Combinatorial Game Theory
who wins, and by how much
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The thread: What a search costs — page 2

A value is worth what it costs to find. These essays price the search rather than quoting the answer: positions visited, states stored, and the size of the board where the counting stops.
One cycle, and how much of it the criterion misses. The matching criterion for undirected geography put to the game where a move uses up the edge it crosses, on every connected graph up to 6 vertices with exactly one cycle. It is right at 77 of 114 starts, and at none of the six on the graph whose cycle has six vertices. What it costs

Two graphs a rule cannot tell apart

The repair proposed for the matching criterion was to read the cycle as well. Over every connected graph with exactly one cycle up to six vertices — 21 graphs, 114 starts — eleven such rules reach at most 91, and the winner is not a function of the matching, the cycle's length, the start's distance from it, its degree and the edge count together: six cells of that table hold both verdicts, the smallest a pair of five-edge graphs.

 The bound, and the cost, on four pools. Hottest-first play against optimal play on boards of two, three and four components drawn from four pools, 4,240 lines in all. The bound on the cost is the largest temperature on the board, which reaches 3; the worst cost measured anywhere is 1, and 100 lines meet the bound exactly. What it costs

The bound names the hottest part and the cost does not

Moving in the hottest component costs at most the largest temperature on the board, and that bound is attained: 100 lines of 4,240 pay exactly it. It is still the wrong quantity. Across four pools and boards of two, three and four parts the cost is nothing on 90.8% of lines and otherwise takes one of two values — half a point or one — on boards whose largest temperature runs to three, and it exceeds the coolest component on 13 lines and twice it on none.

 The tree of {a and b and c}, and the states it costs. The Zielonka tree of one winning condition on which positions a never-ending play recurs at. The root is the whole set of positions; the children of a node are the largest subsets the condition judges the other way. The number of memory states a winner needs is read back up the tree by adding at accepted nodes and taking the largest at rejected ones, and this condition costs 3. How it was found

Two things to hold at once, or three

Whether a condition makes a winner remember has been settled over every condition on three positions; how much it makes them remember has not. A tree built out of the condition alone, with no board in it anywhere, prices all 128: sixty-one cost nothing, fifty-eight cost two states and nine cost three. It also names the property that was nearly right — closure under union of the sets a condition rejects decides it exactly, where being writable as numbers is sufficient and reaches twenty-six.

 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. Out in the world

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.

 The more cheap fights there are, the cheaper the rule is. Hottest-first play against optimal play over 10,410 lines, grouped by how many components share the lowest temperature on the board. Lines costing more than that temperature occur only where one or two components share it; over the 3,230 lines with three or more, none does. What it costs

The cheap fights make the rule cheaper

A conjecture stands that playing the hottest part costs at most the coolest temperature times the number of parts sharing it — proposed on a range where that number never exceeds two. Swept to five-part boards over 10,410 lines it is false, and false the other way round: every line costing more than the coolest part has one or two parts at that temperature, and over the 3,230 lines with three or more, not one does.

 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.

 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.

 A move that leaves no surviving reply. A Sylver Coinage position whose gaps are 1, 2, 3, 4, 6, 7, 8, 9, 12, 13, 14, 17, 18, 23, 28, with the number of the opponent's surviving replies under each move that survives the pairing test. Naming 4 leaves no surviving reply, which proves it wins; the other winner leaves as many replies as the losing moves. Out in the world

A move whose every reply is struck

Read two moves at a time, the Sylver Coinage shortlist is no better ordered than read one at a time: preferring the survivor that leaves the opponent the fewest surviving replies puts a winner first on 24.7% of 583 positions against 22.3% by chance, and places it deeper than chance does. Read as a proof, the same count does what no order could. On 57 positions a survivor leaves no surviving reply at all, and wins by a certificate a few lines long; searched deeper, the survivors prove every position by thirteen moves — at a price that is never below the search that simply finishes.

 6 turns in strict alternation, priced. One quantifier prefix taken apart into its blocks, with the size of a winning strategy computed a term at a time. A choice made after k of the opponent's turns has to be written down once for each of the 2^k lines the opponent can produce, so the total depends on where the opponent's turns sit and not merely on how many there are. What it costs

Twelve turns, and three different prices

The earlier essay prices a universal quantifier at a doubling and leaves it there. Twelve turns with six of them the opponent's cost 6, 63 or 384 decisions to write down, depending on nothing but the order the turns come in — and the cheap arrangements are cheap for only one of the two players. What a claim costs is the number of times the choosing changes hands.

 How many turns are choices. Every position of each game with both sides to move, classified by whether the turn is a choice at all: no move, exactly one move, several moves that all lead to the same verdict, and several that do not. Only the last is a turn at which the alternation is doing any work. What it costs

Eleven moves and one decision

A prefix has one quantifier a turn, so a game of eleven moves is eleven alternations. Counted on the boards themselves, a Toads and Frogs strip of eleven moves has twenty-six turns with exactly one move available and one turn anywhere at which the choice changes the answer; a Clobber board has a hundred and fourteen turns and none. Nim, the game everybody calls solved, decides at four turns in five.

 A win is proved by one move and a loss by all of them. The smallest proof of each position's verdict, averaged by verdict. At a node the mover wins the proof takes the cheapest single option; at a node the mover loses it has to answer every option, which is the existential and universal quantifiers of the prefix showing up as two different objects. What it costs

Proving a loss means answering everything

A win is established by one move and a loss by every move, so the two verdicts are certified by objects of different shapes. Measured over every position of four games, a loss costs between 1.07 and 2.31 times a win — a small constant, never an exponential. The obvious explanation is the branching and it is wrong: Nim answers six options at a losing turn and pays 2.18, not six.

 What the opponent's choosing is worth. Every position answered twice: against an opponent who searches, and against one following a fixed rule with no search in it. Only a loss can change, so the share is taken over the losses. The spread between games is the measurement — in one of them nearly every loss is recovered and in another none is. What it costs

The opponent stops choosing

Replace one player by a rule with no search in it and the question has one chooser left, which is a puzzle rather than a game. Nim recovers five of its six lost positions that way, and six of seven on three heaps of five. Domineering recovers six of a hundred and twenty-two while the fixed rule throws away a winning move eighty-eight times, and one Clobber board recovers none at all — because on that board no rule can misplay.

 A turn is not a bit. The number of turns a game lasts, beside the number of quantified bits those turns amount to. Each ply is measured over the positions actually reachable at it rather than along one line, and the bits are the logarithm of the branching, which is what a quantifier prefix would need one of. What it costs

A turn is not a bit

The prefix a game is read as gives each player one quantifier a turn, and a turn on a board is a choice among however many moves there are. Nim on heaps of 3, 4 and 5 lasts twelve moves and carries 23.6 bits of choice; a Toads and Frogs strip lasts eleven and carries two. Corrected for that, the model predicts a strategy 539 times too large on one board and 67 times too small on another, and the two failures have different causes.

 How much of a board the endgame theory reaches. Every subset of a board's strings, counted by whether the surviving coins fall into chains and loops. The share is taken over the positions with no free box on the table, since a position with a capture available is one a player takes rather than chooses from. Out in the world

The endgame theory arrives late

Every component the chain-and-loop theory names has coins of degree two, so a position it can read is one where every surviving coin holds exactly two strings. Over a six-box board that is 1,033 of the 28,028 positions with no free box on the table — 3.7 per cent — and more than half of them only after twelve of the board's seventeen strings have been cut.

 One misère outcome, searched. The number of positions a misère search visits to decide the outcome of a sum of k heaps of Dawson's chess, each heap at most 9, on a logarithmic scale: the average over the sums and the worst single sum, for k from one to eight. Normal play decides the same sums from 20 stored values. What it costs

A misère sum is searched, not added

Under normal play the outcome of a sum of heaps is a nim-sum of numbers already known: twenty stored values decide every sum of Dawson's chess with heaps up to nine, however many heaps it has. Under misère play each sum is a new position to search. One outcome costs six positions for a single heap, two hundred for four heaps and over five thousand for eight, and a table of every eight-heap outcome costs a hundred thousand. The misère quotient is the only thing that brings the price back down.

 The closure that is enough. A grid for Dawson's chess with heaps up to 9: rows are the largest positions classified, from one heap to four; columns the largest tests, from none to five heaps. Each cell is the number of classes found. The counts stop growing at two-heap tests and three-heap positions. What it costs

Two heaps of testing are enough

A misère quotient is computed by testing positions against positions, and the universe used to find twelve classes of Dawson's chess was every position of up to four heaps tested against every other — 511,225 outcomes. Varied one size at a time, the count stops growing at tests of two heaps and positions of three: 12,100 outcomes find the same twelve classes. The narrower universe the earlier essay drew did not merge anything; it held fewer positions. And the corner that is enough moves: for Kayles at heap twelve, two-heap tests miss a class.

 12 classes, 7 questions. A grid for Dawson's chess with heaps up to nine: rows are the 12 misère classes of positions of at most four heaps, columns the 7 tests a greedy search chose, and each cell the outcome — N for the player to move, P for the other — when the test is added to the class. What it costs

Twelve classes, seven questions

Twelve misère classes of Dawson's chess were found by testing 715 positions against 715 others. Seven of those tests are enough to tell every class from every other — a greedy choice against a floor of four, since each test is one yes-or-no question. Kayles needs nine of 715 and Nim sixteen. The seven cost almost nothing to use and cannot be found without the whole closure, and they do not carry: the tests found with heaps up to seven tell apart only seven of the twelve classes with heaps up to nine.

 A staircase, not a slope. The number of misère classes of Dawson's chess positions as the largest heap allowed rises from three to 16, computed with positions of at most three heaps and tests of at most two. The count stays flat for several heaps at a time and then jumps. What it costs

A staircase, not a slope

With the misère closure cut forty-fold, Dawson's chess can be classified at heaps far beyond nine. The count of classes is a staircase: six from heap three to eight, twelve from nine to twelve, seventeen from thirteen to sixteen. Normal play steps once in that range, from four to eight at heap thirteen, where a Grundy value of four first appears. Misère play steps there too, and once more at heap nine, where normal play does not move at all — the first wild heap. Heaps eleven, fifteen and sixteen are also wild and move nothing.

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