A universe grows by days, not by sums

The same parts, different wholes took the relation defined by 184 contexts — the 22 values born by day two and every sum of two of them — and found it breaking the substitution theorem. Two day-four games the 184 cannot tell apart become tellable once the same day-two value is added to both, and 355 of 567 merged pairs break that way. Every break had the same anatomy: some context $Y$ in the universe separates $G + X$ from $H + X$, so the sum $X + Y$ separates $G$ from $H$, and $X + Y$ is a sum of three day-two values, one more than the universe allowed.

That points at a repair. If a sum of three was missing, allow sums of three; if that leaves pairs that a sum of four would separate, allow sums of four. The question the earlier essay left was whether this regress ends — whether some modest number of summands makes the relation equal to equality on the games it is asked about, or whether it chases its own tail.

It ends, and in the wrong place. And a different way of making the universe larger, which costs far less, ends in the right one.

Each widening is the addition test, run once more #

Call the 22 day-two values $U_1$, and let $U_{k+1}$ be every context in $U_k$ plus every day-two value added to one. Because nought is a day-two value, each universe contains the one before, so the chain is nested: a pair separated at one width stays separated at every larger one, and the count of merged pairs can only fall. The second universe of the chain is exactly the 184 of the earlier essays, and its 567 merged pairs are the same 567.

The nesting does something more useful than guarantee a fall. A pair merged by $U_k$ and separated by $U_{k+1}$ is separated by some context $X + Y$ with $X$ a day-two value and $Y$ in $U_k$ — which says precisely that adding $X$ to both halves of the pair produces two games $U_k$ can tell apart. So the drop from one width to the next is the addition test, and the regress the earlier essay described is this chain read one step at a time. Nothing needs to be run twice: the census of a width is the addition test of the width below it.

Read down the table and the regress is visibly slowing. The 22 day-two values on their own merge 3,044 of the 122,265 pairs in the sample; adding a second summand separates 81 per cent of those. A third summand separates 63 per cent of what is left, which is the 355 of 567 the earlier essay found. A fourth separates only 34 of 212 pairs, 16 per cent, and a fifth 31 of 178. The universes meanwhile grow by a factor of between two and five a step: 866 contexts at three summands, 2,783 at four, 6,880 at five.

The earlier essay put the alternatives plainly. If the rate falls fast, a universe of modest size is a practical stand-in for equality; if it does not, the difference test is not merely convenient but unavoidable. The rate does fall, and then it stops falling, and the reason it stops is not that it reaches nought.

A regress that stops short #

Run on past five summands and the count does not move. At six summands, 14,176 contexts, 147 pairs are merged. At seven, 25,696 contexts, the same 147. The search past five is not cheap — the seventh universe took five minutes and fifteen gigabytes, almost all of it spent holding the sums themselves — and it buys nothing at all.

Three equal counts in a row is evidence and not a proof. Whether the 147 survive every number of summands is not established here, and the nesting argument above cannot establish it, because it runs on the sample’s pairs and not on every game a sum could produce. What it does establish is the practical answer: a reader who repaired the relation by adding summands, and stopped when the count stopped falling, would stop at five with 147 false merges and every reason to believe the job was done.

The pairs are not near misses. Forty have the first game strictly larger than the second, 48 strictly smaller, and 59 are confused — their difference is a first-player win, so neither game is at least the other. A confused pair is about as different as two games can be; there are contexts in which each wins and the other loses. Yet every one of 6,880 sums of day-two values sends both games of such a pair to the same outcome class.

That is a stronger statement than “the universe is a little too small”. The earlier essay’s failure was a context one summand out of reach. These pairs are out of reach of every number of summands tried, and there is no sign in the table of the count moving again.

One day deeper #

The other way to make a universe larger is not to allow longer sums of the same values but to allow values born a day later. The values born on day three with at most two options a side, drawn from day two, number 985 — the same population the earlier essay found the 184 contexts separating exactly. Used as a universe on the same day-four sample, the 985 separate all 495 values into 495 classes. No pair is merged. A universe a seventh the size of the fifth width does what the width chain cannot do at any size tried.

It would be natural to read the width chain as a slow way of getting deeper, since a sum is born later than its parts. By the bound in the birthday of a sum, a sum of five day-two values is born by day ten at the latest, and the fifth universe does reach that far: of its 6,880 contexts, 960 are born on day ten and only 47 on day three. The count of day-three values among the sums is 45 at two summands, 47 at three, and 47 again at four and five. Width piles up late-born games without ever acquiring more of the one day the day-four sample needs. A birthday measures how long a game takes to build; it does not measure which games are in reach, and the stall is a fact about the second.

The stubborn pairs are not stubborn under depth. Every one of the 147 has at least one day-three witness, the median pair has fifteen and the best-separated sixty-one. Only one pair is hard to tell apart, in the sense of having exactly one witness among the 985, and it is worth looking at closely because it shows what a sum universe is missing.

The two games are $\{\{0 \mid 0, ↓∗\} \mid 0\}$ and $\{\{0 \mid 0, \{0 \mid -1\}\} \mid 0\}$. They share a shape — a Left option that is itself a small game hanging off nought, and a Right option of nought — and they differ inside the Left option, where one has $↓∗$ and the other $\{0 \mid -1\}$ among Right’s choices. One of the two is strictly larger than the other, so this is not even one of the confused pairs. Beside the single day-three value $\{0, ↑∗ \mid 0\}$, the first sum is a first-player win and the second a win for Right.

And $\{0, ↑∗ \mid 0\}$ is a game no sum universe holds. If it were a sum of up to seven day-two values it would be in the seventh universe and would separate the pair there, and it does not. The one context that tells these two games apart is a game whose options are day-two values — a game one day deeper than its parts — and no arrangement of day-two values side by side produces it.

The witness is an option, turned over #

Why should one day of depth do what any number of summands cannot? The answer is a short argument, and it is the kind of argument that makes a measurement unnecessary on the range it covers.

Take two games $G$ and $H$ that are not equal. Then one of them is not at least the other; name them so that $H$ is not at least $G$. That means $H - G$ is not a game Left can survive with Right moving first — so Right, moving first in $H - G$, has a winning move, and there are only two places it can be. Either Right moves in $H$, to some Right option $H^R$ with $H^R - G$ no longer winnable by Left moving first, which is to say $H^R \le G$. Or Right moves in $-G$, which means moving to the negative of some Left option $G^L$ of $G$, with $H - G^L \le 0$.

In the first case take the context $X = -H^R$. Then $G + X = G - H^R$ is at least nought, so Right moving first loses it; and $H + X = H - H^R$ is won by Right moving first, simply by moving $H$ to $H^R$ and leaving nought. The two sums are in different outcome classes. In the second case take $X = -G^L$, and the same thing happens with the colours exchanged: $G - G^L$ is won by Left moving first, by moving to $G^L$, and $H - G^L$ is at most nought, so Left moving first loses it.

Either way, the context that separates two unequal games is the negative of one of their options. It is the same move that makes every game has a negative work — copy the opponent in the mirror image — used one level down: the mirror image of a single option is already enough to make the two games behave differently, because one of them can reach that option and leave nought behind while the other cannot. If both games are born by some day, their options are born by the day before, and so are the negatives of those options. So a universe that holds every game born by a day, and is closed under negation, separates every unequal pair born by the next day — not as a measurement, but by an argument that covers every pair at once.

The sample shows the argument working without exception: of the 122,265 pairs of distinct values, every one is separated by minus an option of one of its two games, and every option of every sample game is among the 985 day-three values. Equal in this company observed the pattern from the other side — the day below separates the day above, on day three — and stated it as the practical form of the idea. The argument is the reason, and it also says exactly what a sum universe lacks. A day-three option of a day-four game is generally not a sum of day-two values, and a sum of day-two values is generally not one, however many are allowed.

Exact one day on, coarse two days on #

The argument has a clause that cuts the other way. It promises a universe of day-$n$ games exactness on day $n + 1$, because those games’ options are day-$n$ games. It promises nothing on day $n + 2$, whose games have options the universe may not hold.

The staircase is visible at both heights. The 184 contexts built from day two see the 985 day-three values exactly, one class per value, and merge 178 of the 495 day-four values. The 985 day-three values see the day-four sample exactly — the argument guarantees it — and when they are handed a seeded sample of day five, 353 values built with up to two options a side from the day-four sample, they fall back to 284 classes. Eighty-six of the 353 values share a class with another, and the largest class holds seventeen. Depth is a day behind as well; it is simply never more than a day behind what it was built for.

So the honest summary of the two directions is not that width fails and depth succeeds. Width fails to converge on the games it is asked about. Depth converges exactly one day past its own games and no further, and the reason is a proof rather than a pattern. A universe that keeps up with what it is asked has to grow by days.

The family that stays merged #

The pair everything started from — tiny-two against tiny-four, which equal in every company found its 184 contexts agreeing about — is the clearest illustration of where depth stops, because the tinies are born one a day.

Tiny-$n$ is $\{0 \Vert 0 \mid -n\}$, and since $-n$ is born on day $n$, tiny-$n$ is born on day $n + 2$: tiny-one on day three, tiny-two on day four, tiny-six on day eight. The 22 day-two values cannot tell any of the first six apart. The fifth sum universe and the 985 day-three values both separate tiny-one from the rest — tiny-one is a day-three game, and its options are in reach of both — and both merge tiny-two through tiny-six into a single class.

The argument says which context is missing. Tiny-two and tiny-three are born on days four and five, and their options are nought, $\{0 \mid -2\}$ and $\{0 \mid -3\}$. Minus nought and $\{2 \mid 0\}$, the negatives of the first two, are day-three values, and neither separates the pair: beside each, the two tinies land in the same outcome class. $\{3 \mid 0\}$, minus tiny-three’s right option, does separate them — tiny-two beside it is a win for Left, tiny-three a first-player win — and it is born on day four, which is outside the 985. Tiny and miny describes a family whose members differ by less than any number can measure, and the infinitesimals place it below every multiple of $↑$; the measurement here adds that each member is separated from the next by something exactly as old as it is.

What width is good for #

None of this makes sums of small values useless as contexts. They are cheap to list, they are what a board actually adds up to, and on the 985 values of day three the 184 of them are already exact. The company that is closed shows why a company closed under addition is the thing that makes substitution safe, and a universe of all sums of day-two values, if it could be held, would be closed in exactly that sense: no addition of a day-two value could carry a pair out of the relation it defines.

That is also what the plateau at 147 means. Closure under addition and exactness are different properties. The chain of widths is reaching for the first, and if the 147 never break it has found a relation that adding day-two values cannot disturb — a congruence, as far as that operation goes — that is still not equality. The day-three universe has the second property on the games it was built for and not the first: it is exact on day four, and nothing about it is closed under the addition of a day-three value, since such a sum may be born on day six.

A solver that wants a stand-in for equality therefore has to choose what it is standing in for. To recognise whole games born by a day, the universe of the day before is exact and small, and the argument above says so for every pair at once. To add parts safely, the universe has to be closed under the additions it will see, and then it has to be larger than any listing. The cheap thing that does both is the one comparison is a search describes: play the difference $G - H$ out and see who wins. It is the negative-of-an-option argument run all the way down, and it needs no universe.

The convention named #

Normal play throughout, and outcomes computed by the recursion. A universe “separates” a pair when some context in it puts the two games in different outcome classes. The width chain starts from the 22 day-two values and adds one day-two value per step, so its $k$-th universe is every sum of $k$ day-two values, nought included, which is every sum of up to $k$. The day-four sample is the seeded one of the same parts, different wholes — 6,000 forms with up to two options a side drawn from the 985, canonicalising to 495 values — and the day-five sample is built the same way from it with a second seed, 1,500 forms canonicalising to 353. The counts at six and seven summands come from the same procedure run past the five drawn here.

What the measurements cannot show #

That the 147 are merged by every sum universe. Three widths in a row agree, which is what a stopped regress looks like and also what a slow one looks like on a sample. A pair could in principle be separated by some sum of twelve day-two values. The argument above does not reach this case, because it constructs a witness from the pair’s options, and the options here are not sums.

That the plateau is a property of day four rather than of this sample. The sample is 495 values of a day with far more, drawn with a fixed seed. A different seed would give a different number in place of 147. What it could not do is make depth fail on day four, since the argument covers every pair of games whose options lie in the universe.

And how far down the staircase goes. The day-three universe merges 86 of 353 day-five values, and the argument says the day-four universe would separate all of them; it says nothing about how many day-six values the day-four universe merges, and the counts on each step of the staircase need not be alike. What is fixed is the shape: exact one day on, and nothing promised two days on.

Still open: whether the plateau is a congruence #

The 147 pairs that every width merges are candidates for something specific. If no sum of day-two values ever separates them, then the relation “no sum of day-two values tells them apart” is preserved by adding a day-two value, by its definition — any sum added to a sum is a sum. That relation would be a congruence for the addition of small values that is coarser than equality, the same kind of object the misère quotient is for a single game’s positions. The measurement that would test it is negation and forming options on the 147, as the earlier essay ran them on the 567: negation should survive for the reason it survived there, since the sums are closed under it, and forming options is the clause where a relation that is closed under addition may still fail, because an option can be born a day deeper than any sum.