ means, at every $n$ the machine can reach. A reader given only those facts would conclude the two behave alike.

They do not. At six copies the first is worth exactly six — a number, nothing at stake, either player content to move elsewhere for ever. The second is worth twenty-four plus a fight nested six levels deep, in which the players are still very much engaged and in which the order of play matters.

That is the sharpest thing the residue has to say. The bound is the same and the games are not, and a summary that reports only the bound has thrown away the difference between a settled position and a live one. Temperature and mean are two numbers; what a pile actually is, is a game.

Why this is the right question to ask about a mean #

The mean value is defined by the pile. It is the number $\mu$ such that $n \cdot G$ stays within a bounded distance of $n\mu$, and the whole theorem is that such a number exists and is unique.

Stated that way, the mean is an assertion about the existence of a bound, and the residue is what the bound is hiding. Two positions can have the same mean, the same temperature and the same bound, and differ completely in what their piles leave behind — as $\{2 \mid 0\}$ and $\{5 \mid \{4 \mid 0\}\}$ do, at temperature one apiece.

So the residue is the finer invariant, and the sequence of residues is finer still. It is the object a player would want if the question were how does this position behave when there are several of it on the board, which is the situation a real endgame produces constantly.

Read as a board rather than as an arithmetic, three copies of $\{2 \mid 0\}$ are three regions a player would see side by side, and the account of them is three means plus one leftover fight. The leftover is the switch about nothing, worth a move to whoever takes it and nothing on average, and it is what an odd number of equal fights always comes to. A player who has counted the means has counted everything except the thing the game will be decided by.

The connection to a sum that collapses #

How cold a sum of hot games can be found that a sum of two components loses its heat only when their temperatures are exactly equal, and that the commonest way for that to happen is for the components to be identical.

This is the same fact from the inside. Two copies of a plain switch are worth their mean exactly: the sum is a number, its temperature is below the floor, and the residue is zero. Two copies of a fight with a follow-up are not — the residue at two copies is a hot game — and that is why the pairwise sweep found identical components with follow-ups keeping their temperature.

So the residue sequence predicts which piles collapse and which do not, and it does it one pile at a time rather than pair by pair.

What the sweep cannot say #

Eight positions and eight copies. The four sequences that do not settle inside eight might settle at nine, or at a hundred, and nothing here rules it out; what can be said is that they show no sign of it, since each of the eight residues is new and each is larger than the one before.

The second limitation is the pool. Every position swept is hot and short, with at most two levels of follow-up, and all of them are values rather than positions from a game. A Go endgame region of the kind the accounting is really about has a shape none of these has, and whether real regions settle is not a question this table answers.

And there is a third, which is about the arithmetic rather than the pool. $n\mu$ is a dyadic number for every position here because every mean here is dyadic. That is not a coincidence — the mean of a short game always is — but it is worth naming, because it is what makes the subtraction exact and the whole computation possible.

The convention #

Normal play, disjunctive sums, and the mean computed from the thermograph rather than estimated from the pile. The reduction to canonical form is what makes a residue an object at all: two forms of the same value would produce two entries in the table and a reader would take them for two residues.

Reverse the ending convention and the whole construction goes. Misère play has no negatives, so $n \cdot G - n\mu$ is not a well-formed question, there is no mean, and the pile of copies is not analysable at all — which is a good measure of how much this rung is resting on.

Where the ladder goes next #

The temperature anchor reaches seven rungs, and this one answers the question the fourth left standing.

The rung above is the sequence taken as an object. The rule producing $r(n+1)$ from $r(n)$ is on this page and it is $r(n) + r(1)$, which settles the arithmetic and leaves the interesting half open: four of the eight sequences do not repeat, and a sequence of games that never repeats can still have a description. The nesting is plainly regular — the third residue contains the second inside it in a way a reader can see — and what is wanted is a closed form for the $n$-th term rather than a recurrence, of exactly the kind an octal game’s Grundy sequence gets. The reduction to $n \cdot r(1)$ says where to look: it is a question about the powers of one mean-zero game.

Two neighbours are worth the trip. What is at stake is where the mean and the temperature are defined, and reading it beside this page shows how much the pair of numbers is throwing away. And how cold a sum of hot games can be is the same arithmetic done on two different positions rather than on copies of one, where the collapse has to be arranged rather than being guaranteed by parity.