When a switch is not a switch

A switch is the simplest position that is not a number: Left may move to a game worth $a$, Right may move to one worth $b$, and $a > b$, so both players want the move. The base rung of this ladder gives it two numbers — the mean $\tfrac{a+b}{2}$, where it settles, and the temperature $\tfrac{a-b}{2}$, what moving first is worth — and neither is recoverable from the other.

The condition $a > b$ does the whole of that work and is easy to read past. It is a hypothesis, not a description of the notation: $\{a \mid b\}$ can be written down for any two numbers, and once $b$ climbs above $a$ the object it names is a different kind of thing entirely. The simplicity rule takes over, the position becomes a number, and there is nothing to fight about.

That is the textbook account, and it has two regimes in it. Sweeping the boundary with a solver finds three.

Exactly at the wall #

The middle row is the finding, and it is a genuine third regime rather than a rounding of one of the others.

At $a = b$ the position $\{2 \mid 2\}$ is worth $2\ast$: the number two with a star attached. Its mean is $2$ and its temperature is $0$, so on the two numbers a switch is described by it looks exactly like the number it is worth. Asked whether the canonical form is a number, the solver says no. The same thing happens wherever the boundary is placed — $\{0 \mid 0\}$ is $\ast$ itself and $\{\tfrac12 \mid \tfrac12\}$ is $\tfrac12\ast$, every one checked, mean equal to $a$ and temperature $0$ in each case.

Four boundary cases in one picture make the pattern harder to read as an accident of where a particular sweep happened to start.

The four are not all alike in the one way a player would care about. $2\ast$, $1\ast$ and $\tfrac12\ast$ are wins for Left whoever moves, because a positive number carries the position however the star falls; $\ast$ alone is a win for whoever moves. So the boundary case inherits its outcome from the number it is sitting on, and only at $a = b = 0$ is the star the whole of the position.

A star is confused with zero, and $2\ast$ is confused with $2$ in the same way: the difference $\{2 \mid 2\} - 2$ is a first-player win. So the tempting summary — at $a \le b$ there is nothing to fight over — is half right. Nothing is at stake, because the temperature is zero and no points change hands. There is still a reason to touch it, because a player one move short somewhere else can take this one. That is what “cold” means precisely: a number, or a number plus something infinitesimal, and the boundary lands on the second case.

The nicest thing about that star is that it has nothing to do with numbers at all. $\{G \mid G\} = G + \ast$ for any game $G$ whatever, and the three computed values above are instances of it. Nothing in the identity mentions numbers or the ordering of the options; it is a statement about a position whose two players have identical choices.

The check that makes this more than a pattern is the case where $G$ is not a number. Ask the solver for $\{\ast \mid \ast\}$ and it returns 0 — stops $0$ and $0$, no temperature, and a second-player win. That is what the identity predicts, since $\ast + \ast$ is zero, and it is not what “the boundary case is a number with a star on it” predicts. So the third regime is a general fact about symmetric option sets, showing up at $a = b$ because that is where the two players’ choices coincide, and the numbers are along for the ride.

Simplest is not nearest #

Past the boundary the simplicity rule takes over, and the value it produces is generally not where a reader expects it.

With $a = 2$ and $b$ running through the quarters from $\tfrac94$ to $5$, the value agrees with the midpoint of the two options on 4 of 12 cases and differs on the other 8. At $b = \tfrac{11}{4}$ the value is $\tfrac52$ against a midpoint of $\tfrac{19}{8}$. At $b = 5$ the value is $3$ against a midpoint of $\tfrac72$. And at $b = \tfrac{13}{4}, \tfrac72, \tfrac{15}{4}, \tfrac{17}{4}, \tfrac92, \tfrac{19}{4}$ the value is $3$ every time: the answer sits still through six consecutive steps while the midpoint climbs past it.

That standing-still is what “simplest” means made visible. The rule asks for the number born earliest that lies strictly between the options, so an integer beats every fraction and a half beats every quarter, and widening the gap can only make the answer simpler rather than larger.

Drawn against the midpoint rather than against its neighbours, the same behaviour is a gap that opens and closes.

The contrast with the fight side is sharp enough to state as a pair. Below the boundary the mean is exactly $\tfrac{a+b}{2}$ on every row of the sweep. Above it the value is the midpoint two thirds of the time and is not the midpoint the other third, with no correction term available: the simplicity rule does not approximate an average, it answers a different question. A reader carrying “the value is roughly in the middle” across the boundary has carried the one property that does not survive the crossing.

The other boundary, and it is somewhere else #

Here is the part of the sweep that was not expected, and it is the reason the crossing needed measuring rather than deriving. The outcome class of $\{2 \mid b\}$ does not change where the switch stops being a switch. It changes as soon as $b$ passes $0$, which is four rows earlier.

For $b = -2, -\tfrac32, -1, -\tfrac12, 0$ the outcome is $\mathcal{N}$ — whoever moves first wins. For $b = \tfrac12, 1, \tfrac32, 2, \tfrac52, 3$ it is $\mathcal{L}$ — Left wins whoever moves. So the last first-player win in the sweep is $\{2 \mid 0\}$, still very much a fight with temperature $1$, and the first Left win is $\{2 \mid \tfrac12\}$, also a fight, with temperature $\tfrac34$. Both boundaries are crossed inside the same sweep and they are at different places.

The mechanism is not mysterious once separated out. Being a switch is a claim about the relation between the options; the outcome class is a claim about the position’s relation to zero, and zero is not mentioned in the definition of a switch. Move the whole sweep up or down by a constant and the switch boundary moves with it while the outcome boundary stays put, so the two can be placed anywhere relative to each other. That is a claim about translation, and translation is the one operation this diagram makes literal.

What that costs a reader is a habit. “This position is a fight” and “this position is up for grabs” are different statements, and only the second is about who wins.

What the solver computed, and how #

Every number above comes from one sweep, and it is worth saying which routine produced which.

The position is built as game([dyadic(2)], [dyadic(b)]) — one Left option and one Right option, both numbers, with no reduction applied on the way in. Its value is the canonical form, its temperature and mean come from thermograph(g, 8), its stops from stops(g), and its outcome from outcome(g). Four routines over one object, and the crossing is read off by varying $b$ and nothing else.

The thermograph is where mean and temperature both live. Walls are kept as exact piecewise-linear functions rather than sampled, so the temperature is found by solving for where the two walls meet rather than by looking for where two curves appear to touch. On the sweep the walls are straight and the answer is available in closed form as well, which is what makes the sweep a test: the recursion and the formula $\tfrac{a-b}{2}$ agree on all eight rows with $a > b$, and the recursion is what remains when the formula stops applying.

Drawn properly, the hottest row of the sweep is two straight walls: temperature runs up the page and value across it, each wall marks where a player is still willing to move once a tax of that much is charged per move, and the two meet at a temperature of 2 over a value of 0. The feet of the walls are the stops — 2 and −2, what each player gets moving first with no tax charged at all. All of that diagram is machinery behind two numbers in a column, and on this sweep it never has more than two straight lines in it.

Two conventions in the output need naming, because both are traps.

A temperature of $-1$ is a sentinel, not a temperature. It is what the routine returns for a position that is a number and therefore has no temperature at all. The reversed rows of the sweep carry it, and the figures print “the number 9/4” and “the number 5/2” rather than showing the sentinel, precisely so that nobody reads $-1$ as a cold position colder than zero.

Whether a position is a number is a test on the form, not on the value. Applied to $\{\ast \mid \ast\}$ it returns false, and the position is worth exactly zero. Every claim above about a position not being a number is made against the canonical form rather than the form as written, which is the only way to make the question one about the value.

The simplicity figure carries an assertion that can fail: it checks each marked value against the game it claims to be worth, and refuses a case with $a \ge b$ outright, naming the regime the caller has strayed into.

The temperature is called a temperature because of an operation, and the operation makes the crossing look continuous rather than like a case split.

Put $\{2 \mid -2\}$ under a rising tax and the gap runs 4, 3, 2, 1 as the tax goes 0, 1/2, 1, 3/2, closing altogether at a tax of 2 — from which point the position is frozen at its mean of 0 and neither player wants to touch it. Cooling charges a tax on moving and shrinks the fight; heating a number $x$ by $t$ does the reverse and produces exactly the switch $\{x + t \mid x - t\}$. So the fight side of the boundary is the image of the numbers under heating, and cooling a switch by its own temperature returns the number it was heated from.

Read that way, the sweep is a walk in from the hot side, and the walk is worth drawing at every step rather than quoting.

As $b$ rises towards $a$ that descent reaches $0$ at $a = b$, and what is discontinuous is not the temperature but the kind of object: at every positive temperature a fight, at $b = a$ a number plus a star, one step further a number outright. The eight rows above are the same eight the closed formula $\tfrac{a-b}{2}$ is checked against, and the recursion agrees with it on every one.

The stops behave as simply on the fight side. For a plain switch with $a > b$ the left stop is $a$ and the right stop is $b$, without exception across the whole sweep — what each player gets by moving first and playing on to a number. Past the boundary both stops equal the value, because there is nowhere to walk to.

When the walls are not numbers #

Everything so far assumes both options are numbers, which is the definition’s hypothesis. Dropping it costs both halves of the picture at once, in opposite directions.

Take a row of the sweep, $\{2 \mid 0\}$, whose walls are straight because both options are numbers: stops 2 and 0, temperature 1. Now keep Left’s option and give Right one that is itself a fight — $\{2 \mid \{1 \mid -1\}\}$ — and the temperature is 1 again while the stops are 2 and 1 rather than 2 and −1. The right wall has acquired a bend where the inner fight cools out, and the stops have stopped being the options while the arithmetic quietly stopped applying.

$\{1 \mid \ast\}$ is not a switch under the definition, since $\ast$ is not a number and is not below $1$ in the ordinary sense. It has a temperature of $\tfrac12$ all the same, a mean of $\tfrac12$, stops of $1$ and $0$, and it is a win for Left. So a real fight can live outside the hypothesis entirely.

Going the other way, $\{0 \mid \ast\}$ is $\uparrow$ — mean $0$, temperature $0$, stops $0$ and $0$, and a win for Left whoever moves. By every number a thermograph reports there is no fight in it, and it is neither a number nor equal to zero. $\{\ast \mid 0\}$ is $\downarrow$ and mirrors it; $\{0, \ast \mid 0\}$ is $\uparrow\ast$ and is confused with zero. All three carry the same pair of stops and sit in three different outcome classes.

Those three, with $\ast$ and $\ast 2$ beside them, are five values smaller in absolute size than every positive number and none of them zero. Every one is written $\{a \mid b\}$ with something other than a number on at least one side, not one is a switch and not one is a number, and three of the five are confused with zero — every relation among them settled by playing a difference rather than by reading a diagram.

So the tidy three-regime picture is a fact about a hypothesis rather than about brace notation, and the infinitesimals fill the space it does not cover. What survives outside the hypothesis is the thermograph; what does not survive is the arithmetic — the mean stops being the midpoint, the temperature stops being half the gap, and the stops stop being the options.

The boundary read as an ordering #

The three regimes are also a ranking, and that is where the distinction earns its keep at the board.

Set four rows of the sweep side by side as components of one position and sort them by what is at stake: $\{2 \mid -2\}$ at temperature 2, $\{2 \mid 1\}$ at 1/2, $\{2 \mid 2\}$ — the star — at 0, and $\{2 \mid 3\}$, which is the number 5/2 and has no temperature at all. The whole position settles at 6 and the largest single stake in it is 2, so the first move goes in the widest fight.

Read in that order it is the fight, the narrower fight, the star, and the number — the crossing turned into a move order. Playing the hottest says take the biggest stake first, and it is a theorem up to an error of one temperature’s worth rather than a rule of thumb.

The star’s place in that order is what the third regime buys. A component worth $2\ast$ contributes its two points to the total like any number, and it also contributes a move that neither player has spent — so it is played last among the things worth playing and before the pure numbers, which are never played at all while anything else remains. An accounting that collapsed the boundary case into “a number” would put it in the wrong place in that queue and would lose games by a single tempo.

The ordering is by temperature and not by size, and the temperatures of a sum do not add — the ranking is a rule for choosing a move within one position, not a quantity to total up.

Who found it, and when #

The definition and the notation are Conway’s, from the construction that produced the numbers and the games together in the late 1960s. The switch $x \pm t$ and the pair of numbers behind it were built into working shape with Elwyn Berlekamp, for the practical reason that a Go endgame is a sum of small fights and choosing between them is the question a player actually faces.

The boundary case is older than the theory that names it: $\{0 \mid 0\}$ is the position with one move for each player and nothing behind it, which is a single Nim heap of size one, and it was understood as a first-player win long before anybody wrote it in braces. What the partizan theory added was a name and an arithmetic for it — and the observation that $\{G \mid G\}$ is $G + \ast$ for every $G$, which makes the position at $a = b$ an instance of a rule rather than an exception to one.

What the picture cannot show #

The figures in this essay put values on a horizontal line, and a line is exactly the wrong shape for the finding.

A star has no position on a value axis. $2\ast$ is drawn at $2$, because $2$ is what it is worth and there is nowhere else to put it, and the whole content of the third regime is that it is not the number sitting at that point. The figure resorts to a sentence — “worth 2, and not a number” — printed beside the mark, which is text doing the work a picture cannot.

Confusion has no picture either. Being confused with a number means the difference is a first-player win, which is a fact about a search over a second position and not about a location. Every drawing here shows an ordering, and the ordering is partial.

And the outcome flip is invisible in every one of them. The switch diagram has an axis for value and a column for temperature, and none at all for who wins. That $\{2 \mid 0\}$ and $\{2 \mid \tfrac12\}$ are in different outcome classes — the second boundary, and the surprise of the sweep — appears here only in prose.

The convention, named #

Normal play throughout. Whoever cannot move loses. Every value, temperature, stop and outcome above is computed with that base case, and under misère play — the player who cannot move wins — the sweep would be a different sweep. The boundary at $a = b$ is a normal-play fact about a normal-play value, and misère is a different game with the same rules written down.

Both options are numbers. That is the hypothesis under which $\{a \mid b\}$ has a mean of $\tfrac{a+b}{2}$, a temperature of $\tfrac{a-b}{2}$, stops at $a$ and $b$, and a boundary at $a = b$ with a star on it. Drop it and every one of those five statements fails independently of the others.

And $-1$ from the temperature routine means “no temperature”. It is a sentinel for a number, not a value on the same scale as the temperatures beside it — the one number in this essay that is not a measurement.

Where the ladder goes next #

This rung establishes that the switch regime has an edge, that a position sits exactly on the edge, and that the edge is not where the outcome changes. Four rungs follow it directly.

Switches with walls that are not numbers. $\{1 \mid \ast\}$ has a temperature of $\tfrac12$ and no midpoint to speak of; $\{2 \mid \{1 \mid -1\}\}$ has a bent wall and stops at $2$ and $1$ rather than at its options. What the mean and temperature mean once the hypothesis is dropped is the next thing to compute, and the thermograph is the only machinery that survives the drop.

The mean value theorem. That every position has a mean at all — that the value of $n$ copies stays within a bounded distance of $n$ times a fixed number, with the bound independent of $n$ — is what licenses the word, and it is proved rather than computed. Many copies of one game sets up the measurement it is about.

Cooling as the operation that undoes a switch. Heating a number by $t$ makes a switch of temperature $t$ and cooling it back gives the number; whether that is a bijection, and what cooling does to positions whose walls bend, is a rung of its own.

And the endgame as a sum of switches. A whole board late in the game is several of these at once, with different temperatures and a move order to decide, which is where the two numbers are finally spent — and where a component sitting on the boundary stops being a curiosity and starts costing a point.