Combinatorial Game Theory
who wins, and by how much
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The thread: It has to end — page 2

Every value here is defined by a recursion that needs play to stop. Sometimes that is obvious, sometimes it is a theorem, and sometimes the game ends with nothing bounding when.
A game every play of which ends, and no round settles. A game whose first move chooses how long the game will be, cut off at several sizes. Every play of it is finite and no position is drawn, so the fourth outcome class has nothing to do with what goes wrong. What goes wrong is the round counter: the opening is a loss, a loss settles only when the last of its options is known, and there is no last option. Cut the game off larger and the round grows, so no number in the column is the answer for the untruncated game — and the induction that labels it has to run past every finite stage. How it was found

Every play ends and no round settles

Take the finiteness hypothesis away carefully — not by adding a cycle, which has already been priced twice, but by adding infinitely many positions to a game every play of which still ends. Nothing is drawn, every line finishes, and the round the opening settles in grows with every cut: two, four, six, eight, twelve, sixteen, and no number in the column is the answer.

 A shuttle and a loop, judged by what the play returns to. A three-node loopy game drawn as a graph, with Left's moves in blue, Right's in red and position a marked. Beside it, each position-and-mover pair under the backward labelling and under the rule that a never-ending play goes to Left when it returns to a infinitely often. Four pairs are drawn by the labelling; the new rule gives two to Left and two to Right and leaves the decided pairs as they were. How it was found

What the play keeps coming back to

A draw is what the backward labelling never reaches, and handing every never-ending play to one player turns the draws into wins wholesale. Judge an infinite play instead by what it keeps returning to, and every draw gets a winner of its own: over the 262,144 three-node games, 15,432 send some of their draws to one player and some to the other, which no wholesale rule can do. Finding those winners takes a fixed point inside a fixed point.

 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.

 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.

 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.

 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. What it costs

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.

 Every two-position loopy region, as two names. The 256 loopy regions of two positions, placed by the names of their onside and offside as identified against the 1,474 values born by day three. Ten names cover every side: 0, 1, −1, ∗, on, off, over, under, upon + ∗ and −upon + ∗. The largest groups are on & off with 94 regions and off & off and on & on with 53 each; 25 regions need only finite names. How it was found

A loop is written with two names

A region with a cycle in it has no brace expression, and every one of the 256 regions of two positions can be written anyway — as two names, the game it is when a play that never ends goes to Left and the game it is when it goes to Right. Checked against all 1,474 values born by day three, ten names cover every side, 25 regions need only finite ones, and the pair predicts every sum with a finite game, draws included: a draw arrives exactly where the two names disagree.

 Heights from ω² down to nought. Thirteen Sylver Coinage positions with the gcd of the named numbers, its prime factors counted with repeats, the gaps of the semigroup scaled by the gcd, and the ordinal height those give: ω² for the empty position, ω·4 for {16}, ω for {2}, 3 for {3, 5} and 0 for {2, 3}. Where it stops

No number bounds it

Sylver Coinage cannot be given a termination measure in the natural numbers: from {2} the game can last one move, two, or any number the second player chooses. The measure proposed for it — the gap count beside the number of distinct primes the named numbers share — rises on {4} + 6. Count the primes with their repeats and it falls on every one of 42,519 moves checked, and it is exact: the game tree is ω squared tall, and the height of a position is ω times the prime factors of its gcd plus the gaps that are left.

 Every game on three positions. All sixteen directed graphs on three vertices up to relabelling, each with one of its stable sets filled and the count written beneath. Nine have one, six have several and the 3-cycle has none. How it was found

The solution that was a losing set

In 1944 von Neumann and Morgenstern defined the solution of a game by two conditions on a set of outcomes: none overrules another, and each outcome outside is overruled by one inside. They were writing about coalitions. The same two conditions, read with a move in place of an overruling, describe exactly the lost positions of a game that must end. Where play can loop, the same definition gives several answers or none, and three positions are enough to show both.

 Parity on a cycle. Single directed cycles of two to seven positions. Even cycles have two stable sets, the alternating halves, one of which is filled; odd cycles have none. How it was found

An odd cycle is the only way to have none

A cycle of two positions has two stable sets and a cycle of three has none. Moses Richardson proved in 1953 that this is the whole story of absence: a graph with no odd cycle always has a stable set. Run over all 9,608 games on five positions, the theorem holds without exception, sits inside a chain of five nested families, and leaves open the larger question its converse invites, since 7,452 of those games have an odd cycle and a stable set anyway.

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

A win is not a way of winning

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

 How the word is made. The Prouhet–Thue–Morse word built by doubling. Each row of one, two, four up to thirty-two letters is the row above followed by its swap, which is the same word obtained by replacing A with AB and B with BA. How it was found

A rule that let the game run for ever

In 1929 German chess drew a game when the same sequence of moves came three times running, and the rule was meant to make chess finite. Max Euwe, six years from the world title and trained as a mathematician, showed that it does not, using a word that had already been found twice for other reasons. Two round trips are enough. The board returns to one position thousands of times, no block of play ever comes three times running, and the rule has nothing to say.

 How strict a repetition rule can be before it ends every game. For two, three and four available moves, the longest legal game under each repetition exponent from 7/5 to 3. Two moves stop at three letters at exponent two and never stop above it; three moves stop at 38 letters at 7/4; four moves stop at 121 letters at 7/5. How it was found

How strict a rule can be

A repetition rule can be tightened one notch at a time, and each notch either leaves unending games alive or kills them outright. With two moves a block may not come twice, and every game is over by the fourth letter; relax that by a single letter of overlap and unending games appear. Three moves stop dead at 38 letters under a rule that a whisker's relaxation frees. Four stop at 121. Past each boundary the words go on for ever, and the boundary sits at a fraction nobody would have guessed.

 The repeater's plan on two moves. Every line of play in which the repeater forces the rule of 1929 on two moves, 4 lines of at most 7 letters, each ending in a block three times running. How it was found

The opponent makes it repeat

Euwe's unending game needs both players to keep to it. Give the letters to two players in turn, with one of them wanting the draw that a repetition brings, and the rule of 1929 fires within seven letters on a plan of one sentence: always play A. The word that defeats the rule exists and nobody can be held to it. Add a third round trip and the same player can no longer force anything in twenty-six letters. Existence and force turn out to be separate questions, with separate boundaries.

 A draw by repetition changes no result. Over all labelled games on three and four positions and a sample on five, the result from each start with a draw on the second or third arrival at a position, compared with retrograde labelling of the graph. They agree everywhere. How it was found

What counting positions buys

A rule that counts arrivals at a position ends every game, which the rule of 1929 could not do, and it needs no proof beyond arithmetic: with L arrivals barred on n positions, no game is longer than (L − 1)n − 1 moves. What it does not do is change a single result. Over every game on three and four positions and a sample on five, drawing on the second arrival or the third gives exactly the labels retrograde analysis had before any rule was written. The rule pays for the game's ending and for nothing else.

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