Combinatorial Game Theory
who wins, and by how much
Concept

Dominance — where it appears

The reduction that deletes an option no player would choose, because another option of the same side is at least as good. It is one of the two operations that produce a canonical form, and it is the reason a form is narrower than the position it came from.

Named by 20 essays across 3 fields — each of them below, with the objects they name alongside it.

Which description of a surviving option is right. The two candidate readings of what the reduction keeps, scored over every Domineering option list on six boards. Taking the most room is right on under half the lists, which is what a description with no content scores on lists this short. Leaving the opponent fewest replies is right on nine in ten.

Which option the reduction keeps

Domination deletes an option when another is at least as good, so what survives is the top of an order. On a board that order is made of moves, and two descriptions of the surviving move suggest themselves. Over 1,586 Domineering option lists one of them is right 47% of the time and the other 90%, and the one that wins is not the one a player would guess.

values · Dominance No fifth value. Forms of day-three values built by adding day-three gift horses, and the ordinal sums they give. Over eighteen thousand forms and four followers, no value's forms disagree.

No fifth value

The colon reads a form rather than a value, and the rung below found the forms of a value disagreeing at exactly four of them — the values born by day one. It could only check forms whose options came from day two. Built one day deeper, by adding day-three gift horses to day-three values, eighteen thousand forms give no disagreement at all, while the same treatment still splits nought four ways. The class is about the width of the base's form and not the depth of its options.

sums · Ordinal sum The margin the count needs. Every pair of Left options sorted by how many more replies one leaves the opponent than the other. At a margin of three the option leaving fewer replies is never the worse one.

The margin a count needs

Leaving the opponent fewest replies names only surviving options nine times in ten, which leaves the question of what a bound stated in that count would have to be weakened to. It is a margin. Over 57,879 pairs of Domineering options, the one leaving the opponent fewer replies is the worse of the two 1,052 times at a margin of one and 72 times at a margin of two — and at a margin of three, never.

values · Dominance Four ways to count a reply. The plain count of the opponent's replies against three weightings of it, each scored on the same pairs of Domineering options. Every weighting has a larger threshold than the plain count and gets more pairs wrong.

The weight that blunts the count

The rung below found that counting the opponent's replies gets the direction of a comparison right once the gap reaches three, and proposed a repair: weigh each reply by whether it leaves the opponent anything. Weighing it makes the count worse. The threshold goes from three to four, the failures from 1,124 to 1,320, and all seventy-two of the pairs the repair was written for come through it unchanged.

values · Dominance Add, then reduce again. The arithmetic the homomorphism promises, measured: summing two reduced forms gives a reduced form on 88 per cent of pairs and needs a second reduction on the rest.

Add, then reduce again

The homomorphism promises that a sum's reduced form can be computed from its parts', and says nothing about what the operation is. It is addition followed by a second reduction — needed on 431 of 3,600 pairs of day-three values, and on not one of the 1,751 pairs with a cold part. What the second pass removes is an option that only becomes dominated once the two fights are side by side.

sums · Reduced form Every failure is on one board. The eight Domineering boards of the mobility census with the number of failing pairs on each. Seven of them contribute none; every failure at a margin of two is on the largest board, at two depths, and sixteen positions up to symmetry.

The threshold was a fact about the census

Two rungs failed to account for the seventy-two pairs where a mobility count gets the direction of a comparison wrong, and the third looks at them one at a time. They are not a class of shapes. All seventy-two are on the largest board in the census, at two depths, and sixteen positions up to symmetry — and one board larger the count fails at a margin of three, which the ladder has been quoting as the point at which it never does.

values · Dominance Eight ways to name it, and none of them works. Candidate rules for which option the second reduction deletes, scored on every pair where it deletes exactly one. The best reaches four in five and none is exact.

The option nothing names

The rung below found the arithmetic on reduced forms to be add and reduce again, needing the second pass on 431 of its sums, and asked whether the option that pass deletes can be named from the parts. Eight rules were scored and the best reaches four in five — and on a pool closed under negation it falls to under half, which says the near-miss is a property of the population. What the second pass does have is a shape and a cheap test that rules it out.

sums · Reduced form The collapse happens twice. How many subsets, antichains and values there are. Domination takes 1,793 subsets to 96 antichains and the rest of the reduction takes those to 30 values.

What identifies two subsets

Every subset of a day gives a self-negative value by mirroring it, and 1,793 subsets of day two give thirty values. The collapse happens in two stages with different characters: domination takes the 1,793 to 96 antichains and is a theorem, and the rest of the reduction takes 96 to 30 and is concentrated almost entirely on two values — nought, which has an exact description, and star, which has none.

sums · Negation Thirteen sweeps, four thresholds. The mobility rule's failures on every board and depth the sweep can afford, with the threshold each one gives. The thresholds take four different values and no ordering of the boards produces them.

A threshold is a detection limit

The rung below had two points — a margin of three at fifteen squares, four at eighteen — and asked whether the mobility rule's threshold grows with the board. Eleven more sweeps say no property of a board orders the thresholds, that the same board at two depths gives two of them, and that a tenth of the sweep which produced the four reports three instead. What does move, on every board measured twice, is the depth.

values · Dominance The theorem a proof would have needed. The mobility rule's failures on decomposed positions against connected ones, across every board in the depth sweep.

The easy case was not the reason

The rung below found the mobility rule reaching a failure rate of exactly nought near the endgame and named what a proof would need: that a decomposed board's comparable options are ordered by reply count. That statement is false on all five boards, at margins up to two — and split positions go exact two squares of depth before whole ones, so decomposition is the easy case rather than the cause.

values · Dominance A cross in the table. Pairs needing a second reduction, by the colder temperature and the gap between the two.

A cross in the table

Which pairs need a second reduction had never been asked. Sorted by the two temperatures the answer is a cross — the whole gap-of-a-quarter column and the whole colder-is-three-quarters row — and it is exactly necessary on all 431 with no exception, and wrong 477 times in the other direction.

sums · Reduced form The proposed case, scored. The gift-horse theorem and the domination argument proposed for it, each scored over every gift horse added.

The proof needs both reductions

The gift-horse theorem was to be proved by showing the added option dominated. It is, on 97.8 per cent — and the other 232 are reversible instead, with nothing left over. The case the proposal missed is almost entirely one follower: none under a positive number, 190 under a negative one.

sums · Ordinal sum Two of three are subadditive. The birthday, the option count and the written length, each tested for subadditivity under the disjunctive sum.

Two measures bounded, and one not

A sum is born no later than its parts' birthdays together, and it has no more options than they have between them — a bound nobody had checked, and it is attained. What runs away is the length of the written form: 27 pairs of 231 exceed it, the worst by 29 characters, on a sum with exactly as many options as it was entitled to.

values · Numbers The fall stops. Comparability on days two, three and four, the last built and corrected.

A floor, and not a decline

Comparability fell eighteen points from day two to day three and the next day cannot be enumerated. It can be built — and the construction's bias measured one day lower, where the truth is known. Corrected, day four comes to 60.6 per cent against day three's 59.7: the fall was a one-day event.

sums · Comparison A factor, and it grows. The ratio between reducing and deciding, by the number of options the form carries.

A factor, and not an overhead

Deciding who wins a form searches the form's own tree. Reducing it to canonical form searches a difference game for every comparison, and a difference game is a sum. Over 256 forms the reduction expands 5.46 times as many positions — and the ratio runs from 0.58 at one option to 9.80 at eight.

values · Canonical form 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.

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.

applied · Sylver Seven orders, and none of them better than chance. Seven quantities a one-pass scan could compute about each move surviving the pairing test — the number itself, how many gaps it closes, what it leaves behind — each read as an order over the survivors and scored against the winning moves of 583 positions. The best puts a winner first on 24.9% of positions against 22.3% at random, and every one of the seven places the winner deeper in its order than chance would.

A shortlist with nothing at the top

The one-pass test leaves 9.58 moves of 12.27 and names none of them. Seven quantities a scan could compute about the survivors were turned into orders and scored on 583 positions: the best puts a winning move first on 24.9% against 22.3% by chance, and every one of the seven places the winner deeper in its order than chance does. Read as sieves instead, the gentlest keeps half the list and throws the only winner away on 264 positions.

applied · Sylver The reversing move is the follower's. For each follower under which some gift horse escapes domination, the number of escapes, how many are reversed by Right's move inside the follower, and how many by a Right move in the base part. The follower's move reverses every one.

The follower does the reversing

The gift-horse theorem for the ordinal sum needs two cases, and the second — the added option is reversible — was counted and not described. Recorded move by move, the reversing answer is always Right's move inside the follower: on all 410 escapes under five followers, and on every one of the 2,628 gift horses under every follower that gives Right a move at all. The case split is by follower, not by horse.

sums · Ordinal sum What survived, and what did not. The gift-horse theorem, the two-case proof and the one-line description of the reversal case, each scored on the day-three sweep and its mirror and on the day-four sweep and its mirror.

The split slips one day deeper

The reversal case of the gift-horse theorem was described in one line — the follower's own move reverses every gift horse, whenever the follower has one — and tested only where it was found. In the mirror it holds exactly, with 1 and −1 trading places. One day deeper it fails: under ↑ and ½, three gift horses on built day-four bases are not reversed by the follower's move. All three are dominated, so the theorem stands; the clean split by follower does not.

sums · Ordinal sum The day-four figure, twenty draws at a time. Corrected day-four comparability from twenty seeds of each of two constructions, as dots on a percentage axis, with the range of day three's four slices shaded and the single figure the earlier essay reported marked. The seeds spread over about twelve points and the two constructions agree.

Twenty draws and a second recipe

Day four's comparability was reported as 60.6 per cent against day three's 59.7, from one built sample and one calibration. Built twenty times with each of two recipes whose biases differ by six points, and calibrated against all 1,474 day-three values rather than a quarter of them, the corrected figure spreads over twelve points from seed to seed and the two recipes agree within one standard error. The floor survives; the decimal was one draw.

sums · Comparison

Named alongside it

The objects these essays reach for when they reach for this one.

EnumerationCanonical formNormal playComparisonCounterexampleDisjunctive sumApproximationDomineeringHeuristicValueBirthdayBound

All concepts