Field

Out in the world

Games people played before there was a theory, and questions outside game theory that a game answers. Every one of them solved here rather than cited — including the one every child is taught to play wrong.

A 2 × 3 board of boxes, 6 still on the table. A Dots and Boxes position drawn as dots and lines, and — where the figure asks for it — the same position as a strings-and-coins graph: one coin per box, one string per line, and the border lines running to the ground. Lines already played are solid, lines still available are dashed, and a box with no strings left has been pocketed. The footer carries the exact net score the solver computes from here and the normal-play verdict on the same position.

The game in every exercise book

Dots and Boxes is played by more people than every game in this collection put together, and everybody is taught the same rule — take every box available. The rule is wrong. Establishing that takes a solver rather than an opinion, and the solver says how wrong, on which boards, and by how many boxes.

The impartial game inside the scoring one. For every position of a Dots and Boxes board, two questions asked separately: who wins the scoring game, and who wins Nimstring — the same position under the normal-play convention, with no score kept. The bars show how often the two answers agree, grouped by how many boxes are still on the table. Agreement is near-total when there is enough left to be worth controlling and falls away when there is not.

The chains decide it before the boxes do

Under every game of Dots and Boxes there is an impartial game with no score in it, and it settles the question the scoring game keeps asking — who ends up having to open. The rule players learn as folklore falls out of it, and so do the exceptions nobody mentions.

1 ko point, no ko rule. A ko fight drawn as a position graph and labelled by retrograde analysis: blue edges are Black's captures, red are White's, and each position carries the verdict for whichever side is to move. Positions the propagation never reaches are drawn — neither player can force a win and the game does not end — and they appear only where the rules permit a repetition.

The rule that makes Go a finite game

A ko is a point in Go where a capture can be recaptured for ever, and every set of rules forbids it. That prohibition is not etiquette or tidiness — it is the hypothesis that puts Go inside the class of finite games that the theory of values is about, and removing it removes the values.

Every row of 8 squares, and the 36 values they hold. A census of Kōnane rows: how many arrangements of a row of squares carry each value, with the shortest row carrying that value printed beside it. Most arrangements are worth nothing at all; the rest spread over numbers, halves and quarters, stars, switches and infinitesimals — the whole vocabulary of the theory, from a game that predates it.

A game older than the theory

Kōnane was played on carved lava boards in Hawai‘i long before anybody wrote a brace notation, and its rule for losing is the normal-play convention arrived at some centuries early. Evaluate a row of it and the answers are halves, quarters, stars and infinitesimals — the theory's whole vocabulary, out of a game that was not built to display any of it.

The theorem that names a winner and no move

Strategy stealing proves that the first player wins Hex and wins Chomp, on every board, in about four lines. It exhibits no move, contains nothing a move could be extracted from, and is not going to. The moves have to come from somewhere else, and where they come from runs out almost immediately.

a path with every link doubled: the criterion and the game. A Shannon switching graph with the two marked vertices in gold. Short secures links and Cut deletes them; Short wins by joining the two marks. Lehman's criterion says Short wins moving second exactly when some subgraph holding both marks splits into two edge-disjoint spanning trees — drawn here in blue and red where one exists. The verdicts beside the graph come from playing the game out, and the criterion is computed without looking at the game at all.

A winning strategy that is a spanning tree

The Shannon switching game was sold in a box in 1960 and solved in 1964, and the solution is not a claim that somebody wins. It is a property of the graph anybody can check, and the strategy falls straight out of it — whichever link the opponent cuts, take its partner in the other tree.

The gaps of ⟨5, 7⟩, which are the moves. A Sylver Coinage position drawn as the numerical semigroup it is. Gold squares are the numbers already named; plain squares are sums of them, and so cannot be named again; magenta squares are the gaps, which are exactly the legal moves. The largest gap is the Frobenius number, marked F — past it every integer is reachable, which is why the game has finitely many moves left and must end.

The game that is a number system

In Sylver Coinage two players name integers and nobody may name a sum of what has already been named. Its positions are not boards — they are numerical semigroups, its termination is a theorem of Sylvester's from 1884, and the question of who wins after the opening move 16 has been worth a thousand dollars since 2017.

A game where the last move decides nothing. Rows of coins taken from either end, with the exact score for each side moving first. Under the normal-play convention this family is settled entirely by the parity of the row — nobody is ever without a move until the coins run out — so normal-play theory returns the same answer for every row and it is not the answer anybody wants. The scoring answer depends on nothing but the numbers.

Counting at the end changes everything

Go is scored. So are Dots and Boxes, chess and almost everything anybody plays for money — and none of them is the kind of game the whole theory of sums is built for. The simplest scoring game there is shows what that costs — the normal-play theory gives every position of it the same answer, and the answer is useless.

A pawn ending is a sum

In a blocked pawn ending the material is level, the files never speak to each other, and whoever has to move is the one in trouble. Chess calls that mutual zugzwang; this site calls it a P-position; and the two vocabularies were built four decades and one subject apart to say the same thing.

The cold positions, written in Fibonacci base. The first several cold pairs of Wythoff's game with both heap sizes written in Fibonacci base — as sums of non-consecutive Fibonacci numbers, which every integer has exactly one of. Blue is the smaller heap and red the larger. Read as digits, the pair is a shift: the larger numeral is the smaller one with a zero appended, and the smaller one always ends in an even number of zeros.

The digits say which move wins

Wythoff's cold positions are usually given as a pair of golden-ratio formulas. Written in Fibonacci base they are a statement about digits instead — the smaller heap ends in an even number of zeros and the larger is the same numeral shifted up a place — and a rule about digits answers a question about a heap of a trillion.

Where the count and the value part company. Amazons endgames whose arrows have already cut the board into regions, with the territory count beside the computed value. Territory gives every empty square to whichever amazon can reach it in fewer moves, which is what Amazons programs compute. The positions drawn are the ones where that number gets the outcome wrong, and they have something in common: each is worth a switch, so there is no number for the count to have been right about.

When a real board falls apart

Amazons is played competitively, and late in a game the arrows have cut the board into regions no piece can cross. From that moment the position is a disjunctive sum — arrived at by the play rather than assumed — and the territory count every program uses can be measured against what the sum is actually worth.

A set with three descriptions, and a function with none

Wythoff's cold positions can be written three ways that share no arithmetic — an irrational constant, a greedy rule, a condition on Fibonacci digits — and all three are exact. The same game's Grundy values have no closed form at all. Both facts are about one table, and the gap between them is the subject.

Every part on its own is worth nothing. The Grundy value of each chain and loop considered as a game by itself, and three real Dots and Boxes boards used to check the turn-by-turn walk against the win-or-lose solver already used here. Every component alone is a second-player win, which is exactly what makes the nim-sum useless.

The parts are worth nothing and the sum is not

Every chain and every loop in Nimstring, taken alone, has Grundy value nought. So the Sprague–Grundy theorem predicts that every position built from them is worth nought — and ninety-six of the two hundred and seven positions checked here are not. The theorem is not being misapplied; it does not apply, because a capture keeps the turn. What replaces it is smaller and sharper: count the short chains, and one long component of any kind reverses the parity.

A parity with a first exception

Sort every Sylver Coinage position by how many numbers are still unnameable and the game very nearly falls to parity: odd rows are between a fifth and a half positions the mover loses, and the first three even rows hold none at all. The rule has a first counterexample at genus eight, where it is a single position out of sixty-seven, and eleven more at genus ten. It is a tendency wearing away from both ends rather than a law with exceptions.

Which clause of the rules produces which kind of value. Every combination of pawn-file clause in range, sorted by the kind of value it produces. Files where both pawns can advance are all-small and their values are nimbers and infinitesimals. A file where one pawn is stuck behind a friendly piece gives the other side free moves and is worth an integer. A file whose middle square can be held stops the other pawn the moment somebody reaches it, and is worth a switch — a position both players want to move in. The dictionary is read off the evaluation rather than asserted.

What has to break before a pawn is worth a number

Every value the blocked-file model can hold is an infinitesimal, and the reason is one sentence about the move rule rather than anything about pawns. Break that sentence — a pawn stuck behind a friend, a square only one side can hold — and integers, switches and positions worth fighting over arrive at once.

What one king costs a decomposition. Two pawn files and one king a side, solved as a joint position and again as the sum of its files. With the kings unable to move the two answers agree on every configuration, because a king that cannot choose between files is not a shared resource. Give each king a waiting move and the answers come apart, and on some configurations the sum of the parts names the wrong winner rather than merely the wrong value. Independence is a hypothesis about the position and this is the price of assuming it wrongly.

One king, and two files to be in

The whole apparatus needs the files to be independent, and a king is what makes them not. With the kings unable to move the sum of the parts is exact on every configuration; give each king a single waiting move and the sum names the wrong winner on one configuration in six, and on a hundred and twenty-six of two hundred and forty-three with three files.

A fortress, and the counter that gives it a label. A pawn ending where the defender's king shuffles for ever and the attacker needs time. Down the rows, how many moves of preparation the breakthrough needs; across the columns, how many moves the rule allows before declaring a draw. With no breakthrough the position is drawn whatever the rule says, and drawn as a residue the backward induction never reaches. With a breakthrough and no rule the attacker wins despite the cycle. Where the march is longer than the counter allows, the rule turns a won position into a drawn one.

A position with no value, and the rule that gives it one

A fortress is a cycle in the position graph, so the recursion defining a value has nowhere to bottom out and the propagation never reaches it. Chess has a rule for that — count fifty moves and call it drawn — and the rule does not merely tidy the theory up. On eleven cells of the sweep it takes away a win.

A ko fight decided somewhere else on the board. A ko fight against the number of ko threats each side holds. A threat is a play elsewhere that the opponent must answer, and it lifts the prohibition on recapturing, so a player short of threats runs out of ways to come back. Every cell is a full retrograde labelling and the pattern in them is read off afterwards: the fight goes to whoever is ahead on a quantity that is not on this part of the board at all. Where the labelling never settles, the fight is a no-result whatever anybody holds.

A ko is won somewhere else

The rung below shows the ko rule buying finiteness by deleting one edge. What it buys with the same edge is a fight nobody can settle by looking at it — the prohibition forces a player to spend a threat, threats are counted on the rest of the board, and every decided cell of the sweep goes to whoever is ahead on a quantity that is not in the picture.

$Where the two conventions come apart, counted. Every small Go endgame solved under both scoring conventions. The scores agree exactly when the number of neutral points is even and never when it is odd, which is the parity of the stones each side ends up placing. A counted fraction name different winners. And a smaller fraction are played differently, which is the half of the finding a rules argument does not predict: a neutral point is a one-point play under one convention and worth nothing under the other, so the two rule sets disagree about the order of the endgame and not only about its total.$

Two ways to count a finished board

Territory scoring and area scoring are both in daily use and they are not variants of one rule. Over seventy-nine small endgames they agree exactly on the thirty-nine with an even number of neutral points and on none of the forty with an odd number — sixteen name a different winner, and twelve are played differently, which is not something a convention is supposed to do.

Why a square holds fewer values than a line. The same number of squares laid out two ways, with the number of places a hop can start, the longest chain one can run, and the number of distinct values every arrangement of that shape produces. A hop needs three squares in a line, so a long row supplies more of them than a compact rectangle of the same area — and the value counts follow. The second dimension is not the way to reach the deeper values, which is the opposite of what the rung below expected.

The second dimension is not the deep end

The rung below says a row of eight reaches every corner of the vocabulary and goes far into none of them, and that the narrowness is a fact about the board. So the obvious next move is a rectangle — and nine squares in a square hold twenty-five values where nine squares in a line hold fifty-eight. The geometry says why before any stone is placed.

The only two moves in Kōnane that are not captures. A filled Kōnane board with every opening Black may play marked on it, and the value of the position White's best reply leaves. The opening is the one place in the game where a player removes a stone rather than capturing with one, so it is played under a different rule from everything after it — and the choice is worth a computed amount rather than nothing.

The two moves that are not captures

Kōnane begins from a full board and the first two moves lift stones rather than take them, which is the only time in the whole game anybody does. Nothing in the theory of values applies to them, and the choice is not free — on a 3 × 5 board two of Black's eight openings leave a position a whole move worse than the other six, and on a 3 × 3 board none of the five does.

What a pass buys, and what it costs. Rows of coins solved with and without a pass. Milnor's mean-value theory needs a non-negative incentive to move, and rows containing a coin nobody wants break that condition — a player forced to take is a player who would rather have passed. Allow a pass and the condition is not merely satisfied but unbreakable, on every row in range. The price is that a player who may pass is never stuck, so the last-move convention has nothing to attach to and the game needs a separate rule to end at all.

What a pass is worth to a theory

Fifteen of twenty-seven coin rows turn out to be ones where having the move is a disadvantage, and those are exactly the rows Milnor's mean-value theory has to assume away. Allow a pass and the hypothesis stops being a hypothesis — nought violations, on every row in range. What it costs is the convention the rest of the theory is built on.

$The condition has to hold underneath, not on top. Pairs of coin rows sorted by where the incentive condition holds, with Milnor's bound checked on each pair. Rows that satisfy the condition at every subposition never break the bound. Rows that satisfy it only at the top break it on a counted fraction — and a reader who tested the row rather than the row's insides would have called those safe. The distinction is invisible from the position and decides whether the theorem applies to it.$

A hypothesis has to hold all the way down

Milnor's bound is proved by induction over the play, so the condition it needs has to hold at every position the play can reach. Checked on the row instead, ninety-two pairs pass the test and twenty-four of them break the bound. Checked at every subposition, twenty-eight pairs pass and none breaks it.

A game beside its own mirror, and what is left over. Every coin row added to its own negative, played out exactly, with the resulting scores counted. Under the last-move convention every such sum is worth nothing, because the mirroring strategy guarantees the second player the last move. Here the same strategy is available and the score it produces is not nothing: the mirror of a coin conceded is another coin conceded. Gold is the sums that do come to nothing, which are a minority.

Nothing to subtract with

Comparison is defined by contexts and computed by subtraction, and the equivalence between the two is a theorem about groups. A scoring game is not one — sixty-six of eighty-one coin rows do not cancel against their own negatives — and the difference test then fails on a row compared with itself, which every context accepts and nothing certifies.

A bridge circuit, with a link that is not there. The switching graph drawn as a bridge circuit, with an imaginary link from A to B dashed in gold. The graph alone does not split into two edge-disjoint spanning trees, so Short moving second loses; with the imaginary link it does, drawn in blue and red, so Short moving first wins. The green links are the links of the red tree that cross between the two halves the blue tree falls into without the imaginary link — the first moves the trees name.

The first move is a link that is not there

Lehman's criterion answers one question about a switching game — who wins when Short moves second. The other question has the same answer asked of a different graph: add one link from A to B, and Cut is forced to spend its first move deleting it. The trees of that larger graph then name Short's opening, and on every subgraph of seven graphs they name a winner.

The Bridg-It board of size 3, both players at once. A Bridg-It board of size 3: blue dots in 4 rows of 3, red dots in 3 rows of 4, interleaved. Every bridge blue can usefully build is drawn in blue and every bridge red can usefully build in red, and each blue bridge crosses exactly one red one. Blue's switching graph and its planar dual have the same numbers of points and links, because the dual is red's board turned a quarter.

Cut is Short on another graph

Everything proved about the switching game is proved from Short's side, and Cut appears only as the player whose moves get enumerated. On a graph drawn without crossings Cut does not need a theory of its own: deleting a link is securing the link that crosses it in the dual, so Cut's game is Short's game on a different graph. Bridg-It is the board that is its own dual — one link short of two trees at every size, which is why its first player wins.

A bridge circuit, with a point on every link. The switching graph drawn as a bridge circuit, with a new point in the middle of every link in green and the original inner points in blue, already belonging to Short. Played as a game on the green points it gives the same verdict as the original game on links, because claiming a middle point is securing its link and deleting it is deleting the link.

A point with three neighbours

The switching game on links is settled by counting — enough links, arranged as two trees. Played on points instead, it is the game Hex belongs to, and the count is gone. The link game turns out to be the point game in which every contested point has exactly two neighbours; give one a third, and two graphs with the same points, the same links and the same number of separate routes can have opposite winners.

Where the needle has a sentence

Strategy stealing proves the first player wins every Chomp bar and names no square to take. On two families the square can be said in a sentence — a square bar and a bar two rows deep — and in both the sentence is a pairing that names every later move too. Three rows deep the needle wanders, and the observation that every bar has exactly one needle survives ninety-four rectangles and fails on the ninety-fifth.

Hex on 3 × 4: the nearer edges win whoever starts. Two copies of a Hex board of 3 rows and 4 columns. On the left each cell is coloured by whether Down, joining top to bottom, wins by taking it first: all 12 do. On the right each cell is coloured by whether Across, joining left to right, wins by taking it first: none do. Down's edges are one row nearer together than Across's, and Down wins whoever moves first.

A board one column wider

Strategy stealing proves the first player wins Hex, and it needs three things: no draws, an extra stone never hurting, and rules that treat the two players alike. Add one column to the board and the third goes. The player whose edges are now nearer together wins whoever moves first — and does it with a table of pairs that names every reply, checked against every line to a board of twenty cells.

The positions that pair their gaps off, and who loses them. Every Sylver Coinage position with at most sixteen unnameable numbers, counted by genus, split into the symmetric and pseudo-symmetric semigroups — the irreducible ones — and the rest, with the positions lost for the player to move in each. Of 584 irreducible positions exactly one is lost, the single position whose only gap is 1; the remaining 11,185 positions include 1,405 losses.

Every move closes the largest gap

A census of Sylver Coinage by genus finds a parity that nearly decides the game and asks whether any known property of a numerical semigroup predicts the outcome. One does, completely: a semigroup whose gaps pair off around the largest one is never lost for the player to move — none of 583 up to genus sixteen. The reason is strategy stealing, and it is the same reason the top-right square decides Chomp: every move from such a position closes the largest gap.

Two replies to the first stone. The 4 by 5 Hex board after Across's first stone in row 1, column 1, with every empty cell labelled by the weight of Across's unblocked chains through it. The potential answers in the heaviest cell, row 2, column 4; the pairing, which wins this board for Down, answers in row 1, column 2.

A potential that names every move

Strategy stealing names no move, and the pairings that do name moves need a board with the right symmetry. The Erdős–Selfridge potential needs neither: Down, moving second in Hex, takes the empty cell through which Across's unfinished chains weigh most. Its guarantee reaches only boards two rows deep. It wins far past the guarantee — on every board of three rows that Down can win — and then, on a four-by-five board that a table of pairs wins for Down with certainty, it answers Across's first stone in a different cell and loses along the bottom edge.

The pairing removes moves it cannot name

Symmetric positions were settled by an argument that names a winner and no move. Turned on the moves instead, the same one-pass test strikes off 27,215 of the 159,728 moves in the census and not one of the 21,234 winning ones — a quarter of a full search — and still names nothing. On 583 paired positions nine arithmetic descriptions of the winning gap reach at most 123, and 367 of those positions have exactly one winning move.

An edge bonus, on every board it could help. The Erdős–Selfridge potential for Hex with the chains along the outer rows weighted more heavily, on six boards. No bonus wins the four-by-five board the plain potential loses, and the bonus costs Down 4 boards it was already holding.

The winning reply is the fourth choice

The repair proposed for the potential was to weigh an edge chain more heavily. Fifty-five weightings later, none holds the four-by-five board, and an edge bonus costs Down four boards it was already holding. The reason is not the numbers: over 393,660 turns of the pairing that does hold that board, the potential would take the same cell 26.1% of the time, and the winning cell is its 3.7th choice on average and as low as its seventeenth.

Seven orders, and none of them better than chance. Seven quantities a one-pass scan could compute about each move surviving the pairing test — the number itself, how many gaps it closes, what it leaves behind — each read as an order over the survivors and scored against the winning moves of 583 positions. The best puts a winner first on 24.9% of positions against 22.3% at random, and every one of the seven places the winner deeper in its order than chance would.

A shortlist with nothing at the top

The one-pass test leaves 9.58 moves of 12.27 and names none of them. Seven quantities a scan could compute about the survivors were turned into orders and scored on 583 positions: the best puts a winning move first on 24.9% against 22.3% by chance, and every one of the seven places the winner deeper in its order than chance does. Read as sieves instead, the gentlest keeps half the list and throws the only winner away on 264 positions.

Four restrictions, and what each one buys. Four candidate classes of scoring game — every row, the incentive condition at the top, the same condition at every subposition, and the rows that cancel against their own negatives — scored on two families of coin rows for the mean-value bound, for comparison by subtraction, and for cancellation.

The restriction that buys the most

Four candidate classes of scoring game, scored on the same two families and the same three questions. The class everyone expects to be tiny — the rows that cancel against their own negatives — is empty on rows of three and the widest restriction on rows of four, where it holds fifteen rows against the hereditary class's twelve and gets all 225 of its comparisons right against 108 of 144. The trade everyone expected does not exist.

A move that leaves no surviving reply. A Sylver Coinage position whose gaps are 1, 2, 3, 4, 6, 7, 8, 9, 12, 13, 14, 17, 18, 23, 28, with the number of the opponent's surviving replies under each move that survives the pairing test. Naming 4 leaves no surviving reply, which proves it wins; the other winner leaves as many replies as the losing moves.

A move whose every reply is struck

Read two moves at a time, the Sylver Coinage shortlist is no better ordered than read one at a time: preferring the survivor that leaves the opponent the fewest surviving replies puts a winner first on 24.7% of 583 positions against 22.3% by chance, and places it deeper than chance does. Read as a proof, the same count does what no order could. On 57 positions a survivor leaves no surviving reply at all, and wins by a certificate a few lines long; searched deeper, the survivors prove every position by thirteen moves — at a price that is never below the search that simply finishes.

The endgame theory arrives late

Every component the chain-and-loop theory names has coins of degree two, so a position it can read is one where every surviving coin holds exactly two strings. Over a six-box board that is 1,033 of the 28,028 positions with no free box on the table — 3.7 per cent — and more than half of them only after twelve of the board's seventeen strings have been cut.

The law, on a board rather than in a bag. The parity law applied to every position of a real board that has fallen into chains and loops, with the verdict computed independently from the board's own strings. The components are the ones the geometry produces rather than the ones a sweep constructs.

A thousand positions and no exception

The parity law was fitted to constructed bags of chains and loops inside a string budget. A board's positions are a different population — the sizes are what the geometry allows, the components come correlated, and a six-box board holds exactly one position that is a loop of six. Tested on all 1,032 of them and all 160 of the four-box board's, the law is right every time, against a verdict computed from the strings by a walk that has never heard of a component.

The components the theory does not name. Grundy values of strings-and-coins components with a branching coin, grouped by the value. A chain or a loop is worth nothing on its own whatever its size; a coin with three strings takes four different values depending on its arms, and a coin with four strings is back to nothing.

A coin with three strings is worth something

Every chain and every loop is worth nought on its own, whatever its size, and that is exactly what makes their nim-sum useless. A coin with three strings on it is worth nought, one, two or three depending on its arms — 31 of the 35 measured are not nought, and the four that are are the ones whose arms are all long. A coin with four strings is back to nought every time.

What control is worth, to a box. The margin the player who does not have to open nets, beside the formula that predicts it: the total less four boxes for every long chain after the first. Computed against the solver on every endgame of long chains in range.

Four boxes for every chain after the first

Nimstring answers who is forced to open and says nothing about the score. The margin has a formula: the controller nets the total less four boxes for every long chain after the first — two surrendered and two not taken, each time control is kept. Checked against the solver on 175 endgames it is exact on 172, never too generous, and exact wherever it promises the controller anything at all. The three it misses are the three where it promises nothing.

Two and four are not conventions

Declining costs two boxes on a chain and four on a loop, and those numbers are read off the geometry rather than chosen: one cut completes the last two boxes of a chain and two cuts complete the last four of a loop. Solved again with the fee changed, 418 endgames give a different winner on up to a third of themselves — so the endgame's law is a law about the fee as much as about the shapes, and the fee is not a free parameter.

Cancelling is not pairing

The coin rows that cancel against their own negatives looked like the rows whose coins pair off as nested equal pairs, and on rows of four they are exactly those. From six coins the description fails in both directions — twenty rows pair off perfectly and do not cancel, and one coin set has a hundred and thirty-six that cancel with no pairing at all — and a row of five coins cancels, though an odd row can never pair off. What does hold, on every row swept, is that the first player in a row plus its negative never finishes behind.

Cancelling rows add to cancelling rows. Every pair of cancelling coin rows of two, four and five coins from minus two, one and three, grouped by their lengths, with the number of pairs whose sum cancels against the sum of their negatives.

A cancelling pair is a zero

Two cancelling coin rows side by side cancel against their two negatives — all 190 pairs from rows of two, four and five coins, the odd row that cancels without pairing off included. And a cancelling row beside its negative is invisible next to any other row: in 741 tests against every row of one to three coins, neither score of the context moves. A non-cancelling pair moves a score in 452 of 780. Scoring games have no inverses in general; this class has them, and they behave as inverses must.

Even rows always reward the move

Milnor's mean-value theory needs an incentive to move — the player to move must do at least as well as if the opponent moved first. On a coin row with an even number of coins that is not a hypothesis but a theorem: the first player can collect one whole parity class of coins, and one of the two classes holds at least half the total. So the condition excludes no even row whatever the coins, the class the earlier table called 'incentive at the top' was every row of four, and the hereditary condition is a condition on odd intervals alone.

A different question at every depth

A Sylver Coinage proof that deepens over survivors was never cheaper than the search that skips struck moves and finishes. Keep everything it proves and refutes and it gets cheaper — 41 per cent at twenty gaps, and more when the opponent's positions are kept too — and it is still never cheaper, on any of 1,766 positions, and the gap widens with the gaps. Told the right depth in advance it wins on seven. The finishing search asks one question of a position; a proof asks a different one at every depth.

Orders, weights and pairs. For five Hex boards Down wins moving second: whether any fixed order of cells holds the board, whether the Erdős–Selfridge potential holds it, and how many sets of pairs meet every one of Across's chains in a whole pair. A fixed order holds only 2 × 4; the potential holds every board but 4 × 5; 4 × 5 has exactly three pairings.

Three pairings and no order

The potential lost four-by-five Hex for Down, and fifty-five re-weightings lost it too, which left the question whether anything that orders the cells can hold the board. The plainest ordering — one fixed list, whatever the position — holds none of the boards the pairing is known for, and on four by five the best it can do is the left column. At the other end, every table of replies that wins a small board is a pairing, and four by five has exactly three pairings, each using all twenty cells.

The seven were the order

A Sylver Coinage proof told its depth beat the search that finishes on seven positions of 1,766, and on all seven the winning move came fifth when moves were tried smallest first. Run both searches under five orders and the seven vanish under every other one; try the largest move first and there are fifty-five new ones instead. Give both searches the winning move first, free, and the proof is cheaper on none. The exceptions were never positions where proving is easy. They were positions where one order made the finishing search stumble — and what is left when the order is perfect is the price of asking a proof's question.

Two even rows, and the incentive to move. Every sum of two coin rows of two or four coins under four coin sets, 4,095 sums each. The first player is never behind; the margin is never below the difference of the two rows' own margins, equals it on 3,147 sums, and exceeds it on the other 948.

Two rows still reward the move

A single coin row with an even number of coins always rewards the player to move, because the first player can take a whole parity class. Two such rows side by side break that argument, and the conclusion survives anyway: on every one of 16,380 sums of two rows the first player is never behind, and there is a proof — open the row worth more, then answer in whichever row the opponent moves in — that guarantees the difference of the two rows' margins, which is what two switches would give. The sum meets that floor on three pairs in four and beats it on the rest, once by both margins together. At three rows the switch arithmetic stops being a floor. The incentive does not.

Start at four

Every order a program could compute had made the Sylver Coinage search that finishes dearer than trying the smallest move first, and the free oracle that puts a winning move first costs under a third of it. Two orders a game program would actually use do better: a history table of numbers that have won before wins back 27 per cent of that gap, and a fixed order learned on small positions the same on larger ones. Both are learning one thing. Trying the smallest move first, but from four, wins back 31 per cent and beats smallest first on 1,695 of 1,766 positions — because naming 2 leaves a two-generator semigroup, which is symmetric and never lost for the opponent, unless 3 is already reachable.

The same rule on a star

Kōnane turned out to hold the theory's whole vocabulary, and one old game is an anecdote. Mū tōrere, a Māori game on an eight-pointed star, has the same rule for losing — a player who cannot move loses — and is small enough to solve whole. Of its 630 arrangements, not one is a win for whoever moves first or a loss for whoever must move, sixteen go to each side whoever starts, and most are drawn. A random player still loses it every time. The vocabulary was never in the rule for losing; it was in the captures.

A win with no pairing. A graph of ten points, each with three neighbours, in which Short claiming points and Cut deleting them, Short wins moving second. No pairing of the eight contested points puts a whole pair inside all 10 minimal separating sets; the closest misses {e, f, h}, the three neighbours of A.

Two neighbours keep a pairing, three lose it

In the switching game on points, Short wins moving second by a pairing when every smallest set of points that cuts A from B holds both points of some pair. On every graph of seven points — 971,104 of them — every Short win has one, at every degree. In random link games, where each contested point has two neighbours, every win has one too, although the winning strategy follows trees that change as the game goes. The first wins with no pairing appear at nine points with four neighbours and at ten points with three: a cubic graph Short wins whoever moves, which no fixed set of answers explains.

A pairing that changes once

The cubic ten-point graph is a win for Short that no fixed pairing explains, and the full strategy that does explain it is a tree of 68 decisions. One change is enough: after Cut's first deletion Short claims a stated point, and from then on a fixed pairing of what is left holds every cutting set. Four such plans, eight pairs between them, answer every opening. Of the 23 pairless wins in the samples, 21 need one change and two need a second, and in both the second is a forced claim. Unless the hard problems collapse, no fixed number of changes can be enough for every graph.

Three, again. Every set of disjoint pairs meeting each of Across's minimal chains in a whole pair, on Hex boards 2 × 3 to 5 × 6: 2, 3, 3 and 3 pairings on 5, 25, 148 and 1,225 chains. Every one uses every cell, is unchanged by a half-turn, and is written by one of three rules.

Two reflections and a corner

Hex boards of n rows and n + 1 columns are won by the player with the nearer edges, by a pairing, and the count of pairings that do it ran two, three, three. It is three again on five by six, against 1,225 chains, and three on six by seven, against 16,097. They are the same three every time, and they are not three strategies: one is the classical reflection, and the other two are a reflection across the board's other diagonal, which joins its two far corners, in two versions that disagree about six cells. Each is a rule a sentence long, and each holds on every board out to ten by eleven.

The same table, read two ways. Two games of king and rook against king from a position 16 moves from mate. A white who reads the distance descends one move at a time to mate. A white who reads only whether a move keeps the win wanders and is still 13 moves from mate after 80 moves.

A win is not a way of winning

Solve a lone king against a rook completely and every one of 175,168 positions with white to move is won. Now let white play any move the table says keeps the win, and black resist as long as it can: in 400 games of up to 2,000 moves, white never mates. On a graph that loops, staying won is not progress. From every won position in the queen and rook endings white can stay inside the won set for ever, and what wins is a second number, the distance, which the theory of finite games never had to ask for.

What a corner adds. The replies to a centre opening at noughts and crosses. As played, corners draw and edges lose. In the game where the second player can only block, every reply loses.

Blocking is not enough

Everybody knows noughts and crosses is a draw, and the usual explanation is that the second player can always block. That explanation is false. Against an opponent who can only block, the first player wins, from the centre or any corner. The game is a draw only because the second player's blocks build threats of their own. On fifteen boards solved outright, it is the only one where those two things come apart: everywhere else a draw is a draw because blocking is enough.

Three certificates and the truth. The drawn m,n,k boards up to twenty cells, with whether a pairing exists, the Erdős–Selfridge potential, the exhaustive Maker–Breaker result, and whether a greedy rule based on the potential holds against every line of play.

Three ways to know a draw

A draw on a board too large to search has to be certified, and there are two standard certificates: a pairing of cells and the Erdős–Selfridge sum over lines. Set against exhaustive search, both miss 4 × 4, and a rule taken from the sum holds there anyway. The sum looks far stronger until pairs are allowed to serve several lines at once. Then across 440 boards up to 12 × 12, the sum certifies no draw that a pairing does not, and a pairing certifies 36 that the sum cannot.

A price in stones per turn. Small boards with the exact number of stones Breaker needs per turn to stop a Maker placing one, the rate the biased Erdős–Selfridge potential guarantees, and the number Maker needs per turn to beat a Breaker placing one.

A win priced in stones

Change the rate of play and a yes-or-no question becomes a number. Against a Maker placing one stone a turn, Breaker needs two stones to stop three in a row on every board up to 4 × 4 and six by three, and three on 5 × 4. Against a Breaker placing one, a Maker needs two stones on most boards drawn at one, and three to make four in a row on 4 × 3. The extra stone is enough to erase noughts and crosses altogether, and the biased sum is exact there and high everywhere else.

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