Combinatorial Game Theory
who wins, and by how much
Field

What it costs

Every theorem here can be true and the answer still out of reach. What a search costs, what a proof of a win looks like, and where the shortcuts are.
Which questions are answerable. The theory is exact and much of it is expensive. Values are computable by definition; computing one for a position of any size is a different matter, and deciding the winner of a generalised board game is complete for PSPACE — as hard as anything solvable in polynomial space.

How hard is it

Every theorem on this site stays true at any size. The answers stop being reachable long before the games get interesting — deciding the winner of a generalised board game is PSPACE-complete, and an exact evaluator gives out after a few dozen moves.

 A formula, drawn as a game. A token on a directed graph. A move slides it along an edge to a vertex not yet visited, and a player who cannot move loses. That is the whole game, and deciding who wins it is as hard as anything decidable in polynomial space — which is the strongest hardness claim anybody makes about a combinatorial game.

Hard, proved

A game is as hard as a logical formula when the formula can be drawn as the game. Here is the drawing — a quantified formula turned into a graph with a token on it — with every formula over three variables played both ways and required to agree.

 What it costs to decide the winner. Three families of game, each at several sizes, with the number of distinct positions an exhaustive solver must evaluate beside the work its closed form does. The bars are logarithmic. Hardness is not about the size of the board or the length of the rules — Domineering has the shortest rule here — it is about whether anybody has found the shortcut.

The game with the shortest rule is the hard one

Deciding a generalised board game is PSPACE-complete, which is a statement about families and encodings rather than about size. Nim in the same subject is settled by one pass over the input at any size, and green Hackenbush by one pass over the edges — while Domineering, whose rules take a single line, has no shortcut anybody has found.

 One node per route, one node per position. For each board, the number of nodes in the recursion tree a solver with no memo table would walk, beside the number of distinct positions that tree contains, beside the longest run of moves in it. The first number is the cost of forgetting; the second is the size of the table that avoids it; the third is the stack, and it stays small however the other two grow.

A position reached eleven ways is one position

A 4×4 Domineering board has 5,700 positions in it and 6,257,129 routes through them. Three heaps of 7, 11 and 13 have 480 positions and 7.6 × 10¹⁶ routes. The gap between those two numbers is not an optimisation — it is the difference between a search that finishes and one that does not.

 A board in pieces costs the sum, not the product. A Domineering board with squares blocked out, so that it falls into regions no domino can span. The number of positions in the whole board is exactly the product of the numbers in its regions — which is why evaluating the regions separately, and adding the values, is an exponential saving rather than a tidier way of writing the same search.

The board falls apart, and the arithmetic changes

A 4×5 Domineering board with a wall down the middle has 2,916 positions in it, and that number is exactly 54 × 54 — the product of its two halves. Solving the halves separately costs 108. Decomposition is the one saving in this subject that turns a product into a sum.

 Folding a 4×4 board by its symmetries. The size of a Domineering solver's table when positions related by a board symmetry are stored once. The saving rises toward the size of the symmetry group and stops there — it is a constant factor by construction, and no board is large enough to make it anything else.

What counts as the same position, and what that is worth

Folding a 4×4 Domineering board by its symmetries takes the table from 5,700 entries to 1,522 — a saving of 3.75, against a ceiling of exactly 4. An orbit cannot be larger than the group acting on it, so this is the one saving in the subject that can never change an exponent.

 Every quantifier is a move. A quantified boolean formula with its quantifiers drawn as turns: an existential is a choice by the player to move, a universal a choice by the opponent. The same formula is put through the reduction to Generalized Geography and the two answers are checked against each other, so the prefix of quantifiers and the game beside it are one claim.

A puzzle asks once, a game asks alternately

Quantifier alternation is the whole difference between a puzzle and a game. One chooser is an existential and its answer is a witness somebody can check; two choosers taking turns is a prefix of alternating quantifiers, and the witness stops being an assignment and becomes a strategy.

 The same position, and two rulers to measure it with. Nim positions with the length of their input under two encodings — the heap sizes in binary, and the counters themselves — beside the work the nim-sum does. The work never changes. Which of the two lengths it is compared against decides whether the same algorithm reads as linear or as exponentially fast, and hardness claims are always made against one particular encoding.

Nim is easy, in binary

Three heaps of a thousand counters take thirty bits to write down and three thousand counters to lay out. The nim-sum does three exclusive-ors either way. Whether that counts as fast depends entirely on which of the two numbers the work is compared against.

 A winning strategy on 3×3, drawn whole. The whole of one player's winning strategy on a small Domineering board: their own move at each of their turns, and every reply the opponent has at each of theirs. The strategy branches only where the loser chooses. Its size is what somebody would have to be handed to check the claim that this player wins, and it is far larger than the claim itself.

"Left wins" has no short proof

A complete solution of Nim on heaps of 7, 11 and 13 is 480 table entries. A winning strategy for the same position — one move of the winner's at each of their turns, and an answer to every reply — has 56,167,022 nodes in it. The answer is smaller than the proof by a factor of a hundred thousand.

 One node per route, one node per position. For each board, the number of nodes in the recursion tree a solver with no memo table would walk, beside the number of distinct positions that tree contains, beside the longest run of moves in it. The first number is the cost of forgetting; the second is the size of the table that avoids it; the third is the stack, and it stays small however the other two grow.

The class is named after memory, and that is not an accident

A 4×4 Domineering board has 6,257,129 routes through it, 5,700 distinct positions, and a deepest line eight moves long. Those three numbers are three different resources, and the smallest of them is the one that gives games their complexity class.

 Three things the word “solved” is used for. The three standard senses of a solved game, priced on positions this solver can settle completely. Ultra-weak names the winner; weak supplies a strategy from the opening; strong supplies one from every position. They differ by orders of magnitude, and a claim that a game is solved is nearly useless until it says which of the three it means.

Three different claims are all called solved

Hex is solved in the sense that the first player provably wins, by an argument that names no move whatever. Nim is solved in the sense that a formula gives the right move from any position at any size. Between them sit strategies for one opening, and databases of a few billion positions. The word covers all four.

 Subtraction of 1, 3, 4 — and the window that proves the period. The Grundy values of a subtraction game, with the window that certifies the period marked. Everything after the window follows from it by induction, because a value is a mex over values at most one move back — so a finite check settles the whole infinite sequence, and the thousands of further values computed here agree with a claim that was already proved.

Four values, and the sequence is settled for ever

The Grundy values of a subtraction game repeat with period 7, and proving it needs a window of exactly four of them — one for each size of move the game allows. Everything past the window follows by induction. A finite computation has settled a claim about every heap there will ever be.

 the temperature that bounds the loss. Temperature runs up the page and value across it. Each wall is where a player is willing to move once a tax of that much is charged per move; above the temperature at which they meet, neither wants to move and the position is worth its mean value. The height of the meeting point is what is at stake.

A rule that is never right and cannot be far wrong

Playing the hottest component is not optimal — over 440 measured lines it costs something on 17 of them. What makes it worth having is that the cost is bounded by the temperature, provably, and that the same test run with the ordering reversed breaks the bound on 54.

 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.

The cost is in the closure, not in the positions

Under normal play, Dawson's chess needs four classes for every heap up to twelve, because its Grundy values stay at three or below there. Under misère play the same game needs six, then twelve, and the number rises with the universe rather than with the position — which is a different kind of expense entirely.

 Finding the parts costs the same whether there are any or not. Domineering boards of 4 squares by 5 with different squares blocked out, and what the decomposition is worth on each. The pass that finds the regions is a flood fill and visits every square once, so it costs the same on all of them. What it buys ranges from nothing — on the boards that do not decompose — to a saving of 2,808 positions, and it cannot tell which case it is in until it has run.

Finding the parts

Decomposition turns a product into a sum and is the largest saving in the subject. Nobody labels the regions. The pass that finds them costs the same on every board of a size — including the boards where there is nothing to find — and what it buys ranges from four orders of magnitude to nothing at all.

 Knowing who wins, and knowing what it is worth. Nine positions, each evaluated twice by an instrumented evaluator that starts with an empty cache. The third column counts what deciding the winner costs and the fourth counts what the canonical form costs, in the currency each question is actually paid in.

Knowing who wins, and knowing what it is worth

Deciding a winner expands positions. Computing a canonical form expands pairs of positions, because a comparison unfolds as a recursion over one subposition of each and the reduction makes many comparisons. Measured on the same nine positions by an evaluator that starts empty every time, the second costs between 1.3 and 279 times the first, and the ratio grows with the tree.

 Three questions about the same board. For each sum of two positions: the cost of deciding who wins each part alone, of deciding who wins the whole sum by search, and of computing what each part is worth. The middle question is in the middle on seven of the eight, and the exception is the sum whose two parts are identical.

The question in the middle

Between knowing who wins each part and knowing what each part is worth sits the question a player actually has: who wins the board. Priced on sums of two it lands between the other two on seven of eight, cheaper than the values by up to eight times. On sums of three, with nothing repeated, it is dearer than the values on five of six — because a component multiplies a search and only adds to a value.

 Where in a game a board falls apart. Every position reachable from an empty Domineering board, grouped by how many dominoes have been placed, with the share that have fallen into two or more live pieces. The share is nought at both ends of the game and around three fifths in the middle.

How often a board falls apart

A decomposition turns a product into a sum, so a solver wants to know how often one arrives. Over every position of a 4 × 4 Domineering board the answer is 47 per cent — nought for the first two moves, three fifths in the middle, and nought again at the end. What one decomposition is worth is the other half of the answer and it is a factor of 1.8.

 When a catalogue starts paying. How many decomposed boards a catalogue of regions has to answer before building it costs less than searching each board directly. Five boards for regions of four squares, two hundred for regions of eight.

When the catalogue starts paying

The rung below priced two questions — who wins one board, and what it is worth — and named the third: a program pays for a family of regions once and answers every board over them by addition. The crossover is between five boards and two hundred, depending on how far the catalogue reaches, and it falls as the board grows. The whole catalogue of every region to eight squares costs one part in seventy-six of one undecomposed five-by-five board.

 What each reach answers. Catalogues of four to ten squares against the regions four sizes of board actually produce. The coverage rises from about 54 per cent to about 74 while the catalogue grows from 15 shapes to 12,871.

Where to stop building

The rung below priced a catalogue of small regions against the search it replaces and found the crossover. What it could not say is how far to build, and the coverage answers that: going from four squares of reach to ten multiplies the catalogue by 860 and lifts the share of regions it answers from 54 per cent to 74. The price of a point of coverage runs from five shapes to five thousand.

 The same coverage, an eighth of the shapes. Catalogues ordered by size against catalogues ordered by frequency, at the same coverage. The frequency order wins at every reach and by more at each one.

A catalogue that knows what it will meet

The rung below priced a catalogue of regions by its reach and found the coverage saturating, and asked what a catalogue ordered by frequency would cost instead. Eight shapes answer half the components a played Domineering board produces; a catalogue by size needs fifteen for the same, and 1,042 for what 119 chosen by frequency reach. Three quarters of a size-ordered catalogue never turns up in play at all.

 Which catalogue is safe. Catalogues built from one style of play and used against another. A catalogue measured on random play over-serves a strong player and not the reverse.

The catalogue a strong player needs

A Domineering catalogue built from random play faces an objection that could overturn it: random play is not play. A player that reads the board produces the same head — eight of the ten commonest shapes — and concentrates far harder: 114 entries answer nine tenths of what it meets, against 2,018. And a catalogue measured on random play over-serves it, while the reverse fails.

 Three catalogues, ten entries each. The catalogue built from a sweep against two self-built ones, on reach and on content.

A catalogue that builds itself

A solver that stores every region it has to evaluate builds a catalogue out of its own games. After 650 games it holds 232 of the 1,042 shapes and is still growing — and the order things arrive in is nearly arbitrary while the order they are consulted in reproduces a census of a strong player's games almost exactly.

 The online rule beats the oracle. The three offline orderings against the two online eviction rules, all at ten entries over the same lookups.

The table that changes its mind

The advice that ten entries chosen by use serve nine lookups in ten was untested: it describes a table sorted after the fact rather than a solver that only ever held ten. A solver that only ever held ten gets 94.2 per cent — beating the best ten chosen with the whole run in view, because there is no best ten.

 One board, two answers to how many pieces it is in. Every position reachable from a small Amazons opening, counted by depth, under two ways of deciding whether two squares are in the same region. Counting only edge neighbours, a third of all positions are in pieces; counting corners too, an eighth are.

A wall an amazon can walk through

An arrow burns a square for good, so an Amazons board that has fallen into pieces should stay in pieces. Over 127,583 positions it does not: fifty-one thousand moves put two regions back together. Every one of them is a single diagonal step, and what is wrong is not the game but the rule used to find the regions — which was borrowed from a game whose pieces lie along the board's own lines.

 What the folding costs to do. The same search over a 4 × 4 Domineering board run twice, once folding positions by symmetry and once not, with everything counted. The fold stores 3.75 times fewer entries and spends 17.5 times more elementary operations to decide where to put them.

What it costs to notice a repetition

Folding a 4 × 4 Domineering board by its symmetries takes the table from 5,700 entries to 1,522. It also spends 559,424 square-mappings to work out where each entry goes — seventeen and a half times the entire cost of not folding. The saving has a ceiling of four and the price has no ceiling at all, and knowing which currency each is paid in is the difference between an optimisation and a habit.

 A count that forgets, at six rates of forgetting. The hit rate of a component table limited to 10, 20 and 40 entries under recency, under a use count halved every 25 to 5,000 lookups, under a use count never forgotten, and against the best fixed table chosen with the whole run in view. At every size some half-life beats recency, and the longest half-lives fall back toward the rule that never forgets.

A count that forgets

A Domineering solver with room for ten component values does better evicting whatever it used least recently than evicting whatever it used least often, and the explanation offered was that a use count never forgets. Halve every count at a fixed interval and the count overtakes recency at every table size — by less than half a point, and only with the right interval. The right interval grows with the table: a quarter of a game's worth of lookups at ten entries, five games' worth at forty.

 The board held fixed, and recency still wins. Each of the four board sizes played on its own for 650 games, with a component table of ten entries under recency, a use count never forgotten and the best of three half-lives, against the best fixed ten shapes chosen with the whole run in view. Counting beats recency only on the 4 × 5 board; on 5 × 5, 6 × 6 and 7 × 7 recency beats both counting and the fixed table, by the widest margin on 7 × 7.

One board, and recency still wins

A Domineering solver's table of component values did best evicting whatever it used least recently, and the explanation was that the run changed board size three times. Take the change away — play all 650 games on one board — and counting wins back its lead only on the smallest board. On 5 × 5, 6 × 6 and 7 × 7 recency still beats both counting and the best fixed table, by the most on the largest. The locality recency exploits is not between boards or between opening and endgame. It is inside a single move.

 How much of the board a who-wins search walks. Positions a memoised who-wins search of 4 × 5 Domineering expands under six move orderings, drawn to scale against the 48,670 positions the board has. Leaving the opponent fewest replies expands 1,125; trying losing moves first expands 30,202; all six find the same winner.

The order a solver tries the moves in

A memoised search asking who wins 4 × 5 Domineering expands 1,125 positions when it tries first the move that leaves the opponent fewest replies, and 30,202 when it tries losing moves first — the same answer at twenty-seven times the price. The ordering that already knows which moves win is not the cheapest. A win needs one move and a loss needs all of them, so the price of an order is paid one level down, in the replies it leaves.

 Right, wrong, and right again. A 4 × 5 Domineering position with Right to move, which Right loses, beside what a search cut at each depth from 0 to 9 says about it when it guesses that the player with more placements wins. The guess alone is right, a search one move deeper is wrong, and every deeper search is right.

A verdict that changes with the depth

A who-wins search of 4 × 5 Domineering cut at a fixed depth, guessing that the player with more placements wins where it stops, is right about 72.5 per cent of positions at depth 0 and about every one of them by depth 7. On the way, 4,697 positions are right at one depth and wrong at a deeper one. With a guess that knows nothing, going one move deeper makes the search worse — and its errors alternate in kind with the parity of the depth, so that half its verdicts are proofs.

 What a short key gets wrong. A memoised who-wins search of 4 × 5 Domineering using a Zobrist key of 10 to 32 bits, run under 60 random keys at each length and checked against the exact answer: how many stored positions share a key, how many runs store a wrong verdict or name the wrong winner, and what checking the whole position would cost instead. At 16 bits 59 runs store a wrong verdict and 19 name the wrong winner.

A key shorter than the position

A who-wins table for 4 × 5 Domineering addressed by a 16-bit Zobrist key stores a wrong verdict in 59 runs of 60 and names the wrong winner of the empty board in 19. The pairs of positions sharing a key follow the birthday count exactly while addresses are scarce, and fall away to nothing once the key has more bits than the board has squares, because a Zobrist key is linear. Symmetry and value identify positions that really are the same; a short key identifies positions that differ, at a rate set by arithmetic.

 Two depths that agree. A 4 × 5 Domineering position with Right to move, which Right wins, beside what a search to each depth from 0 to 6 says under the guess that any mover wins. The verdicts alternate until depths 2 and 3 agree, which certifies the answer 3 moves before the longest line.

Where a search may stop

A search deepened until two consecutive depths agree carries a proof of its verdict, and on 4 × 5 Domineering it stops before the longest line on 17,589 of 48,670 positions. It also costs three times what the search that simply finishes costs. The rule that pays is the other one. Search on wherever the two players' counts of placements are within one, and at depth 2 the wrong verdicts fall from 2,140 to 86 for about a quarter more work per search.

 One family of confused positions. Two 4 × 5 Domineering positions with Left to move that share every bit of an 18-bit Zobrist key, differing only on 4 marked squares whose words cancel. Under that key 100 of 136 confused pairs differ on exactly those squares; its other families hold 25 and 11.

A check bit halves the average and not the key

Real transposition tables keep a few of a key's bits beside each verdict and trust an entry only when they match. On 4 × 5 Domineering each such check bit halves the average number of wrong verdicts, exactly as the birthday count says. It does not halve any one key's. A Zobrist key confuses positions in families — every pair that differs on one set of squares whose words cancel — and a bit removes a family whole or not at all, so from eighteen bits to nineteen thirty of fifty-eight keys lose every confusion and fourteen keep every one.

 Using up edges instead of vertices. An undirected graph of 4 vertices and 4 edges, with the winner at every start of two games on it: vertex geography, where a move uses up the vertex it leaves, and edge geography, where it uses up the edge it crosses. The matching criterion decides the vertex game everywhere and is right about the edge game at 0 of 4 starts.

Using up the edges instead

Undirected geography is decided by a maximum matching when a move uses up the vertex it leaves. Use up the edge it crosses instead and the matching is exact on every tree — on a tree the two games are one game — and on nothing else. Over every connected graph on up to six vertices it names the winner at 480 of 745 starts once there is a cycle, it gets a four-cycle wrong from every start, and the more cycles a graph has, the more of its misses are wins that are really losses.

 One cycle, and how much of it the criterion misses. The matching criterion for undirected geography put to the game where a move uses up the edge it crosses, on every connected graph up to 6 vertices with exactly one cycle. It is right at 77 of 114 starts, and at none of the six on the graph whose cycle has six vertices.

Two graphs a rule cannot tell apart

The repair proposed for the matching criterion was to read the cycle as well. Over every connected graph with exactly one cycle up to six vertices — 21 graphs, 114 starts — eleven such rules reach at most 91, and the winner is not a function of the matching, the cycle's length, the start's distance from it, its degree and the edge count together: six cells of that table hold both verdicts, the smallest a pair of five-edge graphs.

 The bound, and the cost, on four pools. Hottest-first play against optimal play on boards of two, three and four components drawn from four pools, 4,240 lines in all. The bound on the cost is the largest temperature on the board, which reaches 3; the worst cost measured anywhere is 1, and 100 lines meet the bound exactly.

The bound names the hottest part and the cost does not

Moving in the hottest component costs at most the largest temperature on the board, and that bound is attained: 100 lines of 4,240 pay exactly it. It is still the wrong quantity. Across four pools and boards of two, three and four parts the cost is nothing on 90.8% of lines and otherwise takes one of two values — half a point or one — on boards whose largest temperature runs to three, and it exceeds the coolest component on 13 lines and twice it on none.

 The more cheap fights there are, the cheaper the rule is. Hottest-first play against optimal play over 10,410 lines, grouped by how many components share the lowest temperature on the board. Lines costing more than that temperature occur only where one or two components share it; over the 3,230 lines with three or more, none does.

The cheap fights make the rule cheaper

A conjecture stands that playing the hottest part costs at most the coolest temperature times the number of parts sharing it — proposed on a range where that number never exceeds two. Swept to five-part boards over 10,410 lines it is false, and false the other way round: every line costing more than the coolest part has one or two parts at that temperature, and over the 3,230 lines with three or more, not one does.

 Two squares a key never needs. A 4 × 5 Domineering board shaded like a chessboard, with two squares of the same shade in rows 1 and 2 marked. Because a domino covers one square of each shade, a vertical domino covers one square in an odd row and one in an even row, and turns alternate, the other eighteen squares determine both marked squares: a key that leaves them out confuses none of the 48,670 reachable positions.

A key is a code, and two squares come free

The families of positions a Zobrist key confuses are the words of a binary linear code — the sets of squares whose words cancel — so choosing a key is choosing a code. On 4 × 5 Domineering the textbook choice, a code with the largest minimum distance, confuses more stored positions than a random key at fourteen, sixteen and nineteen bits. The choice that reads the board confuses none at eighteen: two squares of one shade, in rows of different parity, are decided by the other eighteen, and no seventeen-bit key is exact.

 Two ways to search on, one position. A 4 × 5 Domineering position with Right to move, which Right wins, beside what deepening says at each depth when it declines to guess where the counts of placements are close. Searching on one move at a time, depths 0 and 1 agree on the wrong verdict; searching on two moves at a time, the search stops at depth 3 with the right one.

Search on in pairs of moves

Deepening until two depths agree gives a proved verdict, and searching on where the counts are close gives a better one; put together the obvious way, they stop on a wrong verdict at 3,231 positions of 4 × 5 Domineering. A guess one move past the cut has the other player to move and flatters the wrong side. Searching on two moves at a time keeps the proof, and the window that suits it is one-sided — but however it is widened, the certificate gets cheaper only by turning into the search that finishes, and on four boards it never gets below it.

 6 turns in strict alternation, priced. One quantifier prefix taken apart into its blocks, with the size of a winning strategy computed a term at a time. A choice made after k of the opponent's turns has to be written down once for each of the 2^k lines the opponent can produce, so the total depends on where the opponent's turns sit and not merely on how many there are.

Twelve turns, and three different prices

The earlier essay prices a universal quantifier at a doubling and leaves it there. Twelve turns with six of them the opponent's cost 6, 63 or 384 decisions to write down, depending on nothing but the order the turns come in — and the cheap arrangements are cheap for only one of the two players. What a claim costs is the number of times the choosing changes hands.

 How many turns are choices. Every position of each game with both sides to move, classified by whether the turn is a choice at all: no move, exactly one move, several moves that all lead to the same verdict, and several that do not. Only the last is a turn at which the alternation is doing any work.

Eleven moves and one decision

A prefix has one quantifier a turn, so a game of eleven moves is eleven alternations. Counted on the boards themselves, a Toads and Frogs strip of eleven moves has twenty-six turns with exactly one move available and one turn anywhere at which the choice changes the answer; a Clobber board has a hundred and fourteen turns and none. Nim, the game everybody calls solved, decides at four turns in five.

 A win is proved by one move and a loss by all of them. The smallest proof of each position's verdict, averaged by verdict. At a node the mover wins the proof takes the cheapest single option; at a node the mover loses it has to answer every option, which is the existential and universal quantifiers of the prefix showing up as two different objects.

Proving a loss means answering everything

A win is established by one move and a loss by every move, so the two verdicts are certified by objects of different shapes. Measured over every position of four games, a loss costs between 1.07 and 2.31 times a win — a small constant, never an exponential. The obvious explanation is the branching and it is wrong: Nim answers six options at a losing turn and pays 2.18, not six.

 What the opponent's choosing is worth. Every position answered twice: against an opponent who searches, and against one following a fixed rule with no search in it. Only a loss can change, so the share is taken over the losses. The spread between games is the measurement — in one of them nearly every loss is recovered and in another none is.

The opponent stops choosing

Replace one player by a rule with no search in it and the question has one chooser left, which is a puzzle rather than a game. Nim recovers five of its six lost positions that way, and six of seven on three heaps of five. Domineering recovers six of a hundred and twenty-two while the fixed rule throws away a winning move eighty-eight times, and one Clobber board recovers none at all — because on that board no rule can misplay.

 A turn is not a bit. The number of turns a game lasts, beside the number of quantified bits those turns amount to. Each ply is measured over the positions actually reachable at it rather than along one line, and the bits are the logarithm of the branching, which is what a quantifier prefix would need one of.

A turn is not a bit

The prefix a game is read as gives each player one quantifier a turn, and a turn on a board is a choice among however many moves there are. Nim on heaps of 3, 4 and 5 lasts twelve moves and carries 23.6 bits of choice; a Toads and Frogs strip lasts eleven and carries two. Corrected for that, the model predicts a strategy 539 times too large on one board and 67 times too small on another, and the two failures have different causes.

 One misère outcome, searched. The number of positions a misère search visits to decide the outcome of a sum of k heaps of Dawson's chess, each heap at most 9, on a logarithmic scale: the average over the sums and the worst single sum, for k from one to eight. Normal play decides the same sums from 20 stored values.

A misère sum is searched, not added

Under normal play the outcome of a sum of heaps is a nim-sum of numbers already known: twenty stored values decide every sum of Dawson's chess with heaps up to nine, however many heaps it has. Under misère play each sum is a new position to search. One outcome costs six positions for a single heap, two hundred for four heaps and over five thousand for eight, and a table of every eight-heap outcome costs a hundred thousand. The misère quotient is the only thing that brings the price back down.

 The closure that is enough. A grid for Dawson's chess with heaps up to 9: rows are the largest positions classified, from one heap to four; columns the largest tests, from none to five heaps. Each cell is the number of classes found. The counts stop growing at two-heap tests and three-heap positions.

Two heaps of testing are enough

A misère quotient is computed by testing positions against positions, and the universe used to find twelve classes of Dawson's chess was every position of up to four heaps tested against every other — 511,225 outcomes. Varied one size at a time, the count stops growing at tests of two heaps and positions of three: 12,100 outcomes find the same twelve classes. The narrower universe the earlier essay drew did not merge anything; it held fewer positions. And the corner that is enough moves: for Kayles at heap twelve, two-heap tests miss a class.

 12 classes, 7 questions. A grid for Dawson's chess with heaps up to nine: rows are the 12 misère classes of positions of at most four heaps, columns the 7 tests a greedy search chose, and each cell the outcome — N for the player to move, P for the other — when the test is added to the class.

Twelve classes, seven questions

Twelve misère classes of Dawson's chess were found by testing 715 positions against 715 others. Seven of those tests are enough to tell every class from every other — a greedy choice against a floor of four, since each test is one yes-or-no question. Kayles needs nine of 715 and Nim sixteen. The seven cost almost nothing to use and cannot be found without the whole closure, and they do not carry: the tests found with heaps up to seven tell apart only seven of the twelve classes with heaps up to nine.

 A staircase, not a slope. The number of misère classes of Dawson's chess positions as the largest heap allowed rises from three to 16, computed with positions of at most three heaps and tests of at most two. The count stays flat for several heaps at a time and then jumps.

A staircase, not a slope

With the misère closure cut forty-fold, Dawson's chess can be classified at heaps far beyond nine. The count of classes is a staircase: six from heap three to eight, twelve from nine to twelve, seventeen from thirteen to sixteen. Normal play steps once in that range, from four to eight at heap thirteen, where a Grundy value of four first appears. Misère play steps there too, and once more at heap nine, where normal play does not move at all — the first wild heap. Heaps eleven, fifteen and sixteen are also wild and move nothing.

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