Combinatorial Game Theory
who wins, and by how much
The collection

Every essay — page 2

One idea per essay, ordered so that the earlier ones set up the later ones — but nothing here depends on being read in sequence.

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.

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.

6 figures · Chess 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.

6 figures · Chess 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.

6 figures · Go 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.

6 figures · Go 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.

6 figures · Kōnane 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 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.

6 figures · Kōnane 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

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.

6 figures · Scoring 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.

6 figures · Scoring 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.

6 figures · Scoring 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.

6 figures · Switching 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.

7 figures · Switching 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.

6 figures · Switching Chomp to 12 × 8: one needle on every bar but one. Every Chomp rectangle up to 12 columns by 8 rows, with the number of winning opening moves in each cell, found by search. All but one have exactly one; the 10 × 8 bar has 2. Cells in blue belong to the families whose winning move can be stated in a sentence — a single row, two rows, or a square; cells in gold are found only by searching.

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.

6 figures · Strategy stealing 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.

7 figures · Strategy stealing 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.

6 figures · Sylver

All ladders · Every object named here · The position index · Figures that play back · Search