The bonus an integer is owed

Why cooling has to add ended with an explanation and a test of it. The explanation was that cooling and heating both carry an exception — a clause that stops the recursion before it reaches the bottom of the tree — and that both exceptions fire unevenly inside a sum, where an option of the sum is one part’s option with the whole of the other part beside it. What separates them, it said, is what each exception hands back. Cooling’s freezing clause returns the mean, and means add, so the side of the equation that did not freeze is forced by the arithmetic to arrive at the same number. Heating’s clause returns a number untouched, and nothing on the other side is obliged to match it.

The test it proposed follows from that reading. If what matters is what the exception returns, then a heating whose exception returned what the recursion would have reached ought to add. For a number x and a bonus t, the obvious candidate is the switch {x + t | x − t} — a fight worth x on average and t to move in, which is exactly what heating does to a position that is already a fight. That operator was named in cooling adds and heating does not and deliberately kept apart from heating. The question was whether the recursion built on it adds on the 263 pairs ordinary heating fails, and the answer turns out to be sharper than either outcome the question anticipated.

One recursion with seven stopping rules #

The way to make the comparison honest is to hold everything but the clause fixed. Every operator in the table is the same recursion: a position G that the clause does not catch becomes $\{\, f(G^L) + t \mid f(G^R) - t \,\}$, every Left option heated and paid a bonus of t, every Right option heated and charged one — and for the two cooling rows the signs are turned, so that every move is taxed instead. The rows differ only in what the clause catches and what it returns. Ordinary heating catches every number and returns it unchanged. Cooling catches every number and, in addition, every position colder than the tax, which it replaces by its mean. The others are variations, each chosen because it isolates one half of that description.

The pools and taxes are the ones that produced the original count, so the 263 can be read straight off the table and every other number is comparable with it: each of the 22 values born by day two against each other, and each of a pool of twenty hot positions against each other, at bonuses of one and two. That is 1,768 ordered pairs per operator. For each pair the operator is applied to the sum, then applied to each part and the results added, and the two answers are compared as values, so a failure is a genuine inequality rather than a difference of printed forms.

Ordinary heating reproduces its own count exactly — 62 and 46 failures at a bonus of one, 72 and 83 at two. Cooling reproduces its own: none. The other five rows are the measurement.

Nought has to stay nought #

The switch clause fails on every pair in the sweep: 484 of 484 on day two at each bonus, 400 of 400 on the hot pool, 1,768 in all. The smallest failure is the first pair the sweep tries, and it explains every other one.

Nought is a number, so the clause sends it to {1 | −1}, the switch usually written ±1. Nought plus nought is nought, so the heated sum is ±1 as well. But the heated parts are ±1 and ±1, and ±1 + ±1 is nought: a switch about nought is its own negative, because exchanging Left and Right in {1 | −1} gives {1 | −1} back, and anything added to its own negative is nought. The two sides of the equation are ±1 and 0, and they are not equal.

There is nothing special about the switch in that argument. Any operator that distributes over sums must send nought to nought, because f(0) = f(0 + 0) = f(0) + f(0), and the only game equal to its own double is nought. A clause that turns nought into a fight of any temperature fails that test before any other pair is tried, and the switch clause also fails the next one: on two numbers it adds two switches about x and y, which give the number x + y with the fights cancelled, against one switch about x + y on the other side. Every row of the figure misses by the same ±1, the trace of one switch left over.

So the repair the explanation proposed is not merely unhelpful; it breaks additivity on the 130 pairs of numbers where ordinary heating was exact. That does not refute the explanation. It refutes one reading of “what the recursion would have reached”, and the operators that do add show which reading was meant.

What the operators that add do to a number #

Two rows of the table come out at nought, and neither is a modification of the clause the proposal had in mind. One removes the clause altogether: only the empty game is a base case, and every other position, numbers included, is heated through its own canonical form. The other is Norton’s product, which keeps a clause but moves it — it catches only integers, and returns an integer n as n copies of t∗, the number t with a star attached.

The integers say what a clause has to do. An integer is a position in which one player holds moves in reserve and the other has none: 2 is {1 | }, Left’s two spare moves. Heating pays a bonus for every move made, so a heated game ought to pay for those reserve moves too, and the two additive operators both do. With no clause, the recursion runs down 2 = {1 | } and 1 = {0 | } and pays the bonus at each step, so at a bonus of one an integer n comes back as 2n — each spare move worth its own one and the bonus of one on top. Norton’s product makes each spare move worth 1∗, so that two of them are 2 and three are 3∗. Heating’s own clause pays nothing at all: it leaves n at n, which treats the spare moves as moves that earn no bonus while every other move on the board earns one.

The fractions say the other half. A half is {0 | 1}, and unlike an integer it is a position in which both players have a move — moves that neither wants to make, which is what makes it a number, as nobody wants to move sets out. Heating exempts it. Both additive operators heat it: the clauseless recursion turns {0 | 1} into {1 | 1}, which is 1∗, and Norton’s product turns it into {1 | ∗}, a position with a temperature. They pay the bonus on moves that were never worth making, because inside a sum those moves are not exempt — an option of a sum is a part’s option with the other part beside it, and a half with a star beside it is no longer a number.

The two additive operators are closer than they look. With no clause, the recursion on an integer n = {n − 1 | } gives {(n − 1)(t + 1) + t | }, and when t is a whole number that is the integer n(t + 1). So at bonuses of one and two the clauseless operator sends integers to multiples of t + 1 and heats everything else with the bonus t — which is Norton’s product by the unit t + 1. Norton’s product by a positive unit U is exactly this recursion with the payment set at U plus each of U’s incentives, and Norton’s theorem says it distributes over every sum of short games. The incentive of the integer t + 1 is −1, so its payment is t, the bonus the clauseless recursion pays. Both zeros in the table are the same theorem, once for the unit t∗ and once for the unit t + 1. It was proved by Simon Norton and set out in On Numbers and Games, and it is not reproduced here.

The option heating leaves unpaid #

The smallest pair ordinary heating gets wrong shows the reserve moves going unpaid in a single option. It is ∗ and {1 | ∗}, at a bonus of one.

The sum of ∗ and {1 | ∗} is {1∗ | 0}. Its Left option 1∗ is the second part’s Left option 1, with the first part’s star still beside it — the integer, reached through the sum, with something hot attached. That is the whole of the failure. Heating the second part on its own reaches the integer 1 and exempts it, so Left’s move there is worth 2: the bonus for moving into the 1 and nothing for the move the 1 holds in reserve. Inside the sum the same integer arrives as 1∗, which is not a number, so heating does not exempt it; it heats it to {2 | 0}, adds the bonus for the move into it, and offers Left {3 | 1} — a position worth 2 on average with the reserve move now treated as a hot move and paid.

The two sides of the equation therefore price the same reserve move differently, according to whether it was reached alone or in company, and the difference does not cancel. Heated part by part the sum keeps an extra Left option worth 1 that the heated whole never offers, and the heated whole less the heated parts is { { {4 | 2} | 0} | 0}, a game Left loses whoever moves first: the parts are worth strictly more to Left than the whole. Exempting only integers instead of all numbers changes nothing on this pair, because the number that causes the trouble is an integer, and across the sweep it makes matters worse: 335 failures against 263. Heating the fractions while still exempting the integers does not pay the reserve moves; it only moves more of the board onto the side of the equation that is paid.

The two operators that add pay the reserve move the same way in both routes. With no clause the integer 1 alone is 2, so Left’s move in the second part is worth 3, and inside the sum the 1∗ heats to {3 | 1}, so Left’s option there is {4 | 2}; both sides come to { {4 | 2} | −1}. Under Norton’s product the 1 is 1∗, Left’s move is worth 2∗, and both sides come to { {3∗ | 1∗} | −1}. What a number does to a fight shows a number sliding through a hot position unchanged, and that sliding is what heating’s exception relies on. It holds for a number added to a game, and fails for a number that has to be heated, because heating a game changes the price of the moves it contains, and an integer is made of moves.

Cooling without its freeze #

The other half of the original question was whether the asymmetry lies in the exceptions or in the directions — whether cooling adds because its exception is the right one, or because charging a tax is somehow better behaved than paying a bonus. The table answers that by giving cooling heating’s clause. The second row is the same recursion with the signs turned, catching every number and returning it untouched, and charging the tax however cold the position is: no freezing.

It fails on 404 of the 1,768 pairs, more than ordinary heating does. The worked case is ∗ plus a half. The sum is ½∗, which is {½ | ½}: temperature nought and mean a half. Cooling by one freezes it, since the tax exceeds the temperature, and it cools to its mean, a half — which is what its parts add to, because ∗ cools to nought and a half is left alone. Without the freeze the recursion charges a tax of one on each side regardless: the Left option becomes −½ and the Right option 3/2, the two cross, and the position is the simplest number between them, which is nought. The parts give a half, the whole gives nought.

So a tax has no protective virtue of its own. Charged past a position’s temperature, it does not converge on the mean; it widens the gap between the options until the simplest number inside the gap takes over, and the simplest number in an interval is not additive in the interval’s ends — it is whatever has the earliest birthday, which is why −½ and 3/2 give nought rather than a half. Cooling is additive because its exception replaces that collapse with the mean, the one quantity of a position that the mean value theorem guarantees will add. Heating has no temperature beyond which it collapses, so it needs a different exception, and the one that works is not a replacement for the recursion at all: it is the recursion carried through the integers, or Norton’s version of that.

The asymmetry between the operators is in their exceptions. What the earlier explanation said — that the difference is what each exception hands back — is right as far as it goes, and the proposal it drew is where it went too far: “what the recursion would have reached” from a number is not a switch about that number, because the recursion never treats a number as a fight. It pays a bonus on every move a number contains, and the clause that adds is the one that does the same.

Deeper pools, and where the theorem stops #

A sweep of 1,768 pairs drawn from day two is small, and the two additive operators were run again on the pools why cooling has to add built to break cooling: 81 switches and numbers each carrying an infinitesimal, and every eighth value born by day three.

Norton’s product adds on all 6,561 pairs of the targeted pool, which is what the theorem says it must, and the run is a check that the implementation agrees with the theorem rather than evidence for it. The second row is different in kind. At a bonus of a half the clauseless recursion sends 1 = {0 | } to {½ | }, which is the integer 1 again, and so it sends every integer to itself and heats only fractions. That makes it exactly the operator in the fourth row of the first table — heating that exempts integers — which fails 335 times at bonuses of one and two. At a bonus of a half it does not fail on any of the 34,225 ordered pairs of the day-three sample.

Nothing here explains that. It is not a Norton product: the unit it multiplies integers by is 1, and one plus the incentive of 1 is nought, not a half. The pairs include every dyadic number born by day three, quarters among them, and the infinitesimals and switches of temperatures up to two, so it is not an empty pool. It may be that a bonus of a half is a coincidence of these pools and a deeper day would show a failure; it may be that integer-exempt heating adds whenever the bonus is no more than a half, for a reason Norton’s condition does not capture. The count is consistent with both, and the second would be a statement about all short games that 34,225 agreements cannot make.

Overheating, and what a count cannot separate #

The unit an integer is multiplied by can be set free of the bonus altogether: send n to n copies of a unit U, and heat everything else with the bonus t. That operator has a name — Berlekamp’s overheating from t to U — and the operator that puts the star back introduced it as the general form of which heating, warming and Norton’s product are cases. Norton’s product by U is the case where U plus each of U’s incentives equals t. Every other choice is overheating without the theorem.

The pattern on these pools is partly the one the integers predicted. A unit that pays a reserve move less than the bonus fails: 1 at a bonus of one, which is integer-exempt heating, fails 136 of 884, and 2 at a bonus of two fails 136 as well. The four units that satisfy Norton’s condition — 1∗ and 2 at a bonus of one, 2∗ and 3 at a bonus of two — fail nothing. Paying too much is not reliably safe: 2∗ at a bonus of one, which pays two on a move that should earn one, fails 80. But 4 at a bonus of two pays three per reserve move and is not Norton’s product, and it passes all 884 pairs.

That last row is the reason the table cannot stand in for the theorem. A sweep over day two and twenty hot positions cannot tell an operator that adds on every short game from one that happens to add on every pair it was shown. Overheating does not add in general — the units of 1 and of 2 show it failing on the very pools where Norton’s units pass — so a unit that passes here is a unit not yet refuted, and nothing more. What the counts do establish is narrower and sufficient for the question asked: the clause that heating needs is not a different value for a number, it is a different set of numbers caught — integers, not fractions — and a payment on each integer’s reserve moves that matches the bonus the recursion pays everywhere else.

The conventions the counts depend on #

Heating by t is Conway’s, as in every essay that heats: a number is unchanged and every other position becomes $\{\, f(G^L) + t \mid f(G^R) - t \,\}$, computed on the canonical form and reduced. Norton’s product by U sends an integer n to n copies of U (−n copies of −U for negative n) and recurses on every other canonical form with the payment U plus an incentive of U; for U = t∗ both incentives are ∗ and the payment is t. Overheating from t to U is the same recursion with the payment fixed at t whatever U is. Cooling is the definition in cooling: a position colder than the tax is replaced by its mean, and a tax equal to the temperature is cooled by the recursion. The pools are the 22 values born by day two and the twenty hot positions used for the original count, at taxes and bonuses of one and two; the deeper pools are the targeted pool and day-three sample of the cooling search. Equality is the comparison of values, and every operator is applied to canonical forms, so each is a function of the value and not of how a position happens to be written.

What seven columns of zeros do not establish #

Two of the zeros are a published theorem, and the sweep agreeing with it says that the implementation is right, not that the theorem is. The third — clauseless heating at a bonus of a half — has no theorem behind it here and is a count. The failures are exact and each one is a genuine inequality; the explanation of them in terms of reserve moves is an account of the worked pair, checked against the integers’ images, and not a proof that every failure in the sweep has that shape. Nothing here measures how often heating’s exception fires unevenly deep inside the trees, as the cooling search did for its own clause at the top. And the pools are short games of small birthday: day two, twenty chosen positions, a sample of day three, and nothing hotter than two.

Still open: whether integer-exempt heating adds below a bonus of one #

The fourth row of the first table fails 335 times at bonuses of one and two, and the same operator at a bonus of a half fails on nothing in 34,225 pairs of a deeper pool. Either a pool of day four, or quarters of a bonus between a half and one, would decide which way that goes. If a failure appears at a half on a deeper pool, the sweep’s agreement was a pool too shallow to show it, as why cooling has to add warned about the opposite case. If none appears at any bonus up to a half and failures begin somewhere between a half and one, then there is a threshold at which exempting an integer stops being free. A half is the obvious candidate, for a reason that is a fact about one position rather than a theory: heating the number {0 | 1} by t gives {t | 1 − t}, which is a number with a star at a half and a genuine fight above it, so a half is exactly the bonus above which the moves a fraction offers become worth fighting over. Whether that is the reason, or a coincidence of the smallest fraction, is what the deeper pool would say — and either answer is a statement about incentives, the price of the moves nobody wants to make, rather than about heating.