Combinatorial Game Theory
who wins, and by how much
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The thread: What a search costs

A value is worth what it costs to find. These essays price the search rather than quoting the answer: positions visited, states stored, and the size of the board where the counting stops.
A pairing no motion of the square gives. The smallest Cram shape carrying a pairing that is not a rigid motion, with its three pairs drawn as lines between the squares they join. Impartial games

A pairing that is not a symmetry

Every pairing strategy this ladder has found is a rigid motion of the square, and the requirement mentions no geometry at all. Searching all 8.8 million fixed-point-free involutions instead of the eight maps more than doubles what a pairing explains — and the smallest new one turns out to be a reflection with its two fixed squares swapped.

 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

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.

 Every position, by how much is left. Sylver Coinage positions counted by genus — the number of integers still unnameable — with the share on which the player to move loses. The parity of the genus very nearly decides the game: odd rows run between a fifth and a half, even rows between nothing and a thirteenth. Out in the world

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.

 How far a plain search gets. An exhaustive search of Sprouts and Brussels Sprouts, run on this site, with the number of positions each size costs. Sprouts settles at three spots and Brussels Sprouts at two crosses; the published results on Sprouts go to forty-seven. How it was found

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.

 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. How it was found

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.

 What the theorem replaces. Every grid a brute-force solve can reach, valued both ways, with the grid the theorem is normally drawn at underneath. Twelve squares is four thousand arrangements against twelve products; sixty-four squares is eighteen quintillion against sixty-four. Impartial games

Four hundred and seventy steps

The tartan theorem replaces a search with a multiplication. Measured on every grid a brute-force solve can reach, the two agree on all of them — and the ratio doubles with every square added. On the 8 × 8 grid the theorem is normally drawn at, the search would have to value eighteen quintillion arrangements; the theorem needs twenty-six different nimber products, and computing all of them by the rule that defines them looks at four hundred and seventy pairs.

 The order one day out. Whether the values born by day three still form a lattice. Twice as many pairs are incomparable as at day two, and every incomparable pair still has a least upper bound and a greatest lower bound — so the order becomes more tangled without becoming ragged. Values

One of four questions

Three rungs of this ladder rest on sweeps of day two — 22 values, 253 pairs. Day three is 1,474 values and over a million pairs, and only one of the four questions can be asked of it. The order can: twice as many pairs are incomparable and every one of 1,606 sampled still has a least upper bound and a greatest lower bound, none of them a value day two already had. The other three compare sums of day-three values, which are born on day six, and sixty of those exhausted an eight-gigabyte heap.

 The third point. Amazons regions on a 4 × 4 board holding one amazon of each colour, by the Chebyshev distance between them. A 3 × 3 board reaches distance two and this one reaches three, and the mean temperature rises across all three. Particular games

The third point on the curve

Four rungs below this one measure a shared Amazons region's temperature against how far apart its two amazons are, and every one does it on a 3 × 3 board, where the distance can only be 1 or 2. Two points make a direction, not a curve. A 4 × 4 board reaches distance 3 — and cannot be evaluated at all until the regions are cut to five free squares. Restricted that far, the answer is neither a sign that flips nor an oscillation: the rise continues and it is running out, the second step being 36 per cent of the first.

 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. How it was found

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.

 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. How it was found

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.

 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. How it was found

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.

 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. Out in the world

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.

 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. How it was found

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.

 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. What it costs

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. What it costs

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. What it costs

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. What it costs

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. What it costs

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. What it costs

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. What it costs

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. What it costs

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.

 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. Out in the world

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 gaps of ⟨5, 7, 9, 11⟩, and which of them the pairing removes. A Sylver Coinage position drawn as the numerical semigroup it is, with every legal move marked by what a single pass says about it. Gold squares are the numbers already named; plain squares are sums of them; magenta squares are the gaps, which are the legal moves. Under each gap is "struck" when the position that move reaches has its own gaps paired around its own largest gap — a position the opponent wins, so the move loses — and "wins" when the search says the move wins. Out in the world

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.

 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. How it was found

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.

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