What a part would have to report

Outcomes do not add filled in the table a player would most like to have — the outcome of a sum, looked up from the outcomes of its parts — and found it four-tenths empty. Over every sum of two values born by day two, six of the ten entries come out one way every time, and they are exactly the six the order forces: two positions Left wins whoever moves add to one Left wins whoever moves, and a second-player win added to anything changes nothing. The other four entries, L + R, L + N, R + N and N + N, come out more than one way. That essay drew the moral the whole subject draws — values exist because outcomes do not add — and moved on to what a value has to be.

Moving on skipped a question in between, and it is the practical one. Nobody analysing a real board reports a canonical form for every region. They report something smaller: roughly what the region is worth, how hot it is, who wins it if it is left alone. The question is whether any summary of that kind — more than the outcome, less than the value — is enough to fill in the four open entries. There are two different ways to ask it, and they get different answers. A summary can be used as evidence, from which the outcome of a sum is deduced where the deduction goes through and left open where it does not. Or it can be used as an index, the key to a table whose entries are simply looked up. The first asks what a summary can prove; the second asks whether any table indexed by it could exist at all.

Four inequalities a sum cannot escape #

The first candidate is the pair of numbers every position carries and every player estimates: its two stops. The left stop is the number reached when Left moves first and both players keep taking the biggest thing on offer until a number is all that is left; the right stop is the same with Right moving first. They are the feet of the thermograph, and the fight never runs backwards established the one thing they always satisfy: the left stop is never below the right.

What the stops of a sum are is not a function of the parts’ stops. What a fight does to a fight measured how far off the obvious guess is — adding the stops is right on fewer than half of the hot sums it tried — and which end a sum lands at found the rule that says where in its allowed range each one falls. What the parts’ stops do give is a range, and it is four inequalities long. Writing $L$ and $R$ for the two stops:

$$R(G) + R(H) \;\le\; R(G+H) \;\le\; \min\big(L(G) + R(H),\; R(G) + L(H)\big)$$

$$\max\big(L(G) + R(H),\; R(G) + L(H)\big) \;\le\; L(G+H) \;\le\; L(G) + L(H)$$

The idea behind them is a strategy each player can always fall back on. A player who moves first in one component and afterwards answers every move in whichever component it was made in collects, in each component, at least what that component gives when moved first in one and second in the other. That is where the mixed terms come from. The outer terms are the same argument from the other player’s side: nobody can get more out of a sum than moving first in both parts would give them, and nobody can be held to less than moving second in both. Every one of the 32,428 sums checked below has its solved stops inside these limits.

A range is enough to decide an outcome whenever it keeps clear of nought. If the least the left stop of the sum can be is above nought, Left moving first reaches a positive number and wins; if the most it can be is below nought, Left moving first loses. The right stop answers the same question for Right moving first. So each part reports two numbers, the sum’s two ranges are computed from them, and each range either settles who wins moving first or says nothing.

The picture reads six sums that way. $\{2 \mid 0\} + \{0 \mid -1\}$ has a left-stop band from 1 to 2, so Left wins moving first, and a right-stop band from −1 to 0, which touches nought and says nothing; the sum is a win for Left whoever moves, and the stops have proved half of that. $\{1 \mid 0\} + \{0 \mid -1\}$ touches nought on both sides and is nought. The sum of $\{1 \mid -1\}$ with itself has two wide bands, from 0 to 2 and from −2 to 0, and is nought too, because the second player copies the first. And the fourth and fifth rows are the pair the whole essay turns on: $-1 + 1{\ast}$ and $-1 + 1$ have identical bands, both pinned exactly on nought, and one is ∗ — a win for whoever moves — while the other is nought. The stops of a number with a star on it are the stops of the number, and nothing a stop can say will ever tell those two sums apart.

Half the open table, and never a wrong answer #

Over all 484 sums of two day-two values, 115 have outcomes the classes already force and 369 fall in the four open entries. The stops settle 176 of the 369, and not one of them wrongly. The settled share is not spread evenly. L + R is settled on 44 of its 72 sums, mostly those whose parts are numbers or close to numbers, where the ranges collapse to a single value. L + N and R + N are settled on 62 of 108 each. N + N is settled on eight sums of 81, and all eight are one pairing: $\{1 \mid -1\}$ beside one of the four values whose stops are both nought — ∗, ∗2, ↑∗ and ↓∗ — in either order. The switch’s bands run from 1 to 1 on the left and −1 to −1 on the right, the other part adds nothing to either, and a first-player win is proved.

That $\{1 \mid -1\}$ is the only day-two first-player win to get settled with anything is not an accident of the pool. A first-player win has its left stop at or above nought and its right stop at or below, and on day two eight of the nine first-player wins have one stop sitting exactly on nought: the four star-like values on both, and $\{0 \mid -1\}$, $\{1 \mid 0\}$, $\{1 \mid 0, \ast\}$ and $\{0, \ast \mid -1\}$ on one each. A sum of two of them has a band anchored at nought by construction. The stops of the young first-player wins are too close to the edge to prove anything.

The more striking fact is what the other 193 sums have in common. Every one of them has a bound lying exactly on nought, and not one has bounds on both sides of it. The stops never fail on day two by leaving the fight undecided — by allowing the sum’s stop to be comfortably positive or comfortably negative depending on how the two fights interleave. They fail only by reaching nought and stopping there.

Nought, two ways #

A bound on nought can mean two quite different things, and the next reading separates them.

On 177 of the 193, the sum’s own stop, solved directly, is nought as well. The fight ends level and the outcome is decided by what is left after it — whether the position at the foot of the fight is ↑ or ∗ or nought or ↓∗ — and that is the part of a position a stop by its definition does not record. What an infinitesimal does to a fight measured the other face of the same fact: adding something smaller than every number moves neither stop and still changes the outcome class, and it changes it at exactly the positions with a stop on nought. Those 177 sums are the positions it was talking about, arrived at from the direction of the sum rather than of the part. Even the sum’s own exact stops would not settle them, never mind the parts’.

On the remaining 16, the bound touches nought and the sum’s stop does not. $\{1 \mid 0\} + \{1 \mid 0\}$ is the plainest: its right-stop band runs from nought to one, and the sum is the number 1, whose right stop is one. The parts’ stops were not enough to pin the sum’s, and the true stop happened to sit at the far end of its allowed range. Those sums are a failure of the inequalities rather than of the stops — a sharper rule for where a sum lands, like the one which end a sum lands at found, could in principle have settled them. The 177 are different in kind. No rule about stops can settle them, because the information is not in the stops at all.

The failure day two never shows #

Day two is small, and its values have stops on the integers and halves. A second pool asks whether the clean picture survives: each of the 22 day-two values added to each of the 1,474 born by day three.

It mostly does. The stops settle 10,497 of the 28,329 open sums and are never wrong; 17,216 have a bound on nought, and of those 15,150 have their solved stop on nought as well. But 616 sums fail in the way day two never does: a range lying across nought, from a negative number to a positive one, so that the order of play inside the two fights decides which side the sum ends on. The last row of the number line is the smallest of them. $\{0 \mid -1\} + \{2 \mid \tfrac12\}$ has a right-stop band from $-\tfrac12$ to $\tfrac12$. Right moving first could hope to take the hotter part first and leave Left a cold fight, or play the cooler part and leave Left the hotter one; the inequalities cannot tell which of those the position allows. The solved answer is $\tfrac12$, and the sum is a win for Left.

A straddle needs both parts to have a gap between their stops, and a wide one: rearranging the three strict inequalities shows that each part’s gap has to exceed the amount by which the two right stops together fall short of nought. The widest gap on day two is two, on $\{1 \mid -1\}$, and two of those fall short by exactly two, so day two cannot produce a straddle at all. Day three has wider gaps and stops on the halves, and 368 of its 616 straddles are two first-player wins, the entry where both parts are genuinely fights. So the stops have two ways to fail, and the second one arrives with temperature: first the infinitesimal on the mast, which no number reports, and then, a day later, the interleaving of two real fights, which the four inequalities cannot resolve.

A summary as a lookup table #

Used as evidence, the stops settle what they can prove and leave the rest. The other way to use a summary is as the index of a table, and the question changes: not what can be deduced but whether a table could exist. A table indexed by some summary of each part is impossible exactly when two sums share their pair of summaries and have different outcomes — whatever the table says for that pair, it is wrong about one of them.

The list starts at the outcome class, which cuts day two into four classes and leaves every one of the 369 open sums undecidable — the finding of the essay below this one, restated. The two stops cut day two into ten classes and leave 191. Adding the outcome to the stops gives fifteen classes and leaves 136. The four headline numbers of a thermograph — both stops, the mean and the temperature — give thirteen classes and leave 150, which is worse than the stops with the outcome, because the mean and the temperature of a day-two value are almost always determined by its stops, while the outcome is the one fact the stops cannot see. All of those with the outcome as well gives seventeen classes and leaves 96. The value itself gives twenty-two and leaves none.

Two things are worth drawing out of that column. The first is that the lookup numbers are not the evidence numbers. As evidence the stops settle 176 of 369 and leave 193; as an index they leave 191. The two are close and are counting different things — a sum can be undecidable by deduction and still sit alone in its table cell, or be deducible and share a cell with a sum it disagrees with — and it would be a mistake to read either as the other.

The second is the shape of the list. Every summary short of the value leaves sums undecided, on day two and beside day three alike, where the richest summary still leaves 14,502 of 32,428. There is no plateau where the table fills in before the value is reached, and the reason is the same as in the evidence reading: every summary on the list is a thermograph reading, and a thermograph cannot see anything on its mast.

Four parts with one thermograph #

The bluntest way to see that is to find parts the richest summary calls identical and watch them disagree.

∗, ↑∗, ∗2 and ↓∗ all have both stops at nought, mean nought and temperature nought, and all four are first-player wins. Their thermographs are the same single mast standing on nought. Add a star to each and the four sums are nought, ↑, ∗3 and ↓ — a second-player win, a win for Left, a first-player win and a win for Right. One entry of the richest table on the list, one partner, and every outcome there is.

That settles where the ninety-six undecided sums live. They are not stragglers at the edges of a table that is nearly right. They are the values whose fight is already over, and the whole remaining question about them is the one who moves last is about: which player runs out of moves first once the numbers have nothing left to say. How many ups is the subject’s measure of that, and it is not on the thermograph.

Four questions that name a value #

At the other end of the list is the value, which decides everything and costs a canonical form. The last reading asks how cheaply the value itself can be reported — and finds that it can be written as a handful of outcomes.

The idea is to report, for each part, the outcomes of a few fixed sums — the part alone, the part beside some chosen companions. If those outcomes differ for every pair of day-two values, then the report is the value on day two, and every sum of two reported parts follows. Each outcome is one of four letters, so twenty-two values need at least three questions. No three companions work: all 1,540 sets of three day-two values were tried, and every one leaves two values with the same answers. Of the 7,315 sets of four, exactly one works, and it is the one a reader might write first: nought, −1, 1 and ∗. A day-two value is named by who wins it alone, beside a minus one, beside a one and beside a star.

The companions have a reading that explains the choice. Who wins $G + H$ is the comparison of $G$ with $-H$, so the four questions ask how the part compares with nought, with 1, with −1 and with ∗ — above, below, equal or confused with each. The numbers place a part on the scale; the star is what the scale cannot see. It is the same division the lookup tables found, turned round. The stops are the numeric half of the name, and the letter beside ∗ is the half no thermograph carries — which is why ∗ and ↑∗ get words that differ only in their last letter.

It is a day-two fact and it says so. The same four questions sort the 1,474 day-three values into only twenty-two words; the name has to grow with the pool. What does hold on day three is the larger statement: all twenty-two day-two values, used as company, give every one of the 1,474 day-three values a different string of outcomes. Equal in every company and the same parts, different wholes study that relation directly — two games are equal when no company tells them apart — and the finding here is its cheapest instance: on day two, four companions are all the company there is.

Where the stops were measured, and against what #

Every figure rests on the same objects. The day-two pool is the 22 canonical values of the 256 forms with day-one options, and the day-three pool is the 1,474 canonical values whose options are antichains of day-two values. A sum is formed from the canonical forms of its parts and its outcome decided by searching it outright. Stops are read from each value’s thermograph at no tax — the left wall’s foot is what Left gets moving first — which is the definition reading a thermograph uses, and the bounds are the four inequalities above, applied to those numbers. A sum is settled only when both of its ranges keep clear of nought, and it is counted as wrong if the outcome that settling implies differs from the searched one; none does.

The lookup counts group every sum by the pair of its parts’ summaries and count the sums in groups holding more than one outcome. The companion search is exhaustive over every set of one, two, three and four day-two values, and it stops at four because a set of four is found.

The convention the stops depend on #

A stop is the number a fight ends on, and it discards who made the last move of the fight. That is a convention, and every count above depends on it. The stops of $-1 + 1{\ast}$ and of $-1 + 1$ are both nought because the first ends on the number nought with a star still to play and the stop does not keep the star. A stop that kept one extra fact — which player was to move when the number was reached — would split some of the 177 sums on nought, and it would be a summary somewhere between the stops and the value on the list above. It is not tried here. The counts are also for normal play, where the player unable to move loses; in misère play the stops are not even the right objects, and nothing here tests the four inequalities there.

What four inequalities cannot carry #

The reading is exact and its reach is short. The pools are day two, and day two beside day three; the sums are of two parts; and the list of summaries is one choice of what a reader of a thermograph might record, not a survey of every function of a value. A cleverer deduction from the same two numbers — one that used which end of its range a sum lands at — would settle some of the sixteen slack sums and cannot settle any of the 177, because there the missing fact is not a number. The four-question name is a statement about twenty-two values and fails, as it should, on the next day. And the lookup counts are relative to their pools: a summary that leaves a sum undecided here might decide every sum within some narrower family, and the page does not look for such families.

Still open: what is left on the mast #

The stops fail in two ways, and one of them has a candidate repair. On the 177 day-two sums where the fight ends level at nought, the outcome is decided by the infinitesimal part of each component — the part a thermograph draws as a bare mast. The subject already has a number for that part: the atomic weight, which counts ups the way a stop counts points and which how many ups introduces. A part that reported its two stops and its atomic weight would be reporting a numeric summary of both halves of its name. Whether that closes the 177, and what it does beside day three, where temperature adds the straddling failure the stops cannot see, is the next thing to measure — the question of whether a second number, taken from the mast rather than from the fight, finishes the job the first two started.