What an infinitesimal does to a fight

What a number does to a fight measured the predictable case. Add a number $x$ to a position and both stops move by exactly $x$, the temperature is untouched, and the outcome follows the stops — $G + x$ is a first-player win exactly while $x$ lies strictly between the negated stops. It closed by naming the other side: what adding something smaller than every number does, where the stops say nothing at all and the outcome moves anyway.

Ten thousand three hundred and eighteen additions. Stops moved: nought.

That is a number worth pausing on, because a sweep of this size reporting a clean zero is usually a sign that the wrong thing has been measured. It is not, and the reason is that the claim is a theorem rather than a tendency: the pair of stops is computed by a recursion that stops at numbers, and every shift in the sweep is a position no number reaches. The count is there to confirm that the code computing the stops is computing the thing the theorem is about.

Why the stops cannot see it #

This is not a surprising result and it is worth being precise about why, because the precision is what makes the rest of the essay a measurement rather than a curiosity.

A stop is defined by a recursion that halts at a number. $LS(G)$ is the largest $RS$ over Left’s options, and $RS(G)$ the smallest $LS$ over Right’s, and both are $G$ itself when $G$ is a number. So the whole computation lives among numbers, and an infinitesimal — a position lying strictly between $0$ and every positive number — is invisible to it by construction.

Adding one to a position therefore leaves the number the fight settles at exactly where it was. The fight is the same fight; the pair of stops is the same pair; and the temperature is almost always the same too, moving in between sixteen and twenty-two of each 1,474 additions.

Every shift used here is checked to be an infinitesimal before it is used, by computing its stops and refusing any whose stops are not both zero. A sweep about infinitesimals that quietly included something else would be reporting nothing at all.

And the outcome moves anyway #

Twenty-seven per cent of the time, for a star. Twenty-five for an up or a down. Forty-six for a double up. Thirty-three for an up with a star on it. Twenty-one for a nimber one step further out.

Two thousand six hundred and twenty-two changes of outcome class, from additions that moved neither stop and rarely moved the temperature.

That is the whole argument for the existence of a second theory. If the stops were the answer, none of these additions could change anything; a quarter of them change who wins.

Where it can happen, exactly #

The changes are not scattered. Sorting the 1,474 values by where their stops sit relative to zero and adding an up to each gives a table with two empty rows.

Stops straddling zero strictly — 198 values, and not one changes. The position is already a first-player win by a clear margin; an infinitesimal is far too small to close a gap the numbers have opened.

Stops clear of zero on one side — 124 values, and not one changes. The position is a comfortable win for one player, and again the infinitesimal cannot reach.

A stop sitting exactly on zero — 1,152 values, of which 374 change.

So the rule is exact: an infinitesimal matters precisely where a stop is zero, which is precisely where the numbers have finished speaking. It is the same boundary the confusion interval has at its two ends, arrived at from the other direction.

Twelve directions, and one that should not exist #

The transitions were tallied to see what an infinitesimal can turn a position into, and there are twelve distinct kinds.

The common ones are unremarkable: a first-player win becoming a win for Left, 1,264 times; a win for Right becoming a first-player win, 539 times. Those are positions balanced on a stop, tipped one way by something small.

Two of the twelve are not what a reader expects. A win for Left becomes a win for Right six times, and a win for Right becomes a win for Left thirty times.

That looks impossible. To turn a win for Left into a win for Right, something has to cross the whole of the numbers — and an infinitesimal is smaller than every number.

The resolution is that it does not have to cross the numbers, because the position was never a comfortable win. A value like $\uparrow$ is a win for Left with both stops at zero: Left wins not by any margin at all, but because an infinitesimal is on Left’s side. Add a double down and the infinitesimal on Left’s side is outweighed by an infinitesimal on Right’s, and Right wins. Nothing has crossed any number, because nothing was ever on the far side of one.

That is the sharpest thing the sweep has to say. Wins can be reversed by quantities smaller than every number, and the positions where it happens are exactly the ones whose margin was infinitesimal to begin with.

The transition nobody predicted, and the one that was predicted wrongly #

Before the tally was run the expectation written down was that two of the twelve would be empty — a win for Left could not become a win for Right, and the reverse — on the argument that an infinitesimal cannot cross the numbers.

Both occur. The expectation was wrong, and the reason it was wrong is the useful part: it confused a win with a win by a margin. Every position whose outcome an infinitesimal can change has a margin of zero, and a margin of zero is not a small number, it is no number at all.

Recording the failed prediction matters more than recording the result. A sweep that only confirmed what was expected would have said nothing about the expectation; this one says the intuition behind it is the wrong intuition, and the wrong intuition is the one a reader arrives with.

Two positions, and the difference between them #

$\uparrow$ is a win for Left. Its stops are both nought, so no number distinguishes it from zero, and every figure in the temperature theory reports it as frozen.

$\ast$ is a first-player win. Its stops are both nought too, so on the same evidence the two positions are identical.

Add a double down to each. $\uparrow + \Downarrow$ is $\downarrow$, a win for Right; $\ast + \Downarrow$ is a win for Right as well. Add a single down instead and the two part company: $\uparrow + \downarrow$ is zero, a second-player win, and $\ast + \downarrow$ is a first-player win.

So a single addition separates two positions that every number-valued measurement on this site reports as the same, and the addition is of something no number reaches. That is one line of the 10,318, and it is the line the whole sweep exists to make countable.

The scale, made visible by the failures #

The seven shifts do not behave alike, and the differences are exactly the scale tiny, miny and the sizes below every size is about.

A double up changes 679 outcomes and a single up changes 374. Twice the infinitesimal is nearly twice as effective, which is what a scale means.

Tiny-two changes one. One, out of 1,474.

Tiny-two is an infinitesimal below every multiple of up — smaller, in the ordering, than $\uparrow$ divided by any number of times a reader cares to name. So it is too small to affect almost every position whose margin is infinitesimal, because those margins are themselves multiples of up. The single value it does move is the one whose margin is smaller still.

That single count is the argument for atomic weight in miniature. How many ups a position is worth is a measurement on a scale, and the reason the scale is needed is that “infinitesimal” is not one size: it is a whole ordered family, and a member of it can be enormous or negligible relative to another.

What the star does that the ups do not #

A star changes 394 outcomes, which is between the counts for an up and a double up — but it changes them in a different pattern.

The ups and downs are ordered: up is positive, down is negative, and adding one moves outcomes towards Left or towards Right respectively. Their transition tables are mirror images and every entry is a move in one direction.

A star is confused with zero, so it is neither positive nor negative, and its transitions come in symmetric pairs: 98 wins for Right become first-player wins and 98 wins for Left do the same; 98 first-player wins become wins for Left and 98 become wins for Right. The count is symmetric because the day is closed under negation and the star is its own negative.

Split every change by which player gains ground and the difference becomes a column rather than a remark.

So an up tilts a position and a star destabilises it. That is the practical difference between the two commonest infinitesimals and it is not visible in either’s size — the star moves more outcomes than the up does, and moves them nowhere in particular.

The row worth reading twice is $\uparrow\!\ast$, which is both at once. It is an up with a star on it, so it is positive and confused, and it moves 435 outcomes towards Left against 55 towards Right — a tilt with an unbalancing on top of it, and the only row in the table that is neither one habit nor the other.

The temperature, which moves a little #

One column of the census is easy to skim past and it is the one that stops the essay being tidier than the truth.

Adding an infinitesimal changes the temperature between sixteen and twenty-two times per shift, out of 1,474. A star does it most often, at twenty-two.

That is a small number and it is not nought, and the difference matters. The stops are the walls of a thermograph at zero tax, so a shift that never moves the stops might be expected never to move anything about the diagram. It moves the temperature because the temperature is where the two walls meet, and the walls are built from the options — an option changed by an infinitesimal can change where a wall bends, and a bend can move the meeting point without moving either foot.

So the honest statement is that an infinitesimal is invisible to the stops and very nearly invisible to the temperature, and the residual visibility is a fact about the shape of the diagram rather than about its base. It is the same distinction a thermograph with two bends is built on.

Reading it as a board #

The whole sweep is a statement about play, and the statement is short.

A board consisting of a fight and a small extra component is worth the fight if the fight is decisive. If the fight ends level — if the stops meet, which is the commonest situation in this day, 1,152 values of 1,474 — then the small component decides the game, and which small component it is matters.

That is why a Go endgame is analysed by temperature down to the point where the temperature is zero and by atomic weight below it. The two theories divide the board exactly where this sweep divides the day: numbers while the numbers say something, infinitesimals once they stop.

What a quarter means #

The headline number — a quarter of the additions change the outcome — is a proportion over a particular pool, and it is worth saying what the pool is doing to it.

The values born by day three are overwhelmingly balanced. Of the 1,474, some 1,152 have a stop sitting exactly on zero, which is 78 per cent, and it is among those that every change happens. So the quarter is really a third of the balanced positions, and the balanced positions are most of the day.

That proportion is a fact about the construction rather than about games. The construction builds each day out of subsets of the previous one and it is symmetric in the two players, so it produces balanced positions in quantity; a real game’s positions are not drawn that way and would give a different figure entirely.

What does transfer is the shape of the finding: the changes are confined to the balanced positions, and within them they are common. A player whose board has come down to a level fight and a handful of small components is in the region where the sweep says the small components decide, and the proportion of such boards in a real game is a question about that game.

What a stop cannot see, and why that is the right amount #

The finding here is a negative — the stops do not move — and it is worth reading as a statement about what a stop is for rather than as a limitation.

A stop is the number a fight settles at when both players play it out. It is computed by a recursion that halts the moment somebody faces a number, so the whole of its content is where the fighting ends. An infinitesimal is by construction smaller than every positive number, so it can never change where the fighting ends: whatever it does, it does after the numbers have been settled and below the resolution at which a stop is measured.

So a stop being blind to an infinitesimal is not a defect in stops; it is what an infinitesimal is. A quantity defined as smaller than every number cannot be visible to a measurement whose output is a number, and any instrument that did see it would not be measuring a stop.

That says exactly where the information goes instead. The infinitesimal survives in the outcome, which is not a number and can turn on anything, and the sweep on this page is the demonstration: thousands of shifts, no stop moves, and the outcome moves constantly. Two quantities, one blind and one not, both correctly reporting what they are for.

And it says which of the two a player should hold in a close endgame. A count of territory is a statement about stops and is complete about the fighting; whether it is enough depends on whether the total lands exactly on a boundary, and on a board of infinitesimals it always does. That is why the atomic weight is the instrument below this one, and why a player who stops at the count has stopped one layer above the answer.

The convention, and what the sweep cannot say #

Normal play throughout, and every count is over the values born by day three, which is a fixed and completely enumerated pool. That is a strength — the counts are not a sample — and a limitation, since the day is closed under negation and under small sums, which makes it unusually well behaved.

Three things are outside it.

The sweep adds one infinitesimal at a time. Adding two is a different question and it is not the sum of two answers, since infinitesimals of the same sign reinforce and of opposite signs cancel.

The pool has no positions with large infinitesimal parts, because a day-three value cannot have one. So the 46 per cent figure for a double up is a fact about small margins, and a pool of positions with margins of ten ups would produce different numbers everywhere.

And nothing here says which infinitesimal to reach for. The sweep measures how often each of seven changes an outcome; a player wanting to know which small component wins a given board needs the comparison rather than the frequency, and that is a search rather than a table.

Where the ladder goes next #

The translation anchor has two rungs to here: what a number does to a fight, and what an infinitesimal does.

What a fight does to a fight is the third addend and the first one with two moving parts, where the stops of a sum stop being a translation of anything. A bound with one number too many then finds the natural bound on that error to be looser than it needs to be: the follow-up’s temperature does not belong in it, and the bound without it is attained 358 times over 1,440 sums, which is what settles the question rather than any count of violations.

The two rungs above that describe the error rather than bounding it. Which end a sum lands at gives a four-line rule, exact on all 1,440, saying which end of the permitted range a given pair falls on — and three of its four cases are decided by the value being translated alone, by whether its wall bends, which is the same feature the switches ladder isolated for a different question entirely.

And the same number in two currencies measures the shortfall for the bent-walled values: it is the value’s own hottest follow-up’s temperature, exactly, on 400 of 408 pairs and twice it on the other eight. So the whole error becomes one expression, and the expression is the switches ladder’s constant re-denominated — a half there and a whole here, because a temperature is half a gap between stops.

Read in order the anchor goes from a number moves both stops and changes nothing else to a fight moves them by an amount computable from one follow-up, and the quantity that turns out to run through all of it is the bend.