The simplicity rule

A position where Left can move to a game worth $0$ and Right can move to a game worth $1$ is written $\{0 \mid 1\}$. It has a value. The question is which number, and the answer is not the one most people guess.

It is $\tfrac12$, which happens to be the midpoint — a coincidence that this position is small enough to make, and that larger positions do not repeat. That coincidence is unfortunate, because it makes the rule look like averaging, and averaging is wrong.

The rule #

If every option of $G$ is a number, and every one of Left’s options is strictly less than every one of Right’s, then

$$G = \text{the simplest number } x \text{ with } G^L < x < G^R \text{ for all options.}$$

Simplest means born earliest, in a sense made precise below. It does not mean smallest, or closest to zero, or having the smallest denominator — though it usually coincides with the last of those.

$\{0 \mid 1\}$ has options $0$ and $1$, so the value is the simplest number strictly between them. That is $\tfrac12$, and it is also the midpoint, which is why the example is a bad first one.

$\{0 \mid 10\}$ has options $0$ and $10$. The midpoint is $5$. The simplest number strictly between is $1$, because $1$ is born long before $5$. The value is $1$.

$\{-5 \mid 5\}$ has options $-5$ and $5$. The midpoint is $0$, the simplest number strictly between is $0$, and here they agree. But $\{1 \mid 5\}$ has midpoint $3$ and value $2$, and $\{-100 \mid -1\}$ has midpoint $-50.5$ and value $-2$ — the simplest number strictly between $-100$ and $-1$, which is not $-99$ and not $-50$.

Four of those are worth drawing together, because they have one thing in common that the opening figure’s five do not: every interval here contains a whole number, and that alone settles the answer. Once an integer is inside, no fraction can beat it — every fraction is born after every integer it lies between — so the value is the integer of the interval nearest to nought, and the arithmetic of the two ends stops mattering entirely.

Notice what the last two rows do to any theory that the value tracks the size or the position of the interval. $\{-6 \mid 2\}$ and $\{-5 \mid 5\}$ have different widths, different centres and different endpoints, and both are worth nothing, because both contain the number that was born first.

Birthdays, which make “simplest” precise #

Numbers are built by the same recursion as everything else, and each one has a birthday — the first day on which it exists.

On day zero, $\{\; \mid \;\}$ is born: that is $0$.

On day one, the only options available are $0$, so $\{0 \mid \}$ and $\{\mid 0\}$ are born: those are $1$ and $-1$.

On day two: $\{1 \mid \}$ is $2$, $\{\mid -1\}$ is $-2$, $\{0 \mid 1\}$ is $\tfrac12$, $\{-1 \mid 0\}$ is $-\tfrac12$.

On day three: $3$, $-3$, $\tfrac32$, $\tfrac14$, $\tfrac34$, and their negatives. And so on.

Drawn out, that construction is a tree: each number is born in a gap between two that already exist, the integers march outwards along the edges, and the fractions fill inwards by halving. The next rung is that tree, day by day, and it is where the word being defined here is given its picture. What matters at this rung is only the consequence — simplest means highest in that tree, and nothing else.

Simplest now means unambiguous: the number with the earliest birthday. Between any two numbers there is exactly one of earliest birthday, so the rule is well defined. Between $0$ and $10$ the earliest-born is $1$, born on day one, while $5$ waits until day five.

The tree also makes the rule visually obvious. Between two numbers, the simplest one is the highest common ancestor’s relevant child — the first node the interval reaches as the tree is descended. Reading the value off the tree is faster than any calculation.

Only the dyadic rationals #

A striking consequence: after finitely many days, only the dyadic rationals — fractions with a power of two as denominator — have been born.

$\tfrac13$ is not among them. It never appears on any finite day, because every finite-day number has the form $p/2^k$. To reach $\tfrac13$ requires infinitely many days, at which point the construction produces it as $\{\tfrac14, \tfrac{5}{16}, \ldots \mid \tfrac12, \tfrac38, \ldots\}$ — a limit rather than an entry.

This is not an artefact of the definition. It is a fact about finite games: a position that ends after $n$ moves cannot be worth a number of birthday greater than $n$, so a finite game is never worth a third. Anybody who expected game values to be arbitrary rationals is expecting the wrong kind of object.

The same construction, run without the finiteness restriction, produces every real number and then continues past them into the ordinals. That is Conway’s surreal numbers, and the point worth taking here is that the number line of game theory arrives already built rather than assumed.

Why simplicity and not the midpoint #

The rule has to be proved, and the proof is where the reason becomes visible.

The claim is that $G = x$ where $x$ is the simplest number in the interval, and to prove it, $G - x$ must be a second-player win.

Suppose Right moves first, either in $G$ to some $G^R > x$, or in $-x$ to $-x^R$. If Right moves in $G$ to $G^R$, then $G^R - x > 0$, so Left wins. If Right moves in $-x$, the resulting number $x^R$ is an option of $x$ — and since $x$ is the simplest number in the interval, $x^R$ is not in the interval, so $x^R \ge G^R$ for some option. Left then moves in $G$ to that option and the difference is at least zero. Left wins.

The mirror argument covers Left moving first. Both cases turn on the same fact: $x$ being simplest means its own options fall outside the interval, so any move to one of them overshoots and can be punished.

That is the whole reason it is simplicity rather than the midpoint. The midpoint has no such property — its options are inside the interval, they do not overshoot, and the argument collapses. Simplicity is precisely the condition that makes the second player’s replies exist.

What that proof is doing at every step is a comparison, and comparisons on this site are decided the same way throughout: two positions are equal when their difference is a second-player win, so proving a value means exhibiting the second player’s replies to everything. The simplicity rule is what makes those replies findable, because it puts $x$’s own options outside the interval where the inequality is already known.

The rule in a game somebody plays #

Abstract braces are unconvincing, so here is the rule doing work on a board.

In Hackenbush, a stalk of one blue edge is worth $1$: Left cuts it and leaves nothing, Right cannot move at all. So it is $\{0 \mid \} = 1$.

A stalk of one blue edge with a red edge on top is worth $\tfrac12$. Left’s only move cuts the blue edge, taking everything, leaving $0$. Right’s only move cuts the red edge on top, leaving one blue edge, worth $1$. So the position is $\{0 \mid 1\}$, and the rule gives $\tfrac12$.

Add another red edge above and the position becomes $\{0 \mid \tfrac12\}$, worth $\tfrac14$. Each further red edge halves it, and the pattern is exactly the binary expansion.

That sequence of positions is worth running through the rule directly, because it is the one place where the answer is a halving and the halving is forced rather than assumed. Left’s option stays at nought throughout; Right’s option is the value of the stalk one edge shorter. So each row’s right-hand end is the previous row’s answer, and the rule is being asked the same question about an interval that keeps shrinking towards nought from above.

The halving is what makes the denominators powers of two, and it is why a string of edges reads as a binary numeral rather than as anything else. Four rows of it are enough to see the rule generating the expansion; the stalks themselves are what the rule is being applied to.

That correspondence — position as binary numeral — is the most satisfying single fact in the elementary theory, and it exists because the simplicity rule is what it is. Averaging would have produced nothing so clean.

What the solver computed #

simplestBetween(lo, hi) finds the earliest-born number strictly between two dyadics. It works by the tree: start at zero, and if zero is not in the interval move towards it by integers until an integer lands inside, and otherwise bisect by halving the denominator until a dyadic lands inside.

That routine was wrong on the first attempt, in a way worth recording. It searched upwards from lo and returned the first number found, giving $\{-5 \mid 5\} \mapsto -4$. That is the first number above the lower option, not the simplest number in the interval, and the two coincide only when the interval excludes zero. Simplicity is about distance from the root of the tree, not about position along the line, and a search that walks the line will get it wrong every time the interval straddles zero.

The corrected routine is checked against the game recursion. For each figure, the game recursion builds the position from its options, reduces it to canonical form, and reads off the value; simplestBetween computes the same value from the rule; the two are asserted to match before the figure is drawn. $\{0\mid 10\} = 1$, $\{1 \mid 5\} = 2$, $\{-100 \mid -1\} = -2$, $\{0 \mid \tfrac12\} = \tfrac14$ — all confirmed by both routes.

When the rule does not apply #

The hypothesis is doing real work, and there are two ways to violate it.

An option that is not a number. $\{0 \mid \ast\}$ has a non-number on the right, so the rule says nothing. The position is $\uparrow$, which is not a number at all — it is positive, and smaller than every positive number.

Left’s option not below Right’s. $\{1 \mid 0\}$ has options that are both numbers, but Left’s is larger. The rule requires $G^L < G^R$ and this fails. The position is not a number; it is a switch, worth $\pm\tfrac12$ offset by $\tfrac12$, and both players are eager to move in it because moving there is better than being moved on.

That second case is the important one, because it is where temperature comes from. A position where Left’s best option beats Right’s best is a position worth fighting over, and the whole theory of hot games lives in the gap the simplicity rule refuses to fill.

Both failures have the same signature in the outcome classes. A number is never confused with zero: it is greater than nought, less than it, or equal to it, and those are three of the four ways a position can sit against zero. The fourth class — confused with zero, where whoever moves wins — contains no number at all, and every position in it has failed one of the two hypotheses above. That is not a second test to run; it is the same test read from the other end.

Reading a value off the tree #

The tree turns the rule into a lookup, and it is worth doing a few by eye.

Between $0$ and $10$: descend from the root. Zero is not strictly inside, so step to $1$ — inside. Answer $1$, on day one.

Between $3$ and $4$: neither integer is inside the open interval, so no integer works. Descend to the halves: $\tfrac72$ is inside. Answer $\tfrac72$, born on day five — five days for a number sitting between two of the earliest, which is what being far from the root costs.

Between $\tfrac{1}{4}$ and $\tfrac{1}{2}$: no integer, no half. The eighths: $\tfrac38$ is inside. Answer $\tfrac38$.

Between $-6$ and $2$: zero is inside. Answer $0$, on day zero, which is as simple as anything gets. The interval is wide and off-centre and the answer is still the root of the tree.

The two middle cases are the ones where the descent has to keep going, because neither interval contains an integer at all. Hold one end still and bring the other one in, and the rule is asked the same question about a gap that keeps shrinking — which is the cleanest way to watch the answer arrive later and later.

The pattern is always the same — take the earliest number the interval admits, and the interval’s width and position are otherwise irrelevant. A wide interval is not more complicated than a narrow one; it is simpler, because it catches an earlier node. The converse is the part that surprises people: two intervals of identical width can have answers born five days apart.

So the operative question is never how big the gap is but what is already inside it, which is why the rule is stated with the word simplest and not with any measure of size at all.

The value is not a score #

A number attached to a position invites a misreading, and it is worth blocking.

A position worth $\tfrac34$ does not mean Left is three-quarters of the way to winning, or has three-quarters of a move in hand, or would score $0.75$ under some scoring rule. It means something exact and stranger: that the position behaves, in every sum it might ever appear in, exactly as three-quarters of a free move would.

The operational content is entirely comparative. If a position is worth $\tfrac34$ and it is added to a position worth $-\tfrac34$, the total is zero and the second player wins. If it is added to something worth $-1$, the total is negative and Right wins. That is what the number asserts, and it asserts nothing else.

Which is why equality of games is defined by a difference and a winner rather than by a magnitude. The numbers are useful because they compose, not because they measure.

Integers, and moves in hand #

The integers deserve a sentence of their own, because they are the one case where the value does have a plain reading.

A position worth $n$, for positive whole $n$, is worth exactly $n$ free moves to Left. Left can move $n$ times before running out, Right cannot move at all, and adding such a position to any other gives Left $n$ moves of slack.

In Hackenbush that is a stalk of $n$ blue edges. Left cuts the top one $n$ times; Right, with no red edge to cut, is helpless from the start.

So integers are moves in hand, and fractions are the interpolation between them that the tree forces. A position worth $\tfrac12$ is not half a move in any operational sense — it is the position that, doubled, is worth one move, which is a weaker and more precise statement.

Laid out on a line, the short stalks land exactly where the tree says they should: whole blue edges step outwards by ones, and every red edge after the first blue one halves the remaining distance. That is the same two motions the construction makes each day — outwards by an integer at each end, inwards by a halving in every gap — performed by an object somebody could cut with scissors.

Numbers avoid numbers #

One theorem sits so close to this rule that it belongs here.

If a position is a number, moving in it is a mistake. More precisely, in a sum where some component is a number and some is not, there is always a move at least as good outside the number.

The intuition is that a number is a position with no urgency — the value is fixed and moving in it only gives ground. Formally, if $x$ is a number and $x^L$ is one of Left’s options, then $x^L < x$, so Left moving in $x$ strictly reduces the total.

This is why numbers are described as “cold”: they are the parts of the position nobody wants to touch, the settled territory, the moves saved for when there is nothing else. In a Go endgame it is exactly the distinction between a position worth points and a position worth playing, and it is the reason temperature theory can order the moves at all.

Who found it, and when #

The simplicity rule is Conway’s, from around 1970, published in On Numbers and Games in 1976. Its context there is unusual: the book constructs the real numbers and the ordinals from the same recursion, in the same chapters, and games appear as the general case of which numbers are a special one.

That ordering is deliberate and is the opposite of how the subject is normally presented. Conway does not start with games and discover that some are numbers. He starts with a construction, observes that it produces numbers, and then observes that it produces a great deal more. The simplicity rule is the theorem that identifies which of the products are numbers and what they are worth.

Donald Knuth’s Surreal Numbers, published in 1974 before Conway’s own book, presented the number construction as a novella. It is the only mathematical result to have been popularised before it was published.

Where the model stops #

Both option sets must be numbers, and Left’s must be below Right’s. Neither is a technicality; the two failure modes are the two most interesting classes of position in the subject.

Finitely born numbers are dyadic. Anybody looking for a game worth $\tfrac13$ is looking for an infinite game.

Simplest is a statement about the tree, not the line. This is where implementations go wrong, and it went wrong here.

A value is not a move. Knowing a component is worth $\tfrac14$ does not say what to play in it — and by the theorem above, the answer is usually to play somewhere else entirely.

The ladder from here #

numbers opens here, with a rule for evaluating a position whose options are numbers, and six rungs stand above it.

The number tree says what the rule’s central word means: simplest is born earliest, which is a fact about a construction rather than about a fraction, and the tree is what makes the phrase the simplest number strictly between name exactly one object where the number line names none. Numbers avoid numbers then turns the whole apparatus into advice a player can carry: in a position with a number in it and anything else, the number is never the right move, and that is a theorem.

The numbers came out of the game corrects the order of presentation everything above uses. The construction is always taught numbers first and games second, and Conway arrived at it from positions — which is why the definition quantifies over sets of previously built objects rather than over cuts, and why it produces a genuinely different collection at every finite stage. The recursion this site cannot run removes the stopping condition and reaches $\omega$, its reciprocal and one third, none of which this site’s evaluator can hold.

The last two rungs test the birthday as a measure. How old a value is shows a form’s depth bounding its value’s birthday with canonicalisation attaining the bound, and twenty-four of day two’s 256 forms older than what they are worth. The birthday of a sum bounds the birthday of $G+H$ by $m+n$, exact 163 times over 231 pairs — and finds that every pair missing the bound by three days or more has a sum that is a number or a nimber, so the slack measures how much cancelled rather than how loose the bound is.

What this rung establishes is the rule the rest of them are about: a position whose options are numbers is worth the simplest number the two options allow, and every later use of simplest on this site means the same thing.