Combinatorial Game Theory
who wins, and by how much
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The thread: One clause decides it — page 4

Change a word of the rule and the values change completely. A cliff or a wall, a jump allowed or forbidden, a pass that may or may not end the game — the same board, and nothing in common.
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. Out in the world

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.

 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. Out in the world

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.

 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. Out in the world

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.

 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. Out in the world

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 on this site 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. Out in the world

What a pass is worth to a theory

The rung below finds fifteen of twenty-seven coin rows 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 this site 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. Out in the world

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.

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

The convention Dawson actually used

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

 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.

 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. Out in the world

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. Out in the world

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.

 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.

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

The capture that has to be made

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

 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.

 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. Out in the world

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.

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

Two things to hold at once, or three

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

 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. What it costs

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.

 The money played out, and it never mattered. The bidding rule played move by move with a countable pool of chips, at every way of splitting it. The verdict is constant across the splits and opposite under the two ways of resolving equal bids, so what settles these positions is the tie-break rather than the money. Where it stops

The auction never gets to the money

The critical fraction is computed and never played. Played out with a countable pool of chips — twelve positions, four pool sizes, every split of the chips, every bid answered — the verdict does not move with the money on a single one of the forty-eight sweeps, and the rule for equal bids settles all forty-eight. The reason is one line long: declining every auction wins, and bidding nothing declines.

 The same number, from a rule that needs no tie-break. The number computed twice: once as the critical share of a pot under the auction, and once as the probability that Left wins when a fair coin decides who moves at each turn. They agree on every position, and only the second derivation survives being played out. Where it stops

A coin needs no tie-break

The same recursion has a second derivation: a fair coin decides who moves at each turn, a player whose turn it is with no move has lost, and both play to win. Written from those rules it comes out identical on every position — and it needs no rule for equal bids, because there are no bids. The number is a probability, it belongs to Left rather than Right, and the empty position is the one where the coin decides everything.

 The fee the geometry charges. The same endgames solved with the cost of declining changed. Two boxes on a chain and four on a loop are what a single cut and a pair of cuts complete; altering them changes the winner of a large share of positions, which is what says the law depends on them. Out in the world

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.

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

The step nobody took for thirty-four years

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

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

The picture Bouton's proof leaves behind

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

 Lasker's Nim in sixteen cells. A four-by-four table. Each row and column is a residue mod 4 of one part of a split heap, with the residue of that part's Grundy value beside it; each cell is the residue mod 4 of the split's value, the nim-sum of the two parts. Every split of every heap to four hundred lands in the cell its residues name. Impartial games

The proof is sixteen cells

Lasker's Nim has a four-clause formula that was checked on two thousand heaps and never proved. The proof fits in a four-by-four table: the last two bits of a split's value are fixed by the last two bits of its parts, so no split can land in its own heap's class — except at 3 mod 4, where it lands exactly on the one value the takes leave missing and pushes the answer up by one.

 One split is enough, and some are not. Lasker's Nim beside five versions of it that allow only some splits, over the first twenty-four heaps, with every cell that leaves the formula outlined. Allowing only the split that takes one counter off reproduces the whole sequence; allowing only equal halves turns it back into Nim. Impartial games

One split is enough

A heap of n in Lasker's Nim offers ⌊n/2⌋ ways to split, and the values use at most one of them. Allow only the split that takes a single counter off and every heap to six hundred keeps its value; of all sixty-three sets of split sizes up to six, a set keeps the formula exactly when it contains 1 or 2. Equal halves alone give back plain Nim, because a split into equal parts is a move to nought.

 Two heaps and a held pass. Every pair of heaps up to 16 with one pass available that may not be the last move. Filled cells are the pairs the player to move loses: the empty board and the pairs one and two, three and four, five and six, and so on. Outlined cells are the equal pairs Nim calls lost, all of which are wins once the pass is there. Where it stops

Three heaps and a pass

Nim with a single pass that may not end the game is easy on one heap and on two: a heap swaps each odd size with the even one above it, and two heaps lose exactly at (2k − 1, 2k). On three heaps the losses are known only as a list. Fix the smallest heap and each slice of the list settles into a pattern after an irregular start — period 4, 8, 10, then 160 at a smallest heap of ten, and nothing visible from eleven.

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