The same fight, eight times over

A fight worth $\{2 \mid 0\}$: whoever moves in it gains two, whoever is left with it gains nothing. Its mean is $1$ and its temperature is $1$, and the two numbers are usually explained as “what it settles to” and “what moving there is worth”.

Those are good glosses and they leave the most important thing unsaid. The mean is not a summary of one position. It is a statement about many copies of it, and the statement has teeth.

For every short game $G$ there is a number $m$ and a bound $b$ such that, for every $n$,

$$n\!\cdot\!m - b \;\;\leq\;\; \underbrace{G + G + \cdots + G}_{n} \;\;\leq\;\; n\!\cdot\!m + b.$$

The number $m$ is the mean. The important word is the one that is easy to skim: $b$ does not depend on $n$.

That is what makes $m$ worth calling a mean rather than an average of something. An approximation whose error grew with $n$ would say nothing about large positions; an approximation whose error is a fixed constant says that a pile of a thousand copies is worth a thousand means, plus or minus something that does not care that there are a thousand.

The cleanest way to see that the constant is doing the work is to remove the arithmetic around it. Take a fight whose mean is nought — $\{1 \mid -1\}$, a point to whoever moves and nothing on average — and the number the pile is being compared against is nought at every $n$. Whatever distance the column reports is then the whole of what the pile is, with no growing quantity subtracted from it.

The bound is the temperature #

The figure’s last column does more than stay bounded. It stops at a specific number, and the number is not arbitrary.

For $\{2 \mid 0\}$, with temperature $1$, the smallest bound found on a grid of sixteenths is $17/16$ — one grid step above the temperature. For $\{2 \mid \{1 \mid 0\}\}$, with temperature $3/4$, it is $13/16$: again one step above. The pattern holds for every position drawn here.

So the true bound is the temperature itself, approached and not attained. The reason it is not attained is a small piece of the partial order rather than an accident of the grid: the difference between one copy and one mean is $\{1 \mid -1\}$, and $\{1 \mid -1\}$ is confused with $1$ rather than below it. It is not greater, not smaller and not equal — so a bound of exactly $1$ fails, while every number above $1$ works.

That is a clean example of the partial order doing something an ordering of numbers could not. The infimum of the working bounds is the temperature; the set of working bounds is open at that end.

Why the copies stop drifting #

The mechanism is the exchange argument, applied to a pile of identical components.

Left moves in one copy and gains two. Right, having nothing to answer with in that copy, moves in another and gains two the other way. The two moves cancel, one copy is settled at $2$ and another at $0$, and the pile is two copies shorter with nobody ahead.

So the copies pair off, two at a time, and each pair contributes exactly $2 = 2m$. If $n$ is even, everything pairs and the pile is worth exactly $n \cdot m$ — a number, with no fight left. If $n$ is odd, one copy is left over, and the leftover copy is the entire discrepancy. That is why the bound column alternates between zero and something the size of one component.

That argument has a hidden assumption in it, and a position exists to expose it: it assumes that a copy, once fought over, is finished. $\{1 \mid \ast\}$ is a fight whose Right option is not a number but a star, so settling a copy leaves a star behind — and stars pair off on their own schedule.

The odd leftover never gets smaller and never gets bigger. It is one copy of the same fight whether the pile is three or three thousand. That is the whole reason the bound is constant, and it also explains why the bound is the temperature rather than anything else: what is left over is one unresolved fight, and the size of an unresolved fight is what the temperature measures.

A position where the pairing is not so tidy #

The alternation is a feature of simple switches, and it would be misleading to leave the impression that the bound is always zero-then-temperature.

What changed is that a move no longer finishes a component, so the pairing takes several moves and the residue at each $n$ is a different partly-settled position. What did not change is the ceiling. The theorem is about the ceiling, not about the pattern beneath it.

It is worth looking at the middle column of that figure as well. The values of the piles are getting steadily more complicated — one copy is a two-line expression, six copies is a nest of braces four deep — while the numbers they are being compared against are simply $5/2$, $5$, $15/2$ and so on. That divergence is the reason a mean is wanted in the first place. The exact value of a large pile is correct, computable and completely unreadable; the mean is a single number, wrong by a bounded amount, and it is the only one of the two a player can hold in their head while looking at a board.

The theory is unusual in supplying both, and in being precise about the price of the readable one.

Two claims that look alike and are not #

The essay next door proves that means add. This one is a different statement, and the difference is easy to lose.

Means add says $\operatorname{mean}(G + H) = \operatorname{mean}(G) + \operatorname{mean}(H)$. Both sides are numbers, both are extracted from thermographs, and the equation is exact. Applied to $n$ copies it gives $\operatorname{mean}(nG) = n \cdot m$ — with no error term at all, because the mean of a pile is the mean of a pile.

The mean value theorem compares different objects. On the left is $nG$, which is a game — a position on a board with moves still in it. On the right is $n \cdot m$, which is a number. Those are not the same kind of thing, and the relation between them cannot be an equation; the best available is a two-sided inequality with a slack that has to be paid for.

The first version of the figure on this page confused the two. It printed the difference between the mean of $n$ copies and $n$ times the mean — a column of zeroes, correct and empty, showing the additivity theorem while claiming to show this one. The column that says something is the one comparing the value against the number, and the two columns look nearly identical in a specification and are about different theorems.

A bound that is an infimum and not a minimum #

The observation that every measured bound comes in one grid step above the temperature, and that the temperature itself does not work, is worth more than a footnote — it is a rare shape in this subject and it has one cause.

Almost every bound on this site is attained. A search visits its worst case; a guarantee is met on some line; a maximum is achieved somewhere. Here the set of working bounds is $\{b : b > t\}$ — open at the bottom, with the temperature sitting just outside it — so the theorem’s constant is an infimum that is never a minimum, and no measurement, at any grid resolution, will ever land on it.

The cause is the fourth relation. Asking whether $t$ works means asking whether $G$ is at most $m + t$, and the comparison comes back confused: $\{1 \mid -1\}$ against $1$ is neither less, nor greater, nor equal. Confusion is not a near-miss — it is a definite answer that happens not to be “at most” — so the bound fails at exactly $t$ and succeeds at every $b > t$, with nothing in between for a finer grid to find.

That is the partial order producing an open set, and it is why every careful statement of the theorem in the literature uses a strict inequality where a reader expects a tight one. A totally ordered setting could not do this: there, the set of upper bounds for a bounded quantity is closed, and the infimum is attained.

So the grid in the figures is not merely a computational convenience with a rounding error attached. It is measuring a quantity that has no exact representative, and reporting “one step above the temperature” is the honest form of “the temperature, not attained”.

That is a claim about every grid, so it is worth running the search on a finer one and watching the answer fail to converge on anything.

The same shape as the sum bound #

The last section of this essay observes that a board of forty components has one leftover rather than forty, with a residue the size of the largest temperature — and that sentence is a bound of a shape this site has now met three times.

$$\text{error for } n \text{ copies} \;<\; t(G) \qquad\qquad \text{error for a sum} \;<\; \max_i\, t(G_i)$$

A maximum, not a sum. That is the ultrametric shape rather than the triangular one, and it is the same inequality the temperature of a sum obeys — which is not a coincidence, because the error bound is the temperature of the sum, and the sum’s temperature is bounded by its hottest part.

Chasing it back one more step: the error is bounded by $t(G_1 + \cdots + G_k)$, that is bounded by $\max_i t(G_i)$, and the pairing argument in the section above is the play-by-play version of the same fact. Three descriptions — a strategy, an inequality on thermographs, and a count of leftovers — of one thing.

Which is the reason the means are usable at all, stated as sharply as it goes. If the errors accumulated, a forty-component board would carry forty temperatures of uncertainty and the whole apparatus would be decorative. They do not accumulate because the quantity bounding them is one that never accumulates, and a quantity that never accumulates is a quantity whose bound is a maximum.

It is worth noticing what would break it. A sum whose temperature exceeded its hottest part would give an error larger than any component’s, and there would be no reason for the bound to stay fixed as components were added. That never happens, and the reason is the one-sentence cooling argument: tax everything by the hottest temperature and every part freezes at once, so the whole freezes too. One tax, applied to everything simultaneously, is why forty errors are one error.

What it is for #

The practical use is the one every Go player makes without the vocabulary.

A board is a sum of many components. Adding up the means gives an exact answer about where the board settles, because means add. But a player wants to know something else too — how much the whole board could still swing — and the mean value theorem is what says the answer is bounded rather than accumulating.

If the error in each component’s mean added up, a board of forty components would have a forty-fold uncertainty and the means would be useless. It does not add up: the residue of a sum is one unresolved fight, not forty, and its size is the largest of the components’ temperatures.

The reason is the same pairing, run across different components rather than across copies of one. Left takes the biggest fight, Right takes the next, Left the one after — and each exchange settles two components at their means. At the end at most one component is unresolved, whichever was left over when the alternation ran out, and it is the residue for the whole board. Forty components produce one leftover, not forty, because they are being consumed in pairs rather than each contributing an independent error.

The order the fights are taken in is what makes that argument work, and it is the move rule the next essay along is about: take the hottest available component, so that a big fight is always answered by a big fight rather than by a small one. Pair a temperature-three component against a temperature-one component and the exchange leaves a discrepancy the size of the gap; pair like with like and it leaves nothing. The bound this essay computes and the rule that essay recommends are the same fact approached from the two ends — one asks how large the leftover can be, the other asks how to keep it that small.

What the solver computed, and how #

Each pile is built by repeated addition, canonicalised at every step. That is not tidiness: an unreduced sum of eight copies is a tree the size of the product of eight trees, and reducing as it goes keeps it the size of its value. The essay on tiny and miny records what happens when a family does not cooperate with that.

The mean and temperature come from the thermograph of the single component, computed as exact piecewise-linear walls rather than sampled.

The bound at each $n$ is found by search rather than derived. The generator builds the difference game $nG - n\!\cdot\!m$, then walks a dyadic grid upward from zero, testing at each step whether the difference is both at least $-b$ and at most $b$. The first $b$ that passes both is reported. Comparison is the site’s ge and le, which play the difference out; nothing is inferred from the numbers.

Three things then have to hold or the figure does not build. The bound must never exceed the temperature by more than one grid step. It must be reached at one copy or two rather than at the end — a bound that first appeared at the eighth copy would be a bound that is growing, and the caption’s claim would be false. And each difference must be bounded at all within eight, which catches a position for which the whole framing is wrong.

Where the model stops #

Eight copies is not infinity. The theorem is for all $n$; the figures check the first eight. Beyond that the sums grow expensive, and the essay reports what was computed.

The bound is found on a grid. Reporting “one grid step above the temperature” is a statement about sixteenths. A finer grid would give a closer number and the same conclusion, and no grid will ever give the temperature itself, because the temperature does not work.

Short games only. Every component here is a short game with a thermograph. A loopy component has no mean in this sense and nothing on this page applies to it.

Identical copies are the easy case. A pile of $n$ copies of one position is a much tidier object than a board of $n$ different ones, and the pairing argument is correspondingly cleaner. The general theorem covers the mixed case too, and the residue there is not simply “one leftover copy” — it is whichever partly-settled component the alternation stopped on, which depends on the order the fights were taken in.

The bound is checked at eight copies and the pattern is read off four values. For the simple switches the column alternates with period two; for the position with an inner fight it moves through four. Whether every short game’s bound column is eventually periodic is not something this page establishes, and it is the kind of question that has turned out to be hard elsewhere in this subject — the Grundy sequences look eventually periodic and nobody can prove that they always are.

The mean is not the average of the options. For a switch it happens to be the midpoint, which is why switches are the standard first example and also why they are misleading. For the position with a fight inside it, the mean is not the midpoint of $4$ and the value of $\{2 \mid 0\}$, and reading it off the picture would give the wrong number.

Who found it, and when #

The mean value theorem is Conway’s, in On Numbers and Games (1976), and it is the result that turns temperature from a diagram-reading convention into a measurement. Winning Ways (1982) develops it into the machinery Berlekamp later used on Go endgames, where the bounded residue is the difference between a score that is exact and a score that is exact to within one move.

The idea has a much older cousin. Approximating a complicated object by a simple one, with an error that stays bounded rather than accumulating, is what makes asymptotic analysis work everywhere it works. What is unusual here is that the error is not a number: it is a game, and the bound on it is a statement in a partial order where “at most” and “at least” can both fail at once.

The Go players got there first, as they usually do in this corner of the subject. Miai counting — the practice of valuing a local position by half the difference between the two players’ results there — is the mean and the temperature computed together, by hand, centuries before anybody wrote a thermograph. What the theorem adds is the guarantee: not that the count is a good heuristic, but that the count is exact to within one fight however many fights there are. A practice that works is worth having; a practice that comes with a bound on its error is a different thing entirely.

The same theorem, six years earlier #

The mean-value idea did not begin here. Milnor and Hanner proved it in the 1950s for scoring games, with Go in mind, and the statement is recognisably this one.

Where the ladder goes next #

This is the fourth rung on the temperature ladder, after what is at stake, cooling and playing the hottest first. It supplies the reason the first of those is a measurement rather than a label.

The next rung takes the bound seriously as a quantity in its own right. The residue at each $n$ is a game, not a number, and this essay has only asked how large it is. What it actually is — which partly-settled position is left over after $n$ copies have paired off — is a question with a much more interesting answer, and it is where the theory of orthodox accounting gets its correction terms.