Dominance — the series · Combinatorial Game Theory

What deleting is worth on its own. The reduction split into its two halves and each measured. Deleting a dominated option removes exactly one option and can do nothing else; bypassing a reversible one substitutes an option list and can widen the form. The counts say how much of the reduction the monotone half accounts for.

The reduction that always shrinks

Canonical form is two reductions and they are not the same kind of operation. Deleting a dominated option removes one option and can do nothing else; bypassing a reversible one substitutes a whole option list. Over the 256 forms born by day two, deleting alone finishes 225 of them and accounts for 480 of the 520 options that come off — and the 31 it cannot finish are almost all the ones with a star in them.

part 1 · values

How much a list of options can lose

Deleting a dominated option is the reduction with no surprises, and how many options it takes is decided by the shape of the order rather than by the values in it: the survivors are the maximal elements, and the count is the length of the list less the number of them. The essay separating the two reductions closed by predicting that the longest chain would give the number. It is a lower bound, exact on 3,859 of the 7,315 four-option lists and wrong on the rest.

part 2 · values

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.

part 3 · values

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.

part 4 · values

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.

part 5 · values

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.

part 6 · values

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.

part 7 · values

One board, all the way down. The mobility rule's failure rate on a three by five board at every depth, with the threshold each depth gives.

A heuristic that becomes a theorem

The mobility rule's failure rate had been measured at two depths on each of five boards and found to fall. Swept at every depth it does not merely fall — it accelerates, and it reaches exactly nought before the endgame. From four to eight empty squares onwards the rule has no exceptions at all, which turns a rule of thumb into a guarantee for the last few moves.

part 8 · values

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.

part 9 · values