An exemption that costs nothing below one

Heating does not add in general, and the search for an operator that does ended on a small puzzle. Seven operators had been built from one recursion, each differing only in the clause that stops it. Heating that leaves all numbers alone fails. Heating with no clause adds. Between them sits heating that leaves only the integers alone, and that operator behaved strangely. At bonuses of one and two it failed 335 times over two small pools. At a bonus of a half, on a pool of 185 values and 34,225 ordered pairs, it had never been tried, because the operator tested there was the one with no clause, and that operator had not failed once.

The question that remained was where the failures start. If they begin somewhere between a half and one, then there is a threshold at which exempting an integer stops being free, and it would be worth knowing what the threshold is about. There was an obvious candidate. Heating the fraction {0 | 1} by t gives {t | 1 − t}, which is a number with a star at a half and a real fight above it. A half is exactly where the moves a fraction offers start being worth fighting over, and it seemed plausible that this is where an operator which treats integers differently from fractions would begin to run into trouble.

The experiment takes seconds, and the candidate is wrong by a clean margin.

Twelve bonuses #

Fix two pools, the 22 values born by day two and the 20 positions of the earlier hot pool, so that every result can be set beside the 335 failures already known. Run every ordered pair of each pool through the test: does the heated sum equal the sum of the heated parts? Do it for each of twelve bonuses from an eighth to two.

No failure appears anywhere below one. A half is an unremarkable row in the table, with five eighths, three quarters and seven eighths after it, all clean. Then, at a bonus of exactly one, 74 pairs of day two fail and 62 of the hot pool. The count holds at those values through five quarters, three halves and seven quarters, and it changes again at two, where it becomes 92 and 107. Those last four numbers add to 335, so this table contains the earlier result as two of its rows.

The shape of that column is striking in its own right. A threshold that was about the arithmetic of a fraction would be expected to depend on the fraction, and the counts do not. They hold perfectly still between integers and move only when the bonus crosses a whole number. Something about the operator is working in whole steps, and the operator has no obvious reason to do that, given that the bonus is a real parameter and fractions are heated into all sorts of different things.

The clause was doing nothing #

The way to find what is happening is to compare the two operators directly, rather than testing either of them on sums. Heating with integers exempt returns n when given the integer n. Heating with no clause runs its recursion on the integer’s own form. How often do they disagree?

They do not disagree at all below one. The operator that exempts integers and the operator that has no clause are the same operator, and the exemption in that range has no effect whatever. That explains the whole table at once. Heating with no clause is the operator that adds, so below one the exempting operator adds, for the simple reason that it is that operator.

The reason they coincide is a short calculation with the simplicity rule, and it is worth following step by step. An integer has the canonical form n = {n − 1 | }, with nothing on the right. The recursion with no clause gives

$$f(n) = \{\, f(n-1) + t \mid \,\}.$$

Start at nought, where the recursion has nothing to work with and returns nought. Then f(1) = {t | }, which is the simplest number above t. The simplest number greater than t is the smallest integer greater than t, and for any t below one that integer is 1. So f(1) = 1, with no help from any clause. Next, f(2) = {1 + t | }, whose simplest number is 2 by the same argument. And so on up the integers. The negative integers follow by symmetry. When t is below one, the recursion sends every integer to itself. An exemption saying “leave integers alone” adds nothing to an operator that already leaves integers alone.

At t = 1 that argument stops working, and it stops all at once. Then f(1) = {1 | }, whose simplest number is 2, not 1.

So the recursion is not leaving integers alone in general. It is multiplying them, by a factor that goes up in whole steps: it is ⌊t⌋ + 1. At any whole number t this is what the earlier essay identified as Norton’s product by t + 1, and the staircase says what happens in between. The factor is constant on each interval between integers, whatever else the bonus is doing to fractions in the same interval. That constant factor is why the failure counts in the first table stand still between integers. The bonus can change as it likes, but what an integer is sent to has not changed, and integers are where the two operators part.

So the threshold is exactly where the factor stops being one. Below that, the exemption asks for what the recursion would do anyway. Above it the exemption asks for factor one while every option of every game is being handled at factor two, and that is a contradiction which has to show up somewhere. It shows up on sums, as the failures in the table.

The first failure, opened up #

The first pair that fails is about as small as it could be, and the difference between the two sides is a single extra Left option, the integer 1. Without the exemption both sides are { {4 | 2} | −1}, so the two routes through the sum agree when every integer is scaled alike. With the exemption, an integer is left at its own value on a route where the clause fires, while the moves around it are still being paid for at the factor of two that the staircase describes. The two routes then reach the sum with different numbers in them, and an option that one side dominates is one the other side keeps.

This is the same phenomenon as heating’s original failure, which arose two levels down where one part’s option meets the whole of the other. There the clause exempted all numbers, and here it exempts only integers. That suggests a reading: a clause does harm when it returns something different from what the recursion would have given. The exempting operator returns the same thing below one and so does no harm there. Ordinary heating returns a bare fraction where the recursion would return a number with a star or a switch, so by this reading it should have no safe range at all.

The obvious test of that reading is to run ordinary heating through the same twelve bonuses, and the first result looks like a refutation.

On the two small pools ordinary heating is perfect below one, exactly like the exempting operator. Its failures on those pools begin at a bonus of one as well, at 62 and 46, and go to 72 and 83 at two, which are the 263 failures this operator was first known for. Had the sweep stopped there, the conclusion would have been that all three operators are safe below one and that the integers have nothing to do with it. A deeper pool removes that conclusion at once. On the pool built to break cooling, ordinary heating fails about one pair in five at a half and at three quarters. A separate run on the 185 values sampled from day three, at a half, finds 11,379 failures among 34,225 pairs. So the reading stands, and the shallow pools were simply unable to show it.

This matters for how the first table should be taken. Its clean rows below one are not by themselves evidence that the exempting operator is safe, because a clean row on those pools says the same about an operator that is not safe. The evidence is the calculation, which holds for every game: the exempting operator below one is the clauseless one. The warning about shallow pools came from the opposite case, where cooling looked as if it might break on a richer pool and did not. Here a richer pool does break something, and it breaks only the operator that the calculation did not cover, which is about as much as one pool can be asked to say.

What the half was doing #

The candidate threshold of a half was not foolish. Something does change at a half, and it is what the candidate said would change.

Below a half, the fraction ½ heats to ½ with nothing added. At a half it picks up a star. Above that it becomes a true switch, {3/4 | 1/4} at a bonus of three quarters, with a real temperature. So the fraction changes character at just the point that was expected. What was wrong was the assumption that a change of character must damage additivity, and the last column of the table shows why it does not. Two heated halves at three quarters are {3/4 | 1/4} + {3/4 | 1/4}, and two equal switches are worth exactly the sum of their means, which here is 1. The heated integer 1 at that bonus is 1. So the sum still comes out right, even though each part is now a position with a fight in it and the whole is a number.

That small sum is worth lingering on. A player offered one half and another half is holding a number. Heat each by three quarters and each has become a hot position in which the first player to move gets a benefit, and yet the two together are worth a plain 1, as they should be, because whichever player takes one fight can be answered in the other. Equal switches cancel to their means, and this is a case where that fact is doing work on behalf of the distributive law. The operator turns a sum of numbers into a sum of switches that cancels back to the right number.

This also says something about what the deeper tests were really testing.

The 34,225 pairs at a bonus of a half were run on the operator with no clause. It was not known at that time that this was also the operator with integers exempt, and so the open question about the exempting operator at a half had in fact already been answered by that very run, under a different name. That is a useful sort of discovery, because it removes a question rather than adding to the count of evidence. No deeper pool is needed at a half for the exempting operator, because anything it shows would be a statement about the clauseless one.

So the remaining question has moved. It is no longer whether the exemption is free below one, which the calculation settles for every bonus and every game. It is whether heating with no clause adds at fractional bonuses at all. At whole bonuses it is Norton’s product, and that is a theorem. At fractional bonuses the measurements are all clean: nothing fails at any of the eighths on these pools, or at a half on the 185-value sample. But no proof has been offered here. The recursion at a bonus of a half sends 1 to 1, which no Norton product does, since the integer 1 plus its incentive is not a half. That is the sense in which the operator between the integers has no known theory, whatever its measurements say.

The convention underneath #

Heating here is the recursion {f(G^L) + t | f(G^R) − t} applied to canonical forms, with the operator’s clause checked first. Applying it to canonical forms is what makes each operator a function of the value rather than of the way the position happens to be written. An integer means a number with no fractional part, and that is what the clause tests. Numbers such as ½ are recursed through their own forms in both operators.

“Adds” means equal as values on every ordered pair of the pool, tested exactly rather than sampled, with a pair and its reverse both counted. The two pools are the ones used earlier, so that the counts can be compared: the 22 values of day two, and the 20 hot positions built by hand. Everything said here about agreement between the two operators below one on general positions rests on the calculation above, which covers every integer, so it covers everything built from integers. The thirteen positions in the second figure illustrate it and could not establish it.

What the table cannot show #

That heating with no clause adds at every fractional bonus. Every entry for it in the first table is clean. Those pools are small, and an earlier essay warned about exactly this situation with a pool that seemed adequate, although there the eventual answer was that nothing was there to find. A proof at fractional bonuses would need to explain what the operator is, in the way Norton’s product explains the whole-number case.

Why the failure count holds still between integers. The staircase says that integers are treated the same throughout an interval, which makes a constant count plausible. It does not prove it, since the fractions in each game are heated differently at each bonus and could have changed which pairs fail. The first table shows that the counts stay at 74 and 62 on four sampled bonuses in the interval from one to two, and that is all it shows.

Which of these operators is useful. A distributive law is a good property and not the only one. Norton’s operator earns its place because it inverts chilling on a class of games that actually arise on boards, as its warming version shows. An operator at a fractional bonus that adds on everything tested has so far no application in any game anyone plays.

Anything about a board. Every position in these pools is a value and not a Go or Domineering position, and the original use of heating depended on what board positions look like when chilled, which none of this examines.

Still open: what the operator is, between the integers #

At whole-number bonuses, heating with no clause is multiplication by a fixed game. At any bonus it is additive on everything tested, and it maps the integers to multiples of a whole number, but between the integers it does not seem to be multiplication by anything: there is no single game that plays the part of the unit. A halved bonus sends ½ to ½∗ and 1 to 1, while a true product would have to send 1 to twice whatever it sends ½ to. And twice ½∗ is 1, as it happens, so that test passes. The question is then whether some generalised product, with a unit that depends on the position’s birthday or on the bonus’s denominator, actually describes it. A comparison of the operator against each candidate unit, at bonuses of a half and a quarter, across the values of day three, would either locate that unit or show that the additive map is not a product of any kind. Either answer would explain why the sweeps have never found a failure, which at the moment nothing does.