Field

How it was found

The theory looks inevitable in retrospect and the record says otherwise. The older arguments are run here rather than recounted — and one of them still answers a question nothing since has answered.

Backward induction on a game that ends, one round at a time. Zermelo's argument as it actually runs. Round zero is the positions where the player to move has no move at all, which is the only thing the procedure knows without being told; each later round is what those settle. Anything still unlabelled when nothing more can be deduced has no label and never will — and on a game with a cycle in it, that leftover is exactly the set of drawn positions. The theorem is a statement about this procedure terminating, and it names the winner of nothing.

The first theorem, and the winner it declines to name

Zermelo proved in 1913 that a finite game with no chance and no hidden information is decided before anybody sits down — every position is a win for one side or a draw, and which one is settled already. The proof is a labelling procedure, and watching it run shows exactly how little it says.

Bouton's invariant, checked over 512 positions. Nim positions in binary, one column per bit. Bouton's 1901 argument is that a position is a loss for the mover exactly when every column holds an even number of marks — and that from such a position every move breaks a column, while from any other position some move repairs them all. Both halves are checked here over every position in the range rather than illustrated once, and the middle row shows the repairing move being made.

The theorem that needed none of the theory

Bouton solved Nim completely in 1901, with an argument that mentions no value, no sum of games and no Grundy number, because none of the three existed. The argument is two closure properties and it is airtight — and run on any other game it fails at the step that does the work.

The mex, and the rules that cannot replace it. Six candidate rules for the value of an impartial position, each a function of its options' values, run over the same subtraction game. The top strip is the truth. Every candidate but the mex assigns zero to a position somebody wins, or a non-zero value to a position somebody loses, and the circle marks the first heap where each one does it — which is why two people reaching for the same rule four years apart is evidence about the rule rather than about them.

Two people, four years apart, one theorem

Roland Sprague proved it in 1935 and Patrick Michael Grundy proved it in 1939, neither knowing of the other. That looks like coincidence until the alternatives are examined — and the rule they both reached turns out to be the only one that can work at all.

A golden ratio in a table that never mentions it. Grundy values for Wythoff's game, computed by the mex rule alone — a queen moving left, down or diagonally toward the corner, and whoever cannot move loses. The circles are Wythoff's 1907 description of the losing positions, which came thirty years before any of this machinery: the pairs formed from the golden ratio. They land on the zeros exactly. Nothing in the computation knows about φ and nothing in Wythoff's argument knows about Grundy values.

A golden ratio thirty years early

Wythoff described the losing positions of his game in 1907 with an argument about partitions of the integers, and no Grundy value anywhere in it. The theory that arrived thirty years later computes the same positions — and has never produced a closed form for the values, which the older argument had for the zeros from the start.

The Grundy values of ·137, and the exceptions to its period. An octal game's Grundy sequence, with the periodic part in gold and the exceptions in magenta. The exceptions are the point: a sequence described as eventually periodic contains values that disagree with the value one period later and always will, so the period is a statement about a tail and not about the sequence. The rule used to identify an exception is printed, because published lists of them differ by which convention was used.

A chess problem that turned out to be an octal game

Dawson posed it in 1934 as a puzzle about pawns. It is the octal game ·137, its Grundy sequence is eventually periodic with period 34 from heap 52 — and the word doing the work in that sentence is eventually, because five values below the start disagree with their repeats and always will.

6 octal games, and which of them settle. Each row is an octal game: its code, the moves it allows, the first two dozen Grundy values, and whether a period was found in the values computed here. Guy and Smith surveyed these by hand in 1956 and conjectured that every finite octal game is eventually periodic. Seventy years and a great deal more arithmetic later, the rows in magenta are the state of that conjecture — not counterexamples, but sequences in which nothing periodic has yet appeared.

The sequence nobody has settled

Guy and Smith surveyed the octal games by hand in 1956 and conjectured that every finite one is eventually periodic. Seventy years and a great deal more arithmetic later, some of them have settled and some have not — and the evidence for the conjecture is entirely that nobody has found a counterexample they were looking for.

$The days this site can compute, and the ones it cannot. Zero on the first day, ±1 on the second, and thereafter the simplest number in every remaining gap — the construction run by the game recursion, which produces only fractions with a power of two underneath however long it goes on. Below it, three objects the same recursion reaches when the stopping rule is removed, each written with its option set and the exact reason this site's machinery cannot hold it. They are named rather than drawn, which is the honest half of a figure-first collection.$

The numbers came out of the game

The construction is always taught numbers first and games second, and the discovery ran the other way. Conway arrived at the number system 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.

One position, three ways of writing it, and only one of them adds. The same positions as a sentence about who wins, as a description of the position itself, and in the notation Winning Ways introduced. The first two columns carry identical information and support no operation whatever. The third column can be added — and the sums below it are values that no manipulation of the first two columns could reach, because two of these pairs start from the same two outcomes and finish differently.

The notation was the argument

Up, star and the brace form are not abbreviations for case analyses. They are the claim that these objects add — and the arithmetic they support is arithmetic that no table of outcomes could ever produce, because two positions with identical outcomes can have different sums.

Classes needed, as the heaps get bigger — Dawson's chess ·137. How many kinds of position there are, against how large a heap the universe allows. Under normal play the answer stops growing as soon as the Grundy values stop growing. Under misère play it does not stop, and every new class is a pair of positions that behave identically under normal play and differently under misère.

"Hopeless" was a claim about a method

Misère analysis was declared intractable in the 1970s, and the verdict was correct about what was being attempted. Quotients did not refute it thirty years later — they changed the question from a value per position to a monoid per universe, and the computed sizes show why the first question has no good answer.

One Sprouts game from 3 spots, counted. One randomly played Sprouts game, with the map counted after every move. A move spends two lives and the new spot brings one, so the lives fall by exactly one every time — and unlike the arms of a Brussels cross they are not replaced. Every move either cuts a face in two or joins two separate pieces of the drawing, and how many of each a game contains is up to the players, which is why the length is not fixed.

A conjecture from hand play

Sprouts was invented over tea and its outcome pattern was guessed from games played with a pencil. Computers have checked it far past where a person could go, and this site's own solver gives out at three spots — so the honest figure states the frontier it reaches rather than the number somebody else published.

The endgame, accounted for. Several independent regions, each a fight with a settled value and a size. The account plays them hottest first: add up what each is worth on average, then add the largest amount at stake, subtract the next, and so on down. The exact value of the whole position is computed beside it, and the figure prints both.

The first time it told somebody something

A theory earns its keep when it produces an answer nobody had. Temperature did that for Go endgames — the orthodox account gives a move order that is provably right and is not the one experience offers, and the position it is right about is small enough to check here completely.

What computing further has bought

Sprouts has been searched harder and longer than almost any game, and the period-six pattern has survived every extension. This site's own exhaustive search settles three spots; the published results reach forty-seven, and the gap is not a gap in hardware — the gentler of the two measured growth factors puts forty-seven spots at ten to the hundred and twenty-fifth positions. Beside it sits Brussels Sprouts, which has five million positions holding a choice and not one choice that changes who wins.

What the notation costs to write. Every game born by each day, written in the brace notation and measured. The expressions are all distinct, which is what the notation is for, and by day three the typical one is twenty-two characters and the longest is fifty.

Where the braces stop

The brace notation names every game exactly — 1,474 games born by day three, 1,474 different expressions, no two alike. It also gets long: the middle one is twenty-two characters and the abbreviations everybody actually writes cover one game in twenty-three. And it has two hard edges. A game with a cycle in it has no finite expression at all, and the equation the minus sign encodes — that a game and its negative cancel — is false under misère play on every one of those 1,474.

Which bit of the rule decides. Four properties of an octal rule table set against whether the game it describes settles into a period. Only one holds on every code that does not: whether a move may leave two non-empty heaps. It is necessary and not sufficient.

Three bits of rule

An octal code is three bits a digit. The Grundy sequence it determines costs anywhere from one bit to a hundred and thirty-six — a factor of two hundred and seventy-two across rules that differ by a single digit — or it cannot be written down at all. Of four properties of the rule table tested against that, exactly one holds on every code that never settles: whether a move may leave two non-empty heaps. It is necessary, it is not sufficient, and nine codes carry it and produce answers smaller than their own rules.

How long a win takes, against how long the argument allows. Ordinary impartial games with the size of their position graphs, the number of rounds the backward labelling takes, and the number of moves the longest win actually lasts. The round a position settles in is the length of the play from it, which is computed here a second way so the two must agree. The rounds are a handful and the positions are many, which is the gap Zermelo's 1913 paper is about — his question was how many moves a forced win needs, and the answer he could prove was the size of the whole graph.

The paper was about how long

Zermelo's 1913 paper is remembered for a theorem it proves in passing. The question it actually asks is how many moves a forced win takes, the answer it can prove is the size of the whole position graph, and the round counter in the procedure is the real answer — a quantity nobody named for another forty years.

Two solutions to one set of equations. The winning condition written as a single predicate and solved twice: once as the least solution of its own equations and once as the greatest. The least says Left can force a win; the greatest says Left cannot be forced to lose, which admits the positions where Left can keep the game going for ever. On a graph with no cycle in it the two coincide and the equations determine an answer. Where they differ, the difference is exactly the set the backward propagation never reaches — so a draw is not a leftover of the algorithm, it is the equations failing to have one answer.

The gap between two answers

A draw is usually described as what the backward labelling never reached, which makes it sound like a shortfall of the algorithm. Written as one predicate the winning condition is an equation, the equation is monotone, and it has a least solution and a greatest one — and the set the two disagree about is exactly the drawn set, on every game checked.

A game every play of which ends, and no round settles. A game whose first move chooses how long the game will be, cut off at several sizes. Every play of it is finite and no position is drawn, so the fourth outcome class has nothing to do with what goes wrong. What goes wrong is the round counter: the opening is a loss, a loss settles only when the last of its options is known, and there is no last option. Cut the game off larger and the round grows, so no number in the column is the answer for the untruncated game — and the induction that labels it has to run past every finite stage.

Every play ends and no round settles

Take the finiteness hypothesis away carefully — not by adding a cycle, which has already been priced twice, but by adding infinitely many positions to a game every play of which still ends. Nothing is drawn, every line finishes, and the round the opening settles in grows with every cut: two, four, six, eight, twelve, sixteen, and no number in the column is the answer.

What a certificate costs, in units of the one Guy and Smith wrote. Octal codes with the period of their Grundy sequence, the window a proof of that period needs, and the arithmetic each costs — counted as mex operations and exclusive-ors, which are the two things a person computing by hand actually performs. Everything is priced in units of the certificate for Dawson's chess, so the column reads as multiples of one hand computation rather than as a number of operations. Some codes cost tens of times as much, and some have no certificate at all.

What the arithmetic cost in 1956

The rung below ends by respecting a hand computation without pricing it. Priced in the operations a person actually performs, ·137's certificate is 7,919 of them — and the same sweep says ·47's is sixty-three times that, that a splitting move is what makes the cost quadratic, and that seventeen of sixty-four codes have no certificate at any price.

The same rules under the convention they were posed in. Dawson's chess under misère play, which is how Dawson posed it. Under normal play every position of the game collapses onto one of a handful of nimbers however large the heaps are allowed to get. Under misère play the positions that behave alike form classes whose number grows with the heap limit, and a heap carries a genus rather than a value. The first wild heap is where the two accounts stop resembling each other, and the classification doubles at exactly the limit that admits it.

The convention Dawson actually used

Dawson published his puzzle as a problem where running out of moves loses you the game, and every compact result about ·137 is about the other convention. Under his own, nine values become a classification that doubles the moment a wild heap enters the range, and a heap stops carrying a number at all.

Three complete solutions, each asked about the others. Bouton's 1901 criterion for Nim, Wythoff's 1907 description of his own cold positions, Moore's 1910 rule for taking from several heaps, and the Grundy criterion that arrived thirty years later, each checked against the truth on every position of four games. Every one of the old criteria is exact about its own game and wrong about the others. The blanks matter more than the numbers: Wythoff's is a description of a pair and has no form for three heaps at all, and the Grundy criterion has no form for a game whose moves touch several heaps at once.

Three complete solutions in nine years

Bouton in 1901, Wythoff in 1907, Moore in 1910 — three airtight solutions of three games, all published before there was any theory of games at all. Asked about each other's games they all fail, and two of them fail by being wrong while one fails by having no form for the question. Only the last kind of failure decides anything.

How far a description of that kind could ever have gone. Subtraction games sorted by whether a Bouton-style column criterion describes their losing positions. His test reads the heap sizes in binary and counts the marks in each column, which works exactly when a heap's value is a function of its own bits — and that is true of a small minority of the family. Below it, the weaker readings: a criterion on the low bits, and a sequence that merely repeats. The method itself is available for every game and says nothing; what 1901 supplied was a set with a description shorter than the game.

A set with a short description

Bouton's argument is a closure argument about a set, and every impartial game has such a set — its own losing positions. So the method is complete and proves nothing. What made 1901 a theorem is that his set had a description shorter than the game, and swept over fifty-six subtraction games, exactly seven have one of his kind.

The sentence that solved the other convention

Bouton's paper solves misère Nim too, in one line, and it is the only misère result in the subject that fits on one. Transplanted the way the normal criterion is, it fails differently — the normal one calls losses wins and never the reverse, and this one errs in both directions on every game tried, because the clause it adds is about counters rather than about moves.

A vocabulary that is not closed under its own arithmetic. Every pair of named values added together, with the answer sorted by whether it has a name. The named vocabulary covers every game born by day two and one in twenty-three born by day three, and coverage is the wrong measurement: the notation exists so that positions can be added. A sixth of the sums of two named values at day three cannot be written without opening a brace, and the first one to escape is a sum of two of the symbols anybody learns first.

Two names that add to nothing nameable

The special symbols reach one game in twenty-three at day three. Coverage is the wrong measurement. The notation exists so that positions can be added, and a sixth of the sums of two named values at day three cannot be written without opening a brace — starting with a sum of two of the six symbols anybody learns first.

What two numbers cannot tell apart

A thermograph summarises a position in a mean and a temperature — nine characters against the brace form's twenty-two, and readable in a way the expression is not. It is also not exact: 1,454 of the 1,474 games born by day three share a pair with some other game, 291 of them share one pair, and adding a star to two of those gives different winners.

A board is written as a sum

Every measurement of the brace notation so far has been of a single position, and nobody writes a single position. A board is several parts, and it can be written as the parts joined by plus signs or as the one value they add up to. Over every sum of up to four games born by day two, the one value is usually the shorter — and the share of boards that need a brace climbs with every part added, until the longest value is four times its sum.

A shuttle and a loop, judged by what the play returns to. A three-node loopy game drawn as a graph, with Left's moves in blue, Right's in red and position a marked. Beside it, each position-and-mover pair under the backward labelling and under the rule that a never-ending play goes to Left when it returns to a infinitely often. Four pairs are drawn by the labelling; the new rule gives two to Left and two to Right and leaves the decided pairs as they were.

What the play keeps coming back to

A draw is what the backward labelling never reaches, and handing every never-ending play to one player turns the draws into wins wholesale. Judge an infinite play instead by what it keeps returning to, and every draw gets a winner of its own: over the 262,144 three-node games, 15,432 send some of their draws to one player and some to the other, which no wholesale rule can do. Finding those winners takes a fixed point inside a fixed point.

A hub with two spokes, and the bit of memory it needs. A three-node loopy game in which Left, at a hub, chooses between two spokes and Right must return from either. Left wins a never-ending play that passes through both spokes infinitely often. With one bit of memory recording which spoke is owed, Left wins from the hub; with a strategy that depends only on the position, Left always takes the same spoke and loses. Three position-and-mover pairs change hands.

One bit of memory

Judge an infinite play by whether one position keeps recurring and every winner can play from a table of one move per position, with nothing remembered. Ask for two positions to keep recurring and that stops being true. At a hub with two spokes a player has to alternate, and a table cannot alternate: over every three-node game, 49,487 position-and-mover pairs are won with one bit of memory and lost without it.

One step and four captures. Dawson's pawns on a board three ranks deep and five files wide. A White pawn steps forward on the middle file, and because a capture must be made when one is available, four captures follow: Black takes, White retakes, Black takes, White retakes. Five moves later the three middle files are finished and the two outer files are untouched, which is the octal move taking three from a heap of five and leaving two heaps of one.

The capture that has to be made

Dawson's chess is quoted as the octal game ·137, and the step from a pawn diagram to a row of counters has been taken on trust. Searched as a chess position, the diagram agrees with ·137 on every board from one file to twelve, under both endings, and every exchange it can start is an odd number of moves that lands on one of ·137's options. The whole reduction rests on one rule of the diagram that the octal code never mentions: a capture, when one is available, must be made. Make it optional and the winner changes on two, three, six and seven files.

A row of files, valued rather than won. Dawson's pawn diagram on a single row of one to 5 files, with the value of the position under each capture rule beside the nimber ·137 gives the corresponding heap. The winners agree throughout; the values agree until five files, where the diagram is worth ∗ and the heap is ∗3.

A wall the pawns cannot cross and the rule can

Two rows of Dawson's diagram separated by a file with no pawn on it: 1,616 moves were examined and not one crosses the gap. With captures optional the rows add on every diagram checked. With captures compulsory they do not, because the compulsion is a rule about the whole board — and the game that is a sum is the one ·137 does not describe.

Which conditions make a winner remember. Every condition on which set of three positions a never-ending play keeps returning to, grouped by two properties of the condition alone, against the arenas swept. 32 of 128 conditions have an arena Left wins and cannot win from a table of one move per position; 19 of those are closed under union.

The rule decides who has to remember

Whether a winner needs memory is a property of the winning condition and not of the board, and the property everybody reaches for is the wrong one. Of the 128 conditions on which of three positions a play keeps returning to, 32 demand memory and 19 of those are closed under union. What separates them is measured two independent ways and the two agree on all 128: a condition needs no memory exactly when it can be rewritten as a number on each position.

One more row, and the correction comes back. Dawson's diagram of three files beside a row of one, then two, then three, each drawn with the difference between the whole board's value and the sum of its rows. The correction is ∗2, then 0, then ∗2: adding a row removes it and adding another restores it.

A difference the rows cannot predict

The diagrams that are not the sum of their rows have been counted and never priced. Priced over 50 diagrams and 63,408,981 positions, the difference takes three values and is a function of nothing a reader can see: seven diagrams whose rows are worth ∗ and ∗ split five to two on it, the third value arrives only at the ninth file, and the one rule that survives is a parity — all twenty-one diagrams of three, five and seven rows add, and every failure carries an even number of rows.

The tree of {a and b and c}, and the states it costs. The Zielonka tree of one winning condition on which positions a never-ending play recurs at. The root is the whole set of positions; the children of a node are the largest subsets the condition judges the other way. The number of memory states a winner needs is read back up the tree by adding at accepted nodes and taking the largest at rejected ones, and this condition costs 3.

Two things to hold at once, or three

Whether a condition makes a winner remember has been settled over every condition on three positions; how much it makes them remember has not. A tree built out of the condition alone, with no board in it anywhere, prices all 128: sixty-one cost nothing, fifty-eight cost two states and nine cost three. It also names the property that was nearly right — closure under union of the sets a condition rejects decides it exactly, where being writable as numbers is sufficient and reaches twenty-six.

Every two-position loopy region, as two names. The 256 loopy regions of two positions, placed by the names of their onside and offside as identified against the 1,474 values born by day three. Ten names cover every side: 0, 1, −1, ∗, on, off, over, under, upon + ∗ and −upon + ∗. The largest groups are on & off with 94 regions and off & off and on & on with 53 each; 25 regions need only finite names.

A loop is written with two names

A region with a cycle in it has no brace expression, and every one of the 256 regions of two positions can be written anyway — as two names, the game it is when a play that never ends goes to Left and the game it is when it goes to Right. Checked against all 1,474 values born by day three, ten names cover every side, 25 regions need only finite ones, and the pair predicts every sum with a finite game, draws included: a draw arrives exactly where the two names disagree.

One substitution, thirty-four years. Bouton's criterion and the Sprague–Grundy theorem run side by side over a family of games. They differ in one quantity: the heap's size against the heap's Grundy value. The exclusive-or that combines them is the same operation in both, and it is the one Bouton published in 1901.

The step nobody took for thirty-four years

Bouton's criterion is that the heap sizes exclusive-or to nothing. The 1935 theorem is that the heap Grundy values do. The exclusive-or is the same operation in both and it is his, so the whole of the intervening thirty-four years is one substitution — and run over eight games and 672 positions, the substituted criterion is exact on every one while the original is exact on Nim and nowhere else.

Bouton's argument, indexed by a value. Bouton's two closure properties stated for every Grundy value rather than for nought alone: no move stays inside a value class, and every class above a value can reach it. Checked on each game and each value in range.

The picture Bouton's proof leaves behind

His argument is two closure properties of one set, and the Sprague–Grundy theorem is the same two sentences with nought replaced by a variable — checked here on five games and every value in range, with no move staying inside a class and no class failing to be reachable from above. What the argument also leaves behind is a picture in which the values descend, and that is false: 99 of 444 moves here raise a value, and none of them is in Nim.

Every region of three positions, counted. The 262,144 graphs on three positions reduced to the regions that are genuinely three positions with a cycle in them, and then split by whether the two-position vocabulary has a name for both of their sides.

Four thousand nine hundred regions with no name

Two positions give 256 regions and ten names cover every side of all of them. Three positions give 262,144 graphs, 110,934 genuine loopy regions — and 4,931 of those have a side that no name in the two-position vocabulary reproduces, with 3,990 of them named on one side and blank on the other. The count the earlier essay left open comes back in the affirmative.

The guess, and what it covered. Two attempts to name the leftover sides out of the old vocabulary: every pair of the six stoppers, and every two-position region that is a stopper, each with small finite games added. Both cover nothing, and the count of distinct leftovers is what remains.

The names are not built out of the old ones

The guess was that a three-position region's missing names would be sums of two loopy ones — on plus over, and that family. Built and tried, every pair of the six stoppers covers none of the 4,931 regions that need one, and so does every two-position stopper there is, all seventy-nine of them with small games added. Thirteen names have to be invented, and forty-eight cover the whole census against ten at two positions.

Four positions, sampled

Ten names write both sides of every loopy region of two positions, and forty-eight every region of three. Four positions are over four billion graphs and cannot be counted, but they can be drawn. Three thousand regions at each of three densities: the forty-eight names cover between 95.9 and 99.5 per cent, the thirteen names invented for three positions come back at four almost all of them, and the sparsest sample meets thirty-five sides nothing earlier reproduces — a floor of eighty-three names, and a curve that grows by accretion rather than collapse.

Half the new sides are sums. The 50 sides of four-position loopy regions that no earlier name reproduces, tested against every sum of two games from the day-two values, the stoppers and the thirteen three-position names. 23 are such sums, 15 of them using a three-position name; 27 are not.

Each size is built from the last

At three positions the sides that needed new names were sums of nothing older: no pair of stoppers, and no two-position stopper with a small game beside it, wrote any of them. At four positions the thirteen names three positions forced are available as parts, and they change the answer. Of the fifty sides the samples met that nothing earlier names, twenty-three are sums of two earlier games — fifteen of them using a three-position name — and twenty-seven are not.

Composition at every size

Twenty-seven of the fifty new four-position sides were written by no sum of two earlier games. Sums of three write sixteen of them — seven plain finite games born after day two, nine built on the names three positions forced — and leave eleven, met seventeen times in nine thousand regions. Run the same test one size down and the story the counts told falls apart: eight of the thirteen names three positions were thought to invent are sums of up to three older parts, two of them just two finite values. Every size composes. What each invents is a residue — five, then eleven.

Four rows of lamps and three winning cuts. A Nim board of four rows of up to seven lamps, lit 7, 5, 3 and 2, with each row's binary beside it. The columns add without carrying to 011, a nim-sum of 3; the three rows carrying the middle bit each have a winning cut, to 4, to 0 and to 1, and the lamps each cut would put out are shaded.

A machine that knew one theorem

In 1940 a machine of relays played Nim against the public at the New York World's Fair, on four rows of up to seven lamps, and is reported to have won about nine games in ten. Everything it knew was Bouton's theorem, which is three column parities and one test per row. The record is the interesting number, and it is a measurement of the visitors rather than of the machine: against a visitor choosing at random the machine wins 98.5 per cent, and nine in ten is what a visitor earns by playing perfectly once seven lamps or fewer are lit. On the board the game is usually set up on, a visitor handed the first move cannot win at all.

A Hex board read as a circuit. A three-by-four Hex board with Down to move, one Down stone and two Across stones. Each empty cell carries a disc whose area is the current through it when both players' edges are held at a unit voltage; the busiest cell is the current rule's move. The three cells that actually win for Down are shaded, and the busiest cell is not among them.

A current through the board

In the early 1950s Shannon and Moore built a machine that played Hex by wiring the board as a network of resistors and reading its move off the electrical potential. There is no theorem behind that rule, and on every Hex board small enough to solve it can be scored against perfect play. On boards to twelve cells it takes a winning cell on more than 98 per cent of the positions where one exists, far above chance. And it loses: every board of three rows or more it has to defend, and four by four moving first, because an opponent needs only one of the two per cent and can steer the game to it.

Twelve openings, four answers. A three-by-four Hex board. Each cell is labelled with the cell the current rule of Shannon and Moore's Hex machine takes for Down when Across has a single stone on it. The twelve answers are only four cells, shaded, each answering three openings.

A circuit asked once

The Hex machine's circuit rule loses every board past two rows that it has to defend, and the strategy that holds those boards is a pairing — a partner for every cell, fixed before the game begins. So make the circuit commit: consult it once per cell on an empty board and never again. Its table answers twelve openings with four cells, and loses every board the running circuit lost. Forced to choose a whole pairing, it picks a losing one under seven of eight ways of reading its scores; the eighth finds the winning pairing of three by four by a margin of rescaling and gives up two by four for it.

One chain given up. Two pairings of the three-by-four Hex board. Left, the pairing the chain-sharing scores choose, with the one chain it misses shaded. Right, the diagonal pairing, which meets all 25 chains.

A pairing is checked, not measured

The Hex machine's circuit chose pairings near the winning ones and never at them, which left a finite question: does any score of a pair of cells, computed from the board alone, have a winning pairing as its best? The obvious scores count the chains a pair shares. All four hold two by three; on three by four the plain count cannot choose between a winning pairing and a losing one, and the corrected counts choose the loser; on four by five every one loses. The reason is arithmetic — a sum will give up one chain to cover two others twice. Told which chains it missed, the count finds the diagonal pairing at one rate in three, and by then it is checking, not measuring.

One rate in twelve. The chain count reweighted by the chains it misses, at 12 rates from 1.1 to 3, on the Hex boards 2 × 3, 3 × 4 and 4 × 5. The small boards are held at every rate by the diagonal pairing; 4 × 5 is held only at × 1.5, round 46, and every other rate stalls with chains still missed.

The weights were not the argument

A chain count corrected by its own misses held the four-by-five Hex board with the diagonal pairing, the one pairing there with a sentence behind it, and the hope was that its weights were the sentence in numbers. They are not. One rate in twelve holds the board and one start in twenty; the weights it ends on survive five per cent of noise and not twenty; and weighting exactly the chains the argument relies on finds no winning pairing at all. Told which pairing to prefer, a measurement reaches any of the three in a dozen corrections. What chain weights can do is rule a pairing out, and on three by four they rule out one that wins.

Every winner but one. A perceptron over chain weights taught to make each winning Hex pairing the heaviest, on boards 3 × 4 to 6 × 7. Every winner is reached in at most sixteen corrections except the cross pairing of 3 × 4, which another winning pairing covers on every chain.

The teacher's game

A machine that weighs Across's chains could be taught every winning Hex pairing on four by five, and on three by four it could never be taught the one pairing another winner covers chain by chain. Whether anything else could hide from it — a winner covered not by one rival but by a mixture of several — is a question about a game between a teacher choosing weights and a rival choosing pairings, and Jean Ville's 1938 theorem of the alternative, written for von Neumann's games, answers it in general. On five by six and six by seven every winner is taught in sixteen corrections or fewer, and nothing hides at all.

Three rows that do not add. Four diagrams of Dawson's pawns valued whole and as the sum of their rows. 3 + 1 is worth ∗ against ∗3 by rows. 4 + 3 + 1, three rows, is worth ∗ against ∗3, a difference of ∗2, over 4,153,158 positions. 5 + 2 + 1 and 3 + 2 + 1 add.

The rows that cancel

Every one of Dawson's diagrams that failed to be the sum of its rows had an even number of rows, and twenty-one with an odd number all added. One file past the edge of that sweep, four files beside three beside one fails: three rows, worth ∗ against ∗3 by its rows. What the parity was tracking is two deletions a sum allows — a row worth nought, and two identical rows — which the compulsory board turns out to allow as well. Each of 44 diagrams is worth exactly what is left after them, and in the fifty-diagram sweep that remainder happened to have as many rows, odd or even, as the diagram it came from.

Three winners, none of them teachable. A graph of ten points on which Short, claiming points to join A to B while Cut deletes them, wins moving second. Three pairings of the contested points win, each holding one whole pair in every one of the six minimal separating sets; one, ce df gh ij, is drawn. No weighting of the separating sets makes any winning pairing outscore every losing one.

Two losers that average to a winner

On Hex boards to 6 × 7, a machine that weighs the opponent's winning sets could be taught every winning pairing except one, which another single winner covered. Ville's alternative allowed a second way to be out of reach — covered only by a mixture — and the switching game on points, where winners are many, is where it turns up. None in all graphs of seven points or in 4,000 of eight; 35 covers at nine and ten points. Every one is the even average of two losing pairings, and on six graphs Short wins and no weighting can teach the machine any winning pairing at all.

Every game on three positions. All sixteen directed graphs on three vertices up to relabelling, each with one of its stable sets filled and the count written beneath. Nine have one, six have several and the 3-cycle has none.

The solution that was a losing set

In 1944 von Neumann and Morgenstern defined the solution of a game by two conditions on a set of outcomes: none overrules another, and each outcome outside is overruled by one inside. They were writing about coalitions. The same two conditions, read with a move in place of an overruling, describe exactly the lost positions of a game that must end. Where play can loop, the same definition gives several answers or none, and three positions are enough to show both.

Parity on a cycle. Single directed cycles of two to seven positions. Even cycles have two stable sets, the alternating halves, one of which is filled; odd cycles have none.

An odd cycle is the only way to have none

A cycle of two positions has two stable sets and a cycle of three has none. Moses Richardson proved in 1953 that this is the whole story of absence: a graph with no odd cycle always has a stable set. Run over all 9,608 games on five positions, the theorem holds without exception, sits inside a chain of five nested families, and leaves open the larger question its converse invites, since 7,452 of those games have an odd cycle and a stable set anyway.

A mirror where there was no solution. The sum of two cycles of 3 positions, drawn as a grid of pairs, with a diagonal filled. It is a stable set of the sum, although a cycle of 3 alone has none. Moves that wrap around are left out of the drawing.

What a stable set knows about a draw

On a graph with cycles, outcome labelling calls a position lost, won or drawn, and stops. The stable sets go further in exactly one place. On every game of up to five positions they hold every lost position, no won one, and among the drawn positions are exactly the stable sets of the drawn part taken alone. So a stable set is a certificate that the player who keeps returning to it never gets stuck. It can distinguish draws that labelling cannot, and two parts with no stable set can have one between them.

A stable marriage is a stable set

Draw a graph whose positions are pairs of people, with a move from a pair to any pair one of its members prefers. Its stable sets are exactly the stable marriages, on all 46,656 profiles of three men and three women. So Gale and Shapley's theorem of 1962 is an existence theorem of the kind Richardson's was, on graphs where 49 per cent have odd cycles and his hypothesis is useless. The reason it holds is that every clique in the graph is one person's list, and four roommates show what happens when it is not.

Peeling a square by stable sets. A three-by-three square with lists of three colours per cell, coloured one colour at a time. In each panel the uncoloured cells wanting the colour are shaded and the stable set among them is filled and takes it.

The list that cannot run out

Give each cell of an n × n square its own list of n colours, and ask for a colouring from the lists with no repeat in any row or column. Dinitz asked in 1979 whether that is always possible, and reading the cells in order fails about four times in ten. Galvin's proof of 1994 needs one page. Use an ordinary Latin square to turn each row and column into a preference, so that every cell has exactly n − 1 moves, and peel off a stable set for one colour at a time. A list of n cannot run out against n − 1 moves.

How the word is made. The Prouhet–Thue–Morse word built by doubling. Each row of one, two, four up to thirty-two letters is the row above followed by its swap, which is the same word obtained by replacing A with AB and B with BA.

A rule that let the game run for ever

In 1929 German chess drew a game when the same sequence of moves came three times running, and the rule was meant to make chess finite. Max Euwe, six years from the world title and trained as a mathematician, showed that it does not, using a word that had already been found twice for other reasons. Two round trips are enough. The board returns to one position thousands of times, no block of play ever comes three times running, and the rule has nothing to say.

How strict a repetition rule can be before it ends every game. For two, three and four available moves, the longest legal game under each repetition exponent from 7/5 to 3. Two moves stop at three letters at exponent two and never stop above it; three moves stop at 38 letters at 7/4; four moves stop at 121 letters at 7/5.

How strict a rule can be

A repetition rule can be tightened one notch at a time, and each notch either leaves unending games alive or kills them outright. With two moves a block may not come twice, and every game is over by the fourth letter; relax that by a single letter of overlap and unending games appear. Three moves stop dead at 38 letters under a rule that a whisker's relaxation frees. Four stop at 121. Past each boundary the words go on for ever, and the boundary sits at a fraction nobody would have guessed.

The repeater's plan on two moves. Every line of play in which the repeater forces the rule of 1929 on two moves, 4 lines of at most 7 letters, each ending in a block three times running.

The opponent makes it repeat

Euwe's unending game needs both players to keep to it. Give the letters to two players in turn, with one of them wanting the draw that a repetition brings, and the rule of 1929 fires within seven letters on a plan of one sentence: always play A. The word that defeats the rule exists and nobody can be held to it. Add a third round trip and the same player can no longer force anything in twenty-six letters. Existence and force turn out to be separate questions, with separate boundaries.

A draw by repetition changes no result. Over all labelled games on three and four positions and a sample on five, the result from each start with a draw on the second or third arrival at a position, compared with retrograde labelling of the graph. They agree everywhere.

What counting positions buys

A rule that counts arrivals at a position ends every game, which the rule of 1929 could not do, and it needs no proof beyond arithmetic: with L arrivals barred on n positions, no game is longer than (L − 1)n − 1 moves. What it does not do is change a single result. Over every game on three and four positions and a sample on five, drawing on the second arrival or the third gives exactly the labels retrograde analysis had before any rule was written. The rule pays for the game's ending and for nothing else.

Recording the turn doubles an odd cycle. On the left a cycle of three positions, which has no stable set. On the right the same game with the player to move recorded in each position, a cycle of six, with two stable sets, each consisting of all the positions with one player to move.

Taking turns removes the hard part

Whether a graph has a stable set is NP-complete, and every hard instance depends on odd cycles. A real game's position graph has none. A position records whose turn it is and every move changes that, so all cycles are even, Richardson's theorem applies, and the set of all drawn positions with one player to move is a stable set of the drawn part, always. Mū tōrere and Three Men's Morris both confirm it. The hard question never reaches a game with turns, and the sets it hands over carry no information at all.

The one colour-blind stable set of Mū tōrere. All 33 classes of drawn position in Mū tōrere under rotation, reflection and colour exchange, each drawn as a small board from the viewpoint of the player to move, with the 16 classes of the unique stable set shaded.

A rule that cannot see whose turn it is

The stable sets a game hands over work by telling the players apart. Forbid that, and read every board as mine and theirs, and the odd cycles come straight back. On Mū tōrere, with rotations and reflections identified too, the 892 drawn positions fall into 33 kinds, and among those there is exactly one stable set. It names sixteen shapes. A player who always hands the opponent one of them holds the draw against labelled play without knowing a colour, an orientation or an outcome, and the opening itself is one of the sixteen.

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