Combinatorial Game Theory
who wins, and by how much
Generator

Folding a 4×4 board by its symmetries

Folding a 4×4 board by its symmetries
Folding a 4×4 board by its symmetries. The size of a Domineering solver's table when positions related by a board symmetry are stored once. The saving rises toward the size of the symmetry group and stops there — it is a constant factor by construction, and no board is large enough to make it anything else.

The size of a Domineering solver's table when positions related by a board symmetry are stored once. The saving rises toward the size of the symmetry group and stops there — it is a constant factor by construction, and no board is large enough to make it anything else.

7 essays call identify. The drawing above is what it returns with no arguments at all; every call below passes it something, because a placement that passes nothing draws whichever member of the family the generator happens to default to rather than the one its essay argues about.

Where it is called

Changing this generator changes every one of these figures.

Folding a 4×4 board by its symmetries. The size of a Domineering solver's table when positions related by a board symmetry are stored once. The saving rises toward the size of the symmetry group and stops there — it is a constant factor by construction, and no board is large enough to make it anything else. What it costs

What counts as the same position, and what that is worth

Folding a 4×4 Domineering board by its symmetries takes the table from 5,700 entries to 1,522 — a saving of 3.75, against a ceiling of exactly 4. An orbit cannot be larger than the group acting on it, so this is the one saving in the subject that can never change an exponent.

 Finding the parts costs the same whether there are any or not. Domineering boards of 4 squares by 5 with different squares blocked out, and what the decomposition is worth on each. The pass that finds the regions is a flood fill and visits every square once, so it costs the same on all of them. What it buys ranges from nothing — on the boards that do not decompose — to a saving of 2,808 positions, and it cannot tell which case it is in until it has run. What it costs

Finding the parts

Decomposition turns a product into a sum and is the largest saving in the subject. Nobody labels the regions. The pass that finds them costs the same on every board of a size — including the boards where there is nothing to find — and what it buys ranges from four orders of magnitude to nothing at all.

 Knowing who wins, and knowing what it is worth. Nine positions, each evaluated twice by an instrumented evaluator that starts with an empty cache. The third column counts what deciding the winner costs and the fourth counts what the canonical form costs, in the currency each question is actually paid in. What it costs

Knowing who wins, and knowing what it is worth

Deciding a winner expands positions. Computing a canonical form expands pairs of positions, because a comparison unfolds as a recursion over one subposition of each and the reduction makes many comparisons. Measured on the same nine positions by an evaluator that starts empty every time, the second costs between 1.3 and 279 times the first, and the ratio grows with the tree.

 What the folding costs to do. The same search over a 4 × 4 Domineering board run twice, once folding positions by symmetry and once not, with everything counted. The fold stores 3.75 times fewer entries and spends 17.5 times more elementary operations to decide where to put them. What it costs

What it costs to notice a repetition

Folding a 4 × 4 Domineering board by its symmetries takes the table from 5,700 entries to 1,522. It also spends 559,424 square-mappings to work out where each entry goes — seventeen and a half times the entire cost of not folding. The saving has a ceiling of four and the price has no ceiling at all, and knowing which currency each is paid in is the difference between an optimisation and a habit.

 What a short key gets wrong. A memoised who-wins search of 4 × 5 Domineering using a Zobrist key of 10 to 32 bits, run under 60 random keys at each length and checked against the exact answer: how many stored positions share a key, how many runs store a wrong verdict or name the wrong winner, and what checking the whole position would cost instead. At 16 bits 59 runs store a wrong verdict and 19 name the wrong winner. What it costs

A key shorter than the position

A who-wins table for 4 × 5 Domineering addressed by a 16-bit Zobrist key stores a wrong verdict in 59 runs of 60 and names the wrong winner of the empty board in 19. The pairs of positions sharing a key follow the birthday count exactly while addresses are scarce, and fall away to nothing once the key has more bits than the board has squares, because a Zobrist key is linear. Symmetry and value identify positions that really are the same; a short key identifies positions that differ, at a rate set by arithmetic.

 One family of confused positions. Two 4 × 5 Domineering positions with Left to move that share every bit of an 18-bit Zobrist key, differing only on 4 marked squares whose words cancel. Under that key 100 of 136 confused pairs differ on exactly those squares; its other families hold 25 and 11. What it costs

A check bit halves the average and not the key

Real transposition tables keep a few of a key's bits beside each verdict and trust an entry only when they match. On 4 × 5 Domineering each such check bit halves the average number of wrong verdicts, exactly as the birthday count says. It does not halve any one key's. A Zobrist key confuses positions in families — every pair that differs on one set of squares whose words cancel — and a bit removes a family whole or not at all, so from eighteen bits to nineteen thirty of fifty-eight keys lose every confusion and fourteen keep every one.

 Two squares a key never needs. A 4 × 5 Domineering board shaded like a chessboard, with two squares of the same shade in rows 1 and 2 marked. Because a domino covers one square of each shade, a vertical domino covers one square in an odd row and one in an even row, and turns alternate, the other eighteen squares determine both marked squares: a key that leaves them out confuses none of the 48,670 reachable positions. What it costs

A key is a code, and two squares come free

The families of positions a Zobrist key confuses are the words of a binary linear code — the sets of squares whose words cancel — so choosing a key is choosing a code. On 4 × 5 Domineering the textbook choice, a code with the largest minimum distance, confuses more stored positions than a random key at fourteen, sixteen and nineteen bits. The choice that reads the board confuses none at eighteen: two squares of one shade, in rows of different parity, are decided by the other eighteen, and no seventeen-bit key is exact.

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