A foot is a menu, not a number

What a part would have to report gave each part of a sum its two stops and asked what those numbers could prove about the whole. Four inequalities settled 176 of the 369 day-two sums the outcome classes leave open, never wrongly, and the rest failed in a way that pointed at a single place. On 177 of them the fight ends level: the sum’s own stop is nought, the bounds from the parts are pinned on nought, and who wins is decided by whatever is left once the numbers have run out. That remainder is an infinitesimal, the part of a position a thermograph draws as a bare vertical mast, and a stop by its definition does not record it.

The repair that essay proposed was a second number, taken from the mast rather than from the fight. The subject already has one. How many ups is the atomic weight, the count of ↑ a small position is worth, and it does for the infinitesimal part what a stop does for the numeric part: it is the coarse reading a player actually makes. A part that reported two stops and an atomic weight would be reporting a number for each half of its value. The question here is whether that finishes the job. It does not, and the way it fails says what a part’s mast actually holds, which is not a number at all.

A table that cannot exist #

Every count on this page uses the same test. A report is some summary of a part. A table indexed by it would take the two parts’ reports and look up who wins the sum. Such a table cannot exist, for a given pair of reports, if two sums share that pair and have different outcomes: whatever the cell says, it is wrong about one of them. A sum is called undecided by a report when it shares its cell with a sum of another outcome. That is the lookup reading what a part would have to report used for its list of thermograph summaries, and it is the strict reading: a report that leaves a sum undecided cannot be rescued by a cleverer table, only by more information.

The pools are the ones that essay used. Day two is the 22 values born by then, every ordered pair of them 484 sums. The larger pool adds each of those 22 to each of the 1,474 values born by day three, 32,428 sums, the outcome of each found by searching the sum outright. The two stops leave 191 day-two sums undecided and 19,036 beside day three. Adding the outcome class of each part, one letter, brings those to 136 and 16,402.

The atomic weight comes next, and it is the surprise. On day two it leaves 159 sums undecided. That is better than the stops alone and worse than the stops with one letter of outcome. Beside day three it leaves 14,924, which beats the outcome letter, but not by the margin a second number ought to buy: the atomic weight cuts the day-three values into 376 classes against the outcome letter’s 45, and decides only 1,478 more sums with eight times the resolution. Whatever decides these sums, the count of ups is not most of it.

Three masts that weigh nothing #

The reason is visible on day two, where the masts can be listed. Seven of the 22 day-two values have both stops on nought, so the whole of each is mast: nought, ∗, ∗2, ↑, ↓, ↑∗ and ↓∗. Each is some number of ups with a nimber added, and the atomic weight is the number of ups.

Three of the seven weigh nothing. Nought, ∗ and ∗2 have atomic weight nought, and they are as different as day-two values get. Alone, nought is a loss for whoever moves and the two stars are wins for whoever moves. Beside a star, nought becomes ∗ and wins for the mover, ∗ becomes nought and loses for the mover, and ∗2 becomes ∗3 and wins again. The weight is the same, and between them the three cover two outcome classes alone and two beside the star, in a different order. What separates them is the nimber, and on day two the nimber is most of what a mast carries: four of the seven masts have a star in them, and only four have any weight.

This is not a defect of the atomic weight. It is what the atomic weight was built to ignore. How many ups found that comparison against multiples of ↑ pins the count exactly when no star is present and only brackets it when one is, because a star is smaller than every up and confused with nought. The weight is a reading on a scale, and a nimber is exactly the thing that is not on the scale. A day-two mast is mostly nimber, so a scale reading of it throws away most of the mast.

Where the weight decides and where it does not #

The two cold parts make this exact. Call a part cold when its two stops agree. A cold part is its stop plus an infinitesimal, and that infinitesimal is its foot, the position the fight ends on with the number taken away. Two cold parts add simply: the stops add, and so do the feet. If the feet are ups and nimbers, the ups add and the nimbers nim-add, so the outcome of the sum is the outcome of some number of ups plus some nimber, and that is a small table.

The table splits cleanly into two regions. Wherever the ups number two or more, or minus two or less, the sign of the count decides and the nimber is irrelevant: ⇑∗ is a win for Left, and so are ⇑, ⇑∗2 and ⇑∗3. That is the two-ahead rule in its smallest instance, and the name records the reason: a star can cancel the advantage of one up — ↑∗ is confused with nought — and no nimber cancels two. In that region the atomic weight is exactly the right report, and a table indexed by it never errs.

The other region is a band three rows wide, at nought ups and at plus or minus one. There the nimber decides and the weight is silent. At nought ups the question is only whether there is a nimber at all. At one up the question is whether the nimber is ∗ exactly. ↑∗ is $\{0, \ast \mid 0\}$, and whoever moves first plays to nought and wins, so it is a first-player win; ↑∗2 and ↑∗3 are wins for Left. So the weight decides when it is at least two, and below that the nimber decides. The day-two masts all weigh between −1 and 1, so a sum of two of them leaves the band only when both weigh the same nonzero amount. Everywhere else it lands inside the band, where the weight cannot see.

Replace the weight by the feet themselves, ups and nimber together, and two cold parts are decided exactly, because the sum’s value is then determined. On day two this decides every one of the 484 sums, and it is not just the cold ones: stops and feet together cut the 22 day-two values into 22 classes, which is to say they name the value. On day two the feet are the whole of the mast and the mast is the whole of the value. The interesting question is what happens a day later.

The same menu, a different waiter #

Beside day three, the feet leave 5,842 sums undecided. None of them is a sum of two cold parts: the 352 cold day-three values add exactly to the 15 cold day-two ones, as the grid says they must. Every failure has a hot part in it, a part whose stops differ, and the difficulty is that a hot part does not have a foot.

When Left moves first in a hot part and both players keep to the moves that hold the left stop, the fight reaches a cold position, and that cold position minus the stop is Left’s foot. But the moves that hold the stop are not always unique. At any point several options may reach the same number by different routes, and they need not reach the same foot. A hot part’s foot from either side is a set: a menu of the feet the fight can end on. Reporting the menu cuts the day-three values into 611 classes and leaves the 5,842.

A menu says what can be ordered and not who orders. The two parts above share stops, nought and −1, and Left moving first in either reaches 0 or ∗, with Left to move when it gets there. In $\{\{1 \mid 0, \ast\} \mid -1\}$, Left’s only move is to the switch $\{1 \mid 0, \ast\}$, and Right then chooses the foot: Right takes the 0, Left is to move on a bare nought and loses. In $\{\{1 \mid 0\}, \{1 \mid \ast\} \mid -1\}$, Left chooses which switch to play to, picks $\{1 \mid \ast\}$, Right is forced to play to ∗, and Left takes it. The menu is identical, the outcome is not, and the difference is a fact about the order of the choices: which player is at the table when the foot is decided. Reporting that, a menu with a chooser at each branch, cuts the day-three values into 671 classes and leaves 4,962 sums.

The same foot, a different player to move #

There is a second fact a foot needs, and what a part would have to report named it in its own section on convention: a stop discards who made the last move of the fight. The proposed repair was to keep one extra bit, the player to move when the number is reached. Taken alone it adds nothing on day two. The stops with that bit leave the same 191 day-two sums undecided as the stops without it, because every day-two fight ends within one move, so the stops already say who is to move. Beside day three it leaves 17,088, which is better than the stops and worse than the outcome letter.

What that bit does do is mark the feet. $\{\{1 \mid \ast\} \mid -1\ast\}$ and $\{\ast \mid -1\ast\}$ have the same stops, nought and −1, the same foot from each side, ∗ and −1∗, and nobody chooses anything on the way. But Left reaching ∗ in the first takes two moves, Left to the switch and Right to the star, so Left is to move on the star and takes it. In the second it takes one, Left to the star, and Right takes it. The foot is the same position, and a foot with the move belongs to whoever has the move. Marking each foot with its player to move cuts the day-three values into 725 classes, and the undecided sums fall to 3,164.

So the mast report that works on every sum of two cold parts, and on most sums with a hot one, is not a number. It is a small tree: the feet each side’s fight can end on, which player chooses at each branch, and which player is to move at each leaf. That is a game, a little one, carved out of the part along the lines that hold its stops. The atomic weight is one reading of the leaves of that game, and the reason it fails is that the tree has two other kinds of information in it, and both decide outcomes.

What the mast closes, and where it stops #

The lead was a specific question about the 177 day-two sums whose fight ends level on nought, and the answer can now be given exactly.

As an index, the stops leave 160 of the 177 undecided. The outcome letter brings that to 136. Who moves at the foot brings it nowhere, since on day two it is already implied. The atomic weight brings it to 134, two sums better than the letter. The feet bring it to nought. So a second number from the mast does very little, and a second report from the mast, the right one, closes all 177. Beside day three, among the 15,150 sums that end on nought, the progression is the same shape and does not reach the floor: 14,570, then 13,744, 14,370 with the move bit, 11,592 with the weight, 4,076 with the menu, 3,490 with the chooser and 2,270 with the player to move.

The two right-hand columns say where the remainder lives. The stops, the outcome letter, the move bit and the weight all leave sums of two cold parts undecided: 2,140 of them for the stops and still 938 for the weight. As soon as a report names the feet, the cold column is zero and stays zero. Every one of the 3,164 sums the full tree leaves has at least one hot part, and 584 of them are the straddling sums the stops could not handle either, where the order of play across two real fights decides which side of nought the sum ends on.

Every report agrees, and the company disagrees #

That remainder is not a gap in the report. It is a limit on what any report computed from a part alone can contain.

$\{2 \mid \{1 \mid \ast\}\}$ and $\{2 \mid \{1 \mid 0, \ast\}\}$ have the same stops, two and one, and played alone their fights run the same line. Left moving first takes the 2. Right moving first plays to the inner switch, Left takes its 1, and the fight ends on 1 with Right to move. The menus, the choosers and the tempi agree, both parts are wins for Left alone, and the full tree calls them identical.

Add $\{0 \mid -1\}$ to each and they come apart. Right moves first and plays to the inner switch, as alone. But Left now has a better answer than taking the 1: it can play $\{0 \mid -1\}$ to nought instead, a cooler move elsewhere, and leave Right to move inside the inner switch. Played alone, that never happens. The part’s own fight has Left move there, because Left has nowhere else to go. In company Right is handed a move at a node the part’s fight never visits. At $\{1 \mid \ast\}$ that move is worth only ∗, Left takes the star, and Left wins. At $\{1 \mid 0, \ast\}$ Right takes the 0, Left is to move on nought and loses. One sum is a win for Left and the other a win for whoever moves.

That is the shape of all 3,164. A report computed by playing a part’s fight out alone records the lines that fight takes, and in a sum the fight does not take those lines: a player can answer in another part, and the opponent is then to move at a position the part alone would only ever have handed to the other side. Whatever happens at those positions is part of the value and not part of any foot. What happens there is exactly what which end a sum lands at and what a fight does to a fight measured from the numeric side, the interleaving of fights, and this is the same thing seen from the mast: the infinitesimal a hot part leaves depends on who arrives at it, and who arrives depends on the company.

A name, in two halves, that is not two numbers #

There is one connection worth drawing out, because it closes a loop the subject has been running for several essays. What a part would have to report found that on day two a part is named by four outcomes, who wins it alone, beside −1, beside 1 and beside a star, and read the name as two halves: the numbers place the part on the scale, and the letter beside ∗ is the half no thermograph carries. This essay has tried to split the same name the other way, as a number for the fight and a number for the mast, and found that the mast half is not a number. On day two it is an up count and a nimber, which is two numbers of quite different kinds, one on a scale and one off it. A day later it is a tree with choices and tempi in it.

The atomic weight is the scale half of the mast half. The nimber that the star in the four-question name detects is the other half, and on day two the four questions and the feet agree for the same reason: both report the nimber, and the atomic weight does not. The stops and the weight are two scale readings of a value whose information is mostly not on a scale.

The convention every count depends on #

All of it is normal play, where the player unable to move loses, and the outcome of a cold sum is read off ups and nimbers only because under normal play ∗ + ∗ is nought and ↑ is positive. Under misère play none of those facts hold, and nothing here says what a misère mast would be. The stops are read from each value’s thermograph at no tax, as reading a thermograph defines them. A foot is defined by following every option that keeps a side’s stop and nothing else, so a move that gives up a fraction of the stop in exchange for a better foot is not on the menu. That is a choice, and it is the right one for the question the lead asked, which was about what sits at the end of the stop-holding fight; a report built from every line of play would no longer be a report about the mast.

The lookup counts are counts against these two pools. A report that leaves a sum undecided here leaves it undecided anywhere the same two sums occur, but a report that decides every sum here could fail against a larger pool, and the full tree decides every day-two sum only because day two is small. Each ups-and-nimber reading was found by testing every combination of up to six ups and the nimbers to ∗7 for equality with the foot, and day three’s feet include infinitesimals that are of neither shape; those are reported by their canonical form, which is exact and is not a reading.

What the tree cannot show #

The tree is drawn from a part’s own fight, and its whole failure is that a sum’s fight is not the sum of the parts’ fights. What it cannot show is the set of positions the opponent can be handed in company. A report that included those would have to include, for each node of a part, what happens when the other player moves there instead, and at that point it is the part’s canonical form under another name. The 3,164 are evidence that no report short of that exists, not proof: a cleverer summary of the off-line nodes might decide them, and nothing here searched for one.

The counts also cannot say how fast the remainder grows. Beside day three it is 3,164 of 32,428, under ten per cent. Whether it is a fixed share of the hot sums, or whether it grows with the depth of the fights until the tree is no smaller than the value, is a question about day four, where neither pool can be enumerated.

Still open: which nodes a partner can reach #

The company failure has a structure that suggests a measurement. In the pair above, the node that mattered was one level below the part’s stop-holding line, the inner switch with Right to move, and the partner that reached it was $\{0 \mid -1\}$, a part exactly as hot as the switch. That suggests the off-line nodes a partner can reach are bounded by the partner’s temperature: a cold partner reaches none, which is why two cold parts never fail, and a partner at temperature t reaches only the nodes where the part’s own fight pauses at a temperature at or below t. If that holds on the 3,164, the report that finishes the job is the tree plus the off-line nodes down to a stated temperature, and the size of the report becomes a function of how hot the company is. The measurement is to take each undecided sum, find the shallowest node where the two parts’ sums first diverge, and set its temperature against the partner’s. A rule there would turn this essay’s negative into a graded positive: what a part has to report, as a function of who it will be added to.