What a number does to a fight

Every other sum on this site is complicated. Two hot positions can add to a cold number; outcomes do not add; temperatures do not add; and the value of a sum is not a function of anything simpler than the values of its parts.

Adding a number is the exception, and it is a complete one.

The two invariants #

A number has nothing at stake in it. Every one of its incentives is strictly negative, so neither player wants to move in it, and adding it to a board contributes no urgency of its own.

That much is easy to accept. What the measurement establishes is stronger: adding a number contributes no urgency to the rest of the board either.

$$\text{temperature}(G + x) = \text{temperature}(G), \qquad LS(G + x) = LS(G) + x, \qquad RS(G + x) = RS(G) + x$$

Three thousand cases, no exceptions — which for a claim of this shape is not a sample but a demonstration, since a single failure would have refuted it. The whole thermograph is picked up and slid sideways: same shape, same height at which the walls meet, same bends in the same places, translated by $x$.

The cancellation is visible in the arithmetic. Left’s incentive for a move to $G^L$ is $G^L - G$, and in the translated position it is $(G^L + x) - (G + x)$, which is the same game. Every incentive is unchanged, so everything computed from the incentives — the temperature, the shape of the walls, which option is best at each tax — is unchanged too.

What the number does decide #

If a number changes nothing about the fight, what is it for?

It decides whether there is anything to fight about. A hot position sitting on a board by itself is a first-player win: whoever moves first grabs the better end of it. Add a large enough number of the right sign and the position is decided before anybody moves.

The rule is exact and it is short. For a position $G$ with stops $LS$ and $RS$:

$$G + x \text{ is a first-player win} \quad\Longleftrightarrow\quad -LS(G) < x < -RS(G)$$

So the stops are literally the edges of the fight. $LS$ is what Left gets by moving first and fighting on; $RS$ is what Right gets. A number bigger than $-RS$ hands the position to Left outright, a number smaller than $-LS$ hands it to Right, and in between neither player can afford to be the one who waits.

The endpoints, where it fails #

The prediction is right on 2,890 of the 3,000 translations and wrong on 110. Every single one of the 110 is on an endpoint.

At $x = -LS$ or $x = -RS$ exactly, the position sits on the boundary and the stops have run out of resolution. A stop is a number, and at the boundary the game is worth its stop plus something too small for a number to see — a star, an up, a down — and that infinitesimal decides.

This is the standing pattern of the subject and here it has a count attached. The open interval is exactly right: 3,000 translations, zero exceptions strictly inside or strictly outside. The closed interval is a different question, its answer depends on an infinitesimal, and the stops are the wrong instrument for asking it.

Saying which of the two questions has been answered is most of the value of the measurement. A statement of the rule that did not distinguish them would be right 95% of the time and would be wrong in exactly the cases a close game turns on.

One endpoint, worked through #

$G = \{\ast 2 \mid -\tfrac12\}$ has $LS = 0$ and $RS = -\tfrac12$, so the rule predicts a first-player win for every $x$ strictly between $0$ and $\tfrac12$, and something decided at the two ends.

At $x = 0$ the prediction holds: $G$ itself is a win for Right whoever moves. At $x = \tfrac12$ it does not. The sum is $\{\tfrac12\ast2 \mid 0\}$, and both players win moving first — Left by taking $\tfrac12\ast2$, which is above zero, and Right by taking $0$ and leaving Left facing an empty board.

The two endpoints behave differently and the difference is where the stop is attained. $RS = -\tfrac12$ is attained at a Right option that is a plain number, so adding $\tfrac12$ puts that option exactly at zero and Right can move to a second-player win — a knife edge. $LS = 0$ is attained at $\ast 2$, which is confused with zero rather than equal to it, and the confusion is on Right’s side of the argument, so the lower endpoint falls the way the rule says.

That asymmetry is not visible in the stops. Both are numbers, both are attained, and nothing about the pair $(0, -\tfrac12)$ records that one of them was reached at a nimber and the other at a number. Which is the general shape of the failure: the stops are a summary, and the summary drops exactly the information the endpoint needs.

Set the two sums beside each other and the class has moved between them. $G$ on its own is a win for Right whoever moves; $G + \tfrac12$ is a win for whoever moves, which is neither of the classes the two numbers $0$ and $\tfrac12$ occupy. Nothing about the stops did anything except shift by the half that was added, and the outcome class changed in a way that shift does not predict — which is the whole of what an endpoint failure is.

Why this is the mechanism behind numbers avoiding numbers #

Numbers avoid numbers says that in a sum containing a number and something that is not one, a player should never move in the number. The translation principle is why.

Moving in the number changes $x$ and leaves $G$ alone. By the invariants above, that cannot change the temperature, cannot change the shape of the fight, and cannot change which move in $G$ is best. All it can do is shift the position along the interval — and shifting away from the interval’s edge is precisely giving ground. The number is not a resource to be spent; it is a fixed handicap, and touching it can only make it worse for the player who touches it.

The number line the position sits on #

There is a picture worth carrying away, and it is the one the interval suggests.

A position that is not a number does not sit at a point on the number line. It sits across an interval — from $RS$ to $LS$ — and is confused with every number strictly inside it. The width of that interval is $LS - RS$, which is twice the temperature for a plain switch and less for a position with follow-ups, and it is exactly the set of numbers the position cannot be compared with.

That is the honest geometric account of what a value is. Numbers are points; everything else is an interval with a fine structure at its ends, and the fine structure is the part the interval cannot show. The bottom two rows above are the sharpest version of it available: they are confused with exactly the same numbers, they are the same width, and they are different positions.

Read the same fact as comparisons and it is a statement about the order rather than about a picture. $\{2 \mid 0\}$ is confused with $1$, below $3$ and above $-1$; $\ast$ is confused with $0$ and with nothing else, since its two stops coincide and its interval is empty. The smallest interval there is still contains a position that no number can be compared with.

Where the interval came from #

The two stops were not introduced as the ends of an interval. They arrived as an answer to a different question: what does a position settle at if both players keep fighting until somebody faces a number?

$LS(G)$ is defined by a mutual recursion — the best Left can do moving first, given that Right will then do their best — and it stops when a number is reached, which is the reason numbers are where the recursion halts. Nothing in that definition mentions comparison with numbers, and the identification of the stops with the ends of the fuzzy interval is a theorem rather than a restatement — one comparison proves and measurement here confirms.

It is also the reason the stops are worth computing at all. A pair of numbers that told a reader only “this is what happens if both players fight” would be a fact about one line of play. The same pair telling them “these are exactly the numbers this position is confused with” is a fact about every sum the position can appear in with a number beside it, and that is a much larger claim.

What the pair still does not do is determine the position. $\uparrow$ and $\ast$ have the same two stops — both zero — and one is a win for Left while the other is a win for whoever moves. The interval is the same and the positions are different, and the difference is the thing that decides the endpoints above.

What the solver computed, and how #

One hundred and twenty values born by day three, taken as the first of the enumeration, each added to each of twenty-five numbers — every quarter from $-3$ to $3$ — giving three thousand sums.

The count is smaller than it could be and the reason is worth saying: every translation interns a new canonical form, and ten thousand of them cost more than a gigabyte inside a build that also has to draw five hundred pages. Three thousand is enough for a claim with no exceptions in it to carry, and the size is printed beside every number that comes out of it.

For each sum four things are computed independently. The thermograph of the sum, whose temperature is compared with the original’s. The stops of the sum, compared with the original’s stops shifted by the number. The outcome class, obtained from the recursion rather than from any of the above. And the prediction, obtained from the original’s stops alone.

Failures are separated by whether the offset is on an endpoint of the interval, and the two counts are reported apart. That separation is the finding: a prediction that fails only where it was stated to be about an open interval has not failed, and merging the counts would have produced a 95% success rate about nothing.

The instrument is blind to infinitesimals by construction #

The endpoint failures are described above as the stops running out of resolution, and there is a more exact account of why they run out exactly there — one that makes the 110 look inevitable rather than unlucky.

Set the two operations side by side. Adding a number slides both stops by that number and leaves the temperature alone. Adding an infinitesimal slides nothing at all: $\{2 \mid 0\}$ has stops $2$ and $0$ and temperature $1$; add a star and the three numbers are $2$, $0$ and $1$; add an up and they are $2$, $0$ and $1$ again.

So the pair of stops is a function that is equivariant under adding a number and constant under adding an infinitesimal. The first property is what makes the rule work at all — the interval slides with the position, so a prediction made once holds everywhere. The second is what makes it fail at the ends.

Read the prediction as a comparison and the failure is immediate. Asking whether $G + x$ is a first-player win is asking how $G$ stands against $-x$, and strictly inside the interval the answer is “confused” for reasons a number can see: the two are separated by an amount with a size. At an endpoint they are separated by nothing a number can see, so the answer turns entirely on the infinitesimal part — and the stops were shown, one paragraph ago, not to depend on the infinitesimal part at all.

The instrument is not merely a summary that happens to drop the deciding information. It is a function that is provably constant along exactly the direction the endpoints vary in. No refinement of the stops will fix it; a better-computed pair of numbers has the same blindness, because the blindness is what being a pair of numbers means here.

That is also why $\uparrow$ and $\ast$ can be indistinguishable to the pair while behaving completely differently. Both have both stops at zero, so both are predicted to be decided at every non-zero offset, and both are; at zero itself the prediction says nothing and the two positions part company, one a win for Left and the other a win for whoever moves.

Numbers act on the games, and the invariants are the orbit #

There is a compact way to hold all of this, and it is worth stating because it says which questions the stops can be asked.

The numbers act on the games by addition, and the action is free: if $G + x$ and $G + y$ are the same game then $x = y$, since games form a group under the sum. So every position sits on an orbit — a copy of the number line — and translating moves it along that line and nowhere else.

The invariants are then exactly the quantities that do not depend on where along the orbit a position sits. The temperature is one; so is the width $LS - RS$ of the interval; so is the shape of the thermograph, its bends and the height at which its walls meet; so is which option is best at each tax, since every incentive is unchanged. What a number decides is the position on the orbit and nothing else, which is the whole content of the three thousand translations in one sentence.

Two consequences fall out. Being a number is a property of the orbit rather than of the position, since an orbit either consists entirely of numbers or contains none — and the outcome of a translation is therefore a question about where a single orbit crosses zero. And the fuzzy interval is not a separate object from the temperature and the stops: it is the segment of the orbit on which the crossing is undecided, with the two endpoints being the places the orbit touches zero without crossing it cleanly.

That is the honest sense in which adding a number “changes nothing”. It changes the only thing it can change.

What the invariants are worth in practice #

Two invariants that hold three thousand times out of three thousand are worth turning into working rules, and both of them are used elsewhere on this site without being stated.

The first is that a board’s numbers can be set aside. A position that has broken into a dozen components, some hot and some numbers, has a temperature equal to the hottest of the hot ones, and the numbers contribute nothing to it. So a player deciding where to play can ignore every numerical component entirely while choosing, and add them up afterwards. That is what makes playing the hottest a rule anybody can follow rather than a rule about a board nobody can hold in their head.

The second is that a score can be carried. Once the fights are over, the board is a number; while they are going on, the accumulated numbers can be kept as a running total and never consulted, because they do not affect any decision. That is exactly the bookkeeping the orthodox account does — total the settled regions, then handle the stakes in order — and the translation invariants are why the two halves of that account can be separated at all.

Neither rule is deep. Both are the kind of thing that looks obvious once stated and is not obvious at all beforehand, given that almost nothing else in this subject can be separated from anything else. A board of hot components has no such decomposition, which is what makes a strategy with a guarantee the best available answer there.

Where the model stops #

One hundred and twenty values of the 1,474 born by day three, and quarters over a range of six. The values are a prefix of an enumeration rather than a random sample, and the offsets are dyadic with denominator four — so a position whose stops are eighths would have its endpoints missed by the grid entirely, and the endpoint count ought to be a lower bound on how often the closed question differs from the open one.

That is an argument, and it is the kind this site prefers to settle rather than leave standing. Halving the step doubles the grid and asks the same four questions again.

So the endpoint count is a lower bound in principle and is exact here, which is a weaker statement than it sounds: it says the pool is coarse, not that the grid is fine enough. A pool reaching day four would carry stops at sixteenths and the argument would have to be run again.

Everything here is also about adding a number. Adding an infinitesimal does something completely different: it moves no stop and no temperature, and it changes the outcome — which is the other half of the same subject and is not measured here.

And day three is a day. The invariants are theorems and would be surprising to see fail; the exactness of the open interval is a measurement, and a value with a stop attained in some way day three does not produce would be the place to look for an exception.

Where the ladder goes next #

This rung establishes what adding a number does and identifies the exact boundary at which it stops deciding. The rung above is the other side: what adding something smaller than every number does, where the stops say nothing at all and the outcome moves anyway.

Two neighbours are worth the trip. Where the fight stops is where the two stops are defined and where the pair is shown not to determine the position. And nobody wants to move here is the reason a number contributes no urgency in the first place — every one of its incentives is a loss, so there is nothing in it for either player to reach for.