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

Every essay — page 4

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

What it costs

Every theorem here can be true and the answer still out of reach. What a search costs, what a proof of a win looks like, and where the shortcuts are.

A formula, drawn as a game. A token on a directed graph. A move slides it along an edge to a vertex not yet visited, and a player who cannot move loses. That is the whole game, and deciding who wins it is as hard as anything decidable in polynomial space — which is the strongest hardness claim anybody makes about a combinatorial game.

Hard, proved

A game is as hard as a logical formula when the formula can be drawn as the game. Here is the drawing — a quantified formula turned into a graph with a token on it — with every formula over three variables played both ways and required to agree.

6 figures · Complexity What it costs to decide the winner. Three families of game, each at several sizes, with the number of distinct positions an exhaustive solver must evaluate beside the work its closed form does. The bars are logarithmic. Hardness is not about the size of the board or the length of the rules — Domineering has the shortest rule here — it is about whether anybody has found the shortcut.

The game with the shortest rule is the hard one

Deciding a generalised board game is PSPACE-complete, which is a statement about families and encodings rather than about size. Nim in the same subject is settled by one pass over the input at any size, and green Hackenbush by one pass over the edges — while Domineering, whose rules take a single line, has no shortcut anybody has found.

7 figures · Complexity One node per route, one node per position. For each board, the number of nodes in the recursion tree a solver with no memo table would walk, beside the number of distinct positions that tree contains, beside the longest run of moves in it. The first number is the cost of forgetting; the second is the size of the table that avoids it; the third is the stack, and it stays small however the other two grow.

A position reached eleven ways is one position

A 4×4 Domineering board has 5,700 positions in it and 6,257,129 routes through them. Three heaps of 7, 11 and 13 have 480 positions and 7.6 × 10¹⁶ routes. The gap between those two numbers is not an optimisation — it is the difference between a search that finishes and one that does not.

7 figures · Search A board in pieces costs the sum, not the product. A Domineering board with squares blocked out, so that it falls into regions no domino can span. The number of positions in the whole board is exactly the product of the numbers in its regions — which is why evaluating the regions separately, and adding the values, is an exponential saving rather than a tidier way of writing the same search.

The board falls apart, and the arithmetic changes

A 4×5 Domineering board with a wall down the middle has 2,916 positions in it, and that number is exactly 54 × 54 — the product of its two halves. Solving the halves separately costs 108. Decomposition is the one saving in this subject that turns a product into a sum.

8 figures · Decomposition 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 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.

7 figures · Identification Every quantifier is a move. A quantified boolean formula with its quantifiers drawn as turns: an existential is a choice by the player to move, a universal a choice by the opponent. The same formula is put through the reduction to Generalized Geography and the two answers are checked against each other, so the prefix of quantifiers and the game beside it are one claim.

A puzzle asks once, a game asks alternately

Quantifier alternation is the whole difference between a puzzle and a game. One chooser is an existential and its answer is a witness somebody can check; two choosers taking turns is a prefix of alternating quantifiers, and the witness stops being an assignment and becomes a strategy.

8 figures · Alternation The same position, and two rulers to measure it with. Nim positions with the length of their input under two encodings — the heap sizes in binary, and the counters themselves — beside the work the nim-sum does. The work never changes. Which of the two lengths it is compared against decides whether the same algorithm reads as linear or as exponentially fast, and hardness claims are always made against one particular encoding.

Nim is easy, in binary

Three heaps of a thousand counters take thirty bits to write down and three thousand counters to lay out. The nim-sum does three exclusive-ors either way. Whether that counts as fast depends entirely on which of the two numbers the work is compared against.

6 figures · Complexity A winning strategy on 3×3, drawn whole. The whole of one player's winning strategy on a small Domineering board: their own move at each of their turns, and every reply the opponent has at each of theirs. The strategy branches only where the loser chooses. Its size is what somebody would have to be handed to check the claim that this player wins, and it is far larger than the claim itself.

"Left wins" has no short proof

A complete solution of Nim on heaps of 7, 11 and 13 is 480 table entries. A winning strategy for the same position — one move of the winner's at each of their turns, and an answer to every reply — has 56,167,022 nodes in it. The answer is smaller than the proof by a factor of a hundred thousand.

7 figures · Complexity One node per route, one node per position. For each board, the number of nodes in the recursion tree a solver with no memo table would walk, beside the number of distinct positions that tree contains, beside the longest run of moves in it. The first number is the cost of forgetting; the second is the size of the table that avoids it; the third is the stack, and it stays small however the other two grow.

The class is named after memory, and that is not an accident

A 4×4 Domineering board has 6,257,129 routes through it, 5,700 distinct positions, and a deepest line eight moves long. Those three numbers are three different resources, and the smallest of them is the one that gives games their complexity class.

6 figures · Complexity Three things the word “solved” is used for. The three standard senses of a solved game, priced on positions this solver can settle completely. Ultra-weak names the winner; weak supplies a strategy from the opening; strong supplies one from every position. They differ by orders of magnitude, and a claim that a game is solved is nearly useless until it says which of the three it means.

Three different claims are all called solved

Hex is solved in the sense that the first player provably wins, by an argument that names no move whatever. Nim is solved in the sense that a formula gives the right move from any position at any size. Between them sit strategies for one opening, and databases of a few billion positions. The word covers all four.

7 figures · Complexity Subtraction of 1, 3, 4 — and the window that proves the period. The Grundy values of a subtraction game, with the window that certifies the period marked. Everything after the window follows from it by induction, because a value is a mex over values at most one move back — so a finite check settles the whole infinite sequence, and the thousands of further values computed here agree with a claim that was already proved.

Four values, and the sequence is settled for ever

The Grundy values of a subtraction game repeat with period 7, and proving it needs a window of exactly four of them — one for each size of move the game allows. Everything past the window follows by induction. A finite computation has settled a claim about every heap there will ever be.

6 figures · Periodicity the temperature that bounds the loss. Temperature runs up the page and value across it. Each wall is where a player is willing to move once a tax of that much is charged per move; above the temperature at which they meet, neither wants to move and the position is worth its mean value. The height of the meeting point is what is at stake.

A rule that is never right and cannot be far wrong

Playing the hottest component is not optimal — over 440 measured lines it costs something on 17 of them. What makes it worth having is that the cost is bounded by the temperature, provably, and that the same test run with the ordering reversed breaks the bound on 54.

8 figures · Approximation Classes needed, as the heaps get bigger — Dawson's chess ·137. How many kinds of position there are, against how large a heap the universe allows. Under normal play the answer stops growing as soon as the Grundy values stop growing. Under misère play it does not stop, and every new class is a pair of positions that behave identically under normal play and differently under misère.

The cost is in the closure, not in the positions

Under normal play, Dawson's chess needs four classes for every heap up to twelve, because its Grundy values stay at three or below there. Under misère play the same game needs six, then twelve, and the number rises with the universe rather than with the position — which is a different kind of expense entirely.

6 figures · Misere cost 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.

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.

7 figures · Decomposition 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.

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.

8 figures · Value cost

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