Which end of the interval is open

A position and a number can stand in four relations, and the pair of stops decides three of them. Above the left stop the number wins; below the right stop the position does; strictly between the two, the number and the position are confused. That is the rule the numbers it is confused with established, over 36,850 comparisons with nothing wrong where the rule speaks.

The rule declines to speak at the endpoints, and it declines symmetrically — the same silence at each end. That symmetry is the thing this rung takes apart.

At its own left stop, a position is above the number thirty-three times in 1,474 tries. At its own right stop it is below the number thirty-three times. Everywhere else the two ends behave completely differently.

Why the left end cannot be open upwards #

The argument is one line of arithmetic and it is worth having before the counts, because the counts are then a check on it rather than a discovery.

Stops translate. For a number $x$, $\mathrm{LS}(G - x) = \mathrm{LS}(G) - x$ and likewise on the right — adding a number to a position shifts both ends of its fight by that number and changes nothing else, which is what what a number does to a fight measured over ten thousand translations.

So set $x = \mathrm{LS}(G)$ and consider the difference $D = G - x$:

$$\mathrm{LS}(D) = 0, \qquad \mathrm{RS}(D) = \mathrm{RS}(G) - \mathrm{LS}(G) \le 0.$$

The second inequality is the fight never runs backwards: the right stop is never above the left one.

Now suppose $D > 0$, which is what $G > x$ means. Then Left wins $D$ moving second. But Right moving first drives the play down to $\mathrm{RS}(D)$, which is at most zero, and if it is strictly below zero then Right has reached a number Left cannot be winning at. So $D > 0$ needs $\mathrm{RS}(D) = 0$ — which needs $G$’s two stops to be the same number.

That is the whole of it. The law is at its own left stop a position is never above the number, and the exception is exactly the class where the left stop is also the right stop and the sentence has stopped saying anything.

The exception class, counted #

Three hundred and fifty-two of the 1,474 values born by day three have $\mathrm{LS} = \mathrm{RS}$. They are the positions with nothing to fight over: every number is a member, since a number’s two stops are itself, and so is every infinitesimal, and so is a large class of positions in between.

Fifteen are equal to the stop, and those fifteen are exactly the numbers born by day three. Thirty-three are above it and thirty-three below, the two counts equal because negation swaps them. And 271 are confused with it.

The thirty-three above are the whole of the anomaly in the hero figure, and they are not exotic. $\{\ast 2 \mid 0\}$ has both stops at zero and is strictly below zero; its negative is strictly above. $\{\uparrow\!\ast \mid \downarrow\}$ has both stops at zero and is below it too.

What the difference actually is #

At a stop where the two coincide, $D = G - x$ has both its stops at nought. A game with both stops at nought is an infinitesimal — smaller in absolute value than every positive number — and the temperature theory has nothing whatever to say about it. Its thermograph is a vertical mast at zero.

So the question which of the four relations arrives has been handed from one theory to another. Stops and temperature answer it for the 1,122 values whose stops are distinct, at both ends, without ever having to look inside. For the 352 they say nothing at all, and something else has to.

The class is the cold end of the collection #

There is a clean structural description of the 352 and it is not the one a reader guesses.

The guess is the all-small games, since those are the positions with no material in them at all. Sixty-seven of the class are all-small, which is well under a fifth of it. Another fifteen are numbers. The remaining 270 are neither: $\{\{0, \ast \mid -1\} \mid -1\}$ has a Right option worth $-1$ and no matching Left option, so it fails the all-small condition outright, and its two stops are nevertheless both $-1$.

What the class is — exactly, and asserted in the code rather than observed — is the values whose temperature is nought, together with the numbers. Every one of the 352 is one of those, and not one of the 1,122 values with distinct stops is. The equivalence is refusable: a value with distinct stops and no temperature, or a cold value with distinct stops, would break the reading and stop the build.

So the one-sided law has a restatement with no arithmetic in it. A position that is worth fighting over is never above its own left stop. The exception class is the positions not worth fighting over, and the reason the law fails for them is that they have only one stop wearing two names.

The description has one clause, not two #

The values whose temperature is nought, together with the numbers is two clauses because a number has no temperature in this site’s convention — the walls of a number’s thermograph never separate, so the routine reports a negative height rather than nought.

Written the other way it is one clause. The class is exactly the values with non-positive temperature, which is to say the cold ones, with the numbers falling in because a number is as cold as a position gets. Checked over the whole day: the values whose stops coincide and the values with temperature at most nought are the same 352, agreeing on all 1,474.

That is worth the tidying because a one-clause description is a boundary and a two-clause one looks like a list. The law is then as short as it goes:

A position with a positive temperature is never above its own left stop, nor below its own right one.

And the exception class is not a set of awkward cases; it is the complement, defined by the same quantity, with nothing left over.

Which is the handover point everywhere else too #

That boundary is worth recognising, because it is not a new one. Temperature nought is where every summary in this field stops speaking.

The stops stop. Above it they name the two ends of a genuine interval and decide three of the four relations. At it the interval has collapsed to a point and the two names denote one number, so there is nothing for the endpoint law to distinguish.

The temperature stops. A position with no temperature has no thermograph worth reading: the mast stands at the stop, the walls never separate, and the diagram has one number in it.

And the two-number summary stops. Every value in this class is summarised as mean nought, temperature nought or as its number with nothing beside it — which is why the summary’s largest fibre is precisely the positions with both quantities at zero, and why it holds a fifth of the day.

So the thirty-three anomalies at each end are not a defect in the endpoint law. They are the point at which the whole temperature apparatus has been handed over, and the object that takes over is the atomic weight — a different scale, with different units, measuring in ups rather than in points.

Which is why the next section’s disappointment is the expected one. The bracket does badly on 329 of the 352 not because the instrument is crude but because it is the second instrument in a sequence, and this class is exactly where the first has finished and the second is only beginning. A reader who expected the up-bracket to resolve the class was expecting the handover to happen once; it happens twice, and the tiny scale is the third instrument, sitting inside the bracket’s own first gap.

That is the honest shape of the field’s measuring apparatus. Points, then ups, then tinies — each one exact on the positions the one before it could not separate, each one blind to the positions the one after it exists for, and each handover happening at the moment the previous scale’s unit reaches zero.

What comparison alone can add, and what it cannot #

The natural instrument for an infinitesimal is the up-bracket: the largest $n$ with $D \ge n\!\cdot\!\uparrow$ and the smallest with $D \le n\!\cdot\!\uparrow$, established by comparison and by nothing else.

It is worth being careful about what that instrument can be asked here, because the obvious question is the one it cannot answer. The bracket’s own comparison at $n = 0$ is is $D \ge 0$ — which is the question. A bracket cannot fail to contain the sign of the game it brackets, and reporting that it “decides” the relation would be reporting the question back with a wider margin.

What it genuinely adds is a size. Two positions both strictly above their stops are not equally above them, and the difference matters as soon as either is added to anything: which part to move in is decided by a total, and the total of two infinitesimals is decided by how many ups each is worth.

On that question the bracket does badly. It pins twenty-three of the 352 to a single multiple of up and leaves 329 as a band — widths of one, two, three and four ups, with 111 of them four wide. The widest is $2\!\cdot\!\downarrow\!\ast$, bracketed between $-4$ and $0$: a position known to be at most zero and at least four downs, which for a player deciding a close endgame is close to no information.

Twenty-three of 352 is a rate, and a rate is only worth reading against another one. The same class one day down is small enough to count by hand.

That is not a failure of the measurement. It is the measurement reporting where its own unit is too large, which is the scale below the reach of the ups and the reason the atomic weight is a calculus rather than a comparison. The same shortfall shows up on every row of Clobber and in every sum of ups, and it is the same shortfall each time.

One value, worked through #

The clearest member of the exception class is $\Uparrow$, two copies of up.

Its two stops are both nought, so the interval rule says nothing about how it compares with nought. Its temperature is nought, so it is in the class by the structural description above. Its difference from its own stop is itself. And it is strictly greater than that stop — Left wins it moving second, because Right’s only move leads to $\uparrow\!\ast$ and Left answers there — which is the relation the left-stop law calls impossible.

The bracket pins it: $\Uparrow$ lies between $2\!\cdot\!\uparrow$ and $2\!\cdot\!\uparrow$, so its atomic weight is two exactly, and it is one of the twenty-three members of the class the bracket resolves to a point. A player holding $\Uparrow$ against a board of infinitesimals therefore knows precisely what they hold, which is unusual here.

Contrast $\{0 \mid \ast 2\}$, also strictly above its own stop and bracketed between $0$ and $2$. Both positions are positive, both have stops at nought, both have temperature nought — and one of them has a size and the other has a range two ups wide. Nothing in the pair of stops, the temperature or the outcome class distinguishes them, and in a sum the distinction is the whole answer.

The reason for the difference is a star. $\Uparrow$ has no star anywhere in it and $\{0 \mid \ast 2\}$ has one at the bottom of its only Right option, and a star is what makes a bracket a band rather than a point — the same clause that decides which Clobber rows are pinned and which are not. So the twenty-three pinned members of this class are, almost by construction, the star-free ones, and the class is three quarters stars.

That is worth stating as a warning rather than as a curiosity. A reader who has learned that both stops at nought means this position is worth nothing has learned something false in three different ways at once: it may be worth a great deal to whoever moves, it may be a strict win for one side with no number attached, and two positions agreeing on every number the temperature theory can produce may still be four ups apart.

All four outcome classes are represented inside the exception class, which is the compact form of that warning: $\Uparrow$ is a win for Left, its negative a win for Right, $\ast$ is a win for whoever moves and $0$ a win for whoever moves second — and all four have both stops at nought and a temperature of nought. Every one of the four things a position can be, behind one pair of numbers.

The 271, and what they mean for a player #

The commonest answer in the exception class is confusion, and confusion with one’s own stop is a strange-sounding thing until it is read as a sentence about play.

$G \parallel x$ means whoever moves first wins $G - x$. So a position confused with its own left stop is one where both players have something to gain by moving into the fight, even though the fight is worth nothing on either wall. That is exactly the situation an all-small game is in: no side is ever ahead in moves, the stops are both nought, and the whole question of who is winning is a question about tempo.

The other two answers are the interesting ones for the same reason. A position strictly above its own stops is one where Left is winning outright and the fight is worth nothing — an advantage that no number measures and that the temperature theory records as zero on both walls.

What the count does not settle #

Three limits.

The comparisons here are against the stops themselves rather than against a grid, which is what makes the case count exactly $2 \times 1{,}474$ with nothing missed. The price is that the essay says nothing about numbers near a stop, and near is where a practical question would live: a player holding a board worth its stop plus a quarter is in a different situation from one holding the stop exactly.

The second is the pool. Day three is where this evaluator stops, and the 352 are a quarter of it. One more day would settle whether that quarter is a fact about the construction, and this machine cannot reach it — but the day below is free, and it is enough to show the proportion is not a constant.

So the quarter is not a constant of the subject. It is a fraction falling as the days go up, because each day builds far more positions with something to fight over than positions without, and the exception class is the ones with nothing.

The third is the one the bracket section is about. Every relation quoted here came out of a search over a difference game, and the instrument that ought to replace the search — a calculus producing an atomic weight rather than an interval containing one — is not built on this site. Saying the theory of the very small takes the question on is accurate about which theory owns it and optimistic about what that theory currently delivers here. Twenty-three exact answers out of 352 is the honest figure.

The convention, named #

Normal play throughout: the player unable to move loses. Every stop above was computed as the foot of a thermograph wall, which is the same recursion the definition gives and one implementation rather than two, and every comparison is the outcome class of a difference reduced to canonical form.

Under misère play none of this survives, and it does not survive in an unusually complete way. The stops are defined by a recursion that bottoms out at numbers, misère play has no numbers to bottom out at, and misère play has no negatives — so $G - x$ is not even an available object. The question this page answers cannot be asked there.

Where the ladder goes next #

The stops anchor now has four rungs: what they are, that they never cross, which numbers sit between them, and now that the two ends of the interval are not the same shape.

The rung above is the one the 329 name. The relation at a coinciding stop is the sign of an infinitesimal, and the instrument that produces it as a number rather than as a band is the atomic weight calculus — a recursion on options rather than a comparison against multiples of up. This site brackets and does not calculate, on this page and on every Clobber row and in every sum of ups, and the gap is the same gap each time. Closing it is one piece of work that four essays are waiting on.

Two neighbours are worth the trip. The numbers it is confused with is the rung below, where the interval is established and the endpoints are set aside as a class; this page is that class opened. And the class where nobody runs out first is where the 271 live, taken on their own terms rather than as an exception to something else.