What a fight does to a fight

The translation anchor asks what adding a fixed game to a position does to the pair of stops — what each player gets by moving first and playing the fight out.

What a number does to a fight answers it for a number: both stops move by exactly that number, over ten thousand translations with no exception, and the outcome changes exactly where the arithmetic says it should.

What an infinitesimal does to a fight answers it for something smaller than every number: neither stop moves at all, and the outcome changes anyway — which is the sharpest available statement of what the pair of stops leaves out. It closed by naming the case neither had touched:

The rung above is the one both leave standing — what happens when the thing added is neither, which is to say a hot game.

Adding a hot game does neither of those things, and the instrument breaks.

The instrument fails #

Six switches — $\pm 1$, $\{2 \mid 0\}$, $\{4 \mid 0\}$, $\pm 2$, $\{1 \mid 0\}$ and $\{3 \mid 1\}$ — added to each of the first 120 day-three values. Seven hundred and twenty sums.

The stops add on 330 of them. Forty-six per cent, and the proportion is the same for every one of the six addends: 55 of 120, six times over.

That uniformity is worth noticing before anything else. Whether the stops add is not a fact about how hot the addend is — the coldest switch here has a temperature of a half and the hottest of two, and both fail on the same number of sums. It is a fact about the pair.

So the pair of stops is not a translation invariant of any kind once the addend has something at stake. A number shifts them, an infinitesimal fixes them, and a fight scrambles them.

The failure is not a near miss either. Of the 390 sums where the stops do not add, 209 are out by a whole unit and 40 by two, against 126 out by a half — so the typical failure is larger than the smallest thing the arithmetic could have got wrong. On a board where the components are worth a point or two, being out by two is being out by the whole position.

What survives #

Something has to, or a sum with a hot part would be unanalysable, and the thing that survives is the quantity the whole temperature apparatus is built on.

The mean adds, on all 720. The mean value of a sum is the sum of the mean values, without exception and without qualification, and that is the theorem every mean-value argument in this subject rests on.

It is worth being clear about how much stronger that is than the stop result it replaces. The stops are two numbers read off the bottom of a thermograph; the mean is one number read off the top, and it is the one that composes. What is at stake is where the pair is defined and shown to be a measurement rather than a label, and this page is where the two halves of that pair part company: the mean is additive and the temperature is not.

And the sum is never hotter than its hotter part, on all 720, which is the standard theorem confirmed. It is exactly the larger of the two temperatures on 695, and the other 25 are colder — which is the phenomenon a sum of hot games can be cold is entirely about, arriving here as a by-product.

The failure has a bound #

A reading that fails on half its cases is usually not worth much. This one is, because the failure is bounded by a quantity that is already in hand.

No stop is ever out by more than twice the smaller of the two temperatures. All 720, no exception, and the census asserts it.

The bound is tight: 222 of the 720 sit exactly on it, which is 31 per cent. A pair whose smaller temperature is a half is out by exactly one, over and over.

That turns the failure into something usable. A player who has the two stops of each part and their two temperatures can write down an interval containing each stop of the sum, of width twice the smaller temperature — and on nearly a third of sums that interval’s endpoint is the answer.

And the direction the bound runs in is the informative half. It is the smaller temperature that bounds the error, not the larger. Adding a switch of temperature two to something cold moves the stops by nothing at all; adding it to something else hot moves them by up to four. The damage is done by the interaction, and an interaction needs both parts to be hot.

Why the smaller temperature #

The mechanism is short.

A stop is what a player gets by moving first and playing the fight out. In a sum, “playing the fight out” means answering in whichever component the opponent moved in, and the two components take turns — so the sequence of moves in a sum is not the two sequences concatenated, it is interleaved.

The interleaving costs a move in one component per move in the other, and the amount that costs is the temperature of the component being left alone. When one part is cold, leaving it alone costs nothing and the stops add. When both are hot, each is being left alone while the other is played, and the price is set by whichever of them is cheaper to leave — which is the smaller temperature.

Twice it, rather than once, because both players do it.

The special cases fall out of the same picture. If the addend is a number, leaving it alone costs nothing at any point — a number is not worth moving in, which is the whole content of numbers avoiding numbers — so the smaller temperature is nought and the bound collapses to nothing, which is the first rung’s exact shift. If the addend is an infinitesimal, the same argument gives the same nought, and the second rung’s result follows.

So the three rungs are one statement with the smaller temperature taking three values: nought for a number, nought for an infinitesimal, and something positive for a fight. That the first two rungs are the degenerate cases of the third is not how they were found, and it is a better way to remember them.

That is a sketch and not a proof, and its weak point is nameable: it assumes optimal play alternates between the components, and the sums where the stops add exactly are precisely the ones where it does not have to.

One sum, drawn #

The whole finding is visible on a single pair, and it is worth drawing rather than describing.

$\{2 \mid 0\}$ has stops at 2 and 0. $\pm 1$ has stops at 1 and −1. If the stops added, the sum would stop at 3 and at −1.

It does not. The sum’s left stop is lower, because Left cannot bank both fights: taking the two in $\{2 \mid 0\}$ invites Right to answer in $\pm 1$, and the answer costs more than the second fight was worth.

The means do add: 1 and 0 make 1, and the sum’s mean is 1. The temperature of the sum is 1, which is the larger of the two, and the whole content of this page is that those two statements survive the addition and the four numbers at the bottom do not.

The three answers together #

The anchor now has three answers and they are worth stating in one place, because they are three different shapes.

A number. Both stops move by exactly the number. The stops are a complete description of what happened.

An infinitesimal. Neither stop moves. The stops are a complete description of what happened and it is the wrong description — the outcome changes and the stops do not see it.

A hot game. The stops move by something between the sum of the stops and that plus twice the smaller temperature, and which it is depends on the pair. The stops are not a description at all, and the mean is.

The progression is not a weakening of one result three times. It is three different relationships between a pair of numbers and a sum, and the third is the one that says why the subject has thermographs rather than a table of stops. A stop is a number; a thermograph is a function; and adding two positions adds their functions in a way that has no expression at the level of the two numbers at the bottom.

What a player does with it #

There is a practical reading, and it is unusually direct for a result at this level of the theory.

Do not add stops. A player who knows what two components stop at cannot add those numbers and expect the sum’s stops, and the failure rate is better than even odds against. That is worth saying because adding stops is exactly what a reader who has met where the fight stops and nothing further would naturally do.

Add means. The means always add, so the value the whole board settles to under a tax is a quantity a player can compute part by part — which is why the mean rather than the stop is the number a Go player is taught to count with.

And use the bound when the stops are what is wanted. If both stops of both parts are known, each stop of the sum lies within twice the smaller temperature of their sum. That is a real interval, it is narrow whenever either part is nearly cold, and it collapses to a point when one of them is a number — which is the number rung’s result recovered as a special case.

What the census does not say #

Four limits.

The uniformity is not explained. Fifty-five of 120 for every addend is a striking number and this page has an argument for why it could be a property of the cold part alone and no argument for why it is exactly 55. A sweep over more addends would say whether the count is a constant or a coincidence of six, and that is one line of the census changed.

Six addends, and all of them switches. A switch is the simplest hot game and the six here span temperatures from a half to two. A hot game with a bent wall — one whose option is itself a fight — is not among them, and that class behaves differently in every other measurement on this site, so there is no reason to assume it behaves the same here.

A hundred and twenty values. The day-three values are taken in the order the enumeration produces them, which is not random and is not chosen. A different 120 would give different counts; what it should not give is a sum outside the bound, and that is the claim.

The bound is checked, not proved. Seven hundred and twenty sums with 222 of them exactly on it is strong evidence of the right quantity, and the argument in the section above is a mechanism rather than an induction.

And the mean adding is a theorem being confirmed rather than discovered. It is stated in every account of the subject; what this census contributes is that the site’s evaluator agrees with it on 720 sums, which is a check on the evaluator as much as on the mathematics.

Two moving parts, and what stops being available #

This is the anchor’s third addend and the first with nothing held fixed, and it is worth listing what the earlier rungs’ arguments used, because all of it goes at once.

Adding a number works because a number moves both stops by itself and changes nothing else. The argument is one line and it is available because the number contributes no fight: there is nothing for the two positions to interact over.

Adding an infinitesimal works because the infinitesimal is below the resolution a stop measures, so the stops do not move at all. Again the addend contributes no fight, and again the argument needs nothing about the position it is added to.

Adding a fight is different in kind rather than in degree. Both addends have moves worth making, so play alternates between them, and where the fighting ends depends on the order — which is a fact about the pair rather than about either. Neither earlier argument survives, because both relied on the addend having nothing to contest.

So the loss of a clean translation rule here is not a failure to find one; it is the hypothesis of the earlier rules being gone. What replaces a rule is a bound, and a bound is the honest shape when a quantity depends on an interaction that neither operand determines.

That also says why the bound’s error turns out to be expressible in one follow-up’s temperature. The interaction is an exchange, an exchange happens or does not, and what it costs is what one exchange is worth — which is the same object the rest of this ladder keeps measuring, arriving here as an error term.

The uniformity, which is the odd part #

One number in the first figure deserves more attention than it got, because nothing predicts it.

Every one of the six addends fails on exactly 55 of its 120 sums. Not approximately — exactly, six times, across addends whose temperatures run from a half to two and whose stops run from $\{1, 0\}$ to $\{4, 0\}$.

That says the question do the stops add is decided by the day-three value and not by the switch. Which is plausible in retrospect — the interleaving argument turns on whether the cold part can be finished without inviting an answer, and that is a property of the cold part — and it is not what a reader would have guessed from the size of the error, which does depend on the switch: the worst a stop moves is 1 for the coldest addend and 2 for the others.

So the two halves of the failure are decided by different things. Whether it happens is a fact about the position; how bad it is when it happens is a fact about the pair. That is a cleaner split than this site usually finds and it was not looked for.

The convention, named #

Normal play. The stops of a position are what each player gets by moving first and playing the fight out with no tax charged, read off the feet of the two walls of the thermograph.

The mean is the value the position settles to once moving is taxed enough — the height at which the walls meet, read across. A number is its own mean.

The temperature is the height of that meeting point, with a number given $-1$ by convention; the bound above uses $\max(t, 0)$ so that adding a number is charged nothing, which is the right treatment and is why the number rung’s result is the $t = 0$ case of this one.

A switch is $\{a \mid b\}$ with $a$ and $b$ numbers and $a > b$, and the six used here are the ones the site’s pool of hot components already contains.

Where the ladder goes next #

The translation anchor has three rungs: what a number does, what an infinitesimal does, and now what a fight does.

The rung above is the bent wall. Every addend here is a switch, whose thermograph is two straight lines, and the interleaving argument above is about two straight walls meeting. An addend whose own wall bends is a position with a follow-up, and the coupon ladder has just found the follow-up’s temperature to be a second temperature in its own right — so the natural conjecture is that the bound becomes twice the smaller of three numbers rather than two. That is the same census with a different pool and it is not run here.

Two neighbours are worth the trip. Temperatures do not add is the companion result at the top of the thermograph, and reading it beside this page gives the whole picture: neither the stops at the bottom nor the temperature at the top is additive, and the mean between them is. And how cold a sum of hot games can be is where the 25 colder-than-either sums are taken on their own terms, which is the one thing this page found and did not pursue.