Combinatorial Game Theory
who wins, and by how much
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The thread: Who moves last — page 2

The player unable to move loses. That single convention generates the whole theory, and reversing it — misère play — destroys almost all of it.
Cram on 4 by 4: the pairing strategy. Cram is Domineering with the orientations shared: either player may place a domino either way up, so both players have exactly the same moves and the game is impartial. Every position therefore has a Grundy value, and this board's was computed by the mex rule over its own placements. Impartial games

Cram

Domineering with one word of the rule changed: both players may place a domino either way up. That makes the game impartial, and the entire partizan apparatus collapses into a single Grundy value — on the 4 × 4 board, Domineering's canonical form runs to 114 characters of nested braces and Cram's answer is the one character 0.

 Mock Turtles on 8 coins: finding the move is decoding. Every row of the game, sorted by what it takes to win from it. The lost rows are the codewords; a won row is a codeword with errors, and the winning move is the error pattern that turns them off. The distance column is a fact about the code and the coins column is a fact about the rules, and the two do not quite agree. Impartial games

The code names the move

If the lost rows of a coin-turning game are a linear code, then a won row is a codeword with errors in it and the winning move is whatever turns the errors off. Over all 256 rows of Mock Turtles on eight coins: 16 codewords, 240 won rows, none more than two coins from a lost one — and 64 of them whose cheapest winning move has to turn three coins anyway.

 Seven infinitesimals, added to a whole day. Each row adds one infinitesimal to every value born by day three. The stops never move — that is what being smaller than every number means — the temperature moves a handful of times, and the outcome class moves in a quarter of the additions. Sums and comparison

What an infinitesimal does to a fight

Adding a number moves both stops by exactly itself. Adding something smaller than every number moves neither — across 10,318 additions to a whole day of values, not once — and the outcome class changes anyway, 2,622 times. It changes at exactly one kind of position: the ones with a stop sitting on zero, which is where the numbers have run out of things to say.

 Seven, ten, thirteen and sixteen. The four values the rung below named, each read off the prime factorisation: one plus the largest prime plus the product of the two largest. Particular games

The size of a cake

Ω gives the sign of a Maundy Cake and says nothing about the size, and the rung below left four values — 7, 10, 13 and 16 — unaccounted for. For a one-row cake they are a formula: write the prime factors largest first and add up their running products. The rule behind it is greedy — cut by the largest prime — and it is exact on every one-row cake to two hundred and wrong on a fifth of the two-sided ones.

 Still a count of squares. One-sided Amazons regions on two-dimensional boards, swept exhaustively. Every one is worth exactly the number of free squares in it. Particular games

A region one player owns

On a strip, a region containing only one player's amazons is worth exactly its free-square count, on all 45,057 positions of the rung below's sweep — and it predicted the exactness would fail in two dimensions, where an amazon can be short of room in one direction and not another. It does not fail. Over 2,412 two-dimensional regions there is no exception, and the reason is one clause: an amazon may shoot back at the square it has just left.

 Sorted by how many heaps are odd. The 2,002 positions of the census grouped by how many of their heaps hold an odd number of counters. Four of the six groups are entirely lost or entirely won. Impartial games

The count of odd heaps

The rung below refused a family of two-part rules for bounded Moore's Nim and asked what the 364 losing positions have in common as a set. They have an invariant, and it is a statistic of the whole position rather than of a heap: how many heaps hold an odd number. Every all-even position is lost, at every width of move, by a restoring strategy — and the count settles every position at one heap a move and at four, and a little over half at two.

 The check fires on positions that lose. How often the pairing check accepts a position, and how often the position is a loss. On every even board in the sweep it is wrong between an eighth and a fifth of the time. Impartial games

The check that was not a check

The rung below asked for a depth-conditioned solver and named the board to measure it on. Building it found two things first. The pairing check is unsound at interior positions — on a four by five Cram board it fires on 8,613 positions and 1,026 of them are losses — and the board it named has twenty-five squares, so the check can never fire there at all. Repaired, the check is right everywhere, and the policy that pays is the root alone.

 Thirty-two words, four of them lost. Every position of five heaps grouped by the parities of its heaps in decreasing order of size. Each word is uniform, and four of the thirty-two are losing. Impartial games

The parities, in size order

The rung below settled four of six parity classes in bounded Moore's Nim and asked whether the sizes pick out the losing positions in the two it could not. They do — but only through the order they put the parities in. Sort the heaps largest first, read off their parities, and that five-bit word settles the whole game at every width of move, with the losing words forming a subspace.

 Four conditions. The four linear conditions whose kernels are the losing sets, with the cases each covers. Impartial games

The parameter was the difference

The losing words of bounded Moore's Nim form a linear subspace and no map was known whose kernel they are. The equations exist, four conditions cover all thirteen cases at three to six heaps, and they are indexed not by the heap count but by the heaps less the width of a move — which turns the failure at six heaps into a prediction about seven.

 The clause that was free. Five requirements on a pairing strategy, with which of them each map meets. Impartial games

A symmetry that is not a pairing

The quarter turn was the last symmetry a Cram pairing argument had not tried, and the one a square board seemed to offer. It fires on the empty four by four and it settles nothing the half turn misses — and the reason is a clause four rungs of this anchor never had to write down, because every map tried so far was its own inverse.

 Five of six. The six predictions made for seven heaps by the difference reading, each scored against the sweep that was declined at the time. Impartial games

The family with two witnesses

Six predictions about seven heaps were written down and deliberately not run. Five of them held. The one that broke is the condition that had been checked against two cases when it was proposed — the fewest of the four — and at seven heaps it does not merely give the wrong answer, it asks a question the parity word has stopped being able to answer.

 One set, described three ways that share no arithmetic. Wythoff's cold positions can be stated as the Beatty pairs of the golden ratio, as a greedy construction over the integers that mentions no constant, and as a condition on Fibonacci numerals. None of the three consults the game. The fourth column is the game — a mex table over the moves — and all four name the same set of cold pairs over the whole square, which is what the figure counts. Out in the world

A set with three descriptions, and a function with none

Wythoff's cold positions can be written three ways that share no arithmetic — an irrational constant, a greedy rule, a condition on Fibonacci digits — and all three are exact. The same game's Grundy values have no closed form at all. Both facts are about one table, and the gap between them is the subject.

 A bridge circuit, with a link that is not there. The switching graph drawn as a bridge circuit, with an imaginary link from A to B dashed in gold. The graph alone does not split into two edge-disjoint spanning trees, so Short moving second loses; with the imaginary link it does, drawn in blue and red, so Short moving first wins. The green links are the links of the red tree that cross between the two halves the blue tree falls into without the imaginary link — the first moves the trees name. Out in the world

The first move is a link that is not there

Lehman's criterion answers one question about a switching game — who wins when Short moves second. The other question has the same answer asked of a different graph: add one link from A to B, and Cut is forced to spend its first move deleting it. The trees of that larger graph then name Short's opening, and on every subgraph of seven graphs they name a winner.

 How much of the board a who-wins search walks. Positions a memoised who-wins search of 4 × 5 Domineering expands under six move orderings, drawn to scale against the 48,670 positions the board has. Leaving the opponent fewest replies expands 1,125; trying losing moves first expands 30,202; all six find the same winner. What it costs

The order a solver tries the moves in

A memoised search asking who wins 4 × 5 Domineering expands 1,125 positions when it tries first the move that leaves the opponent fewest replies, and 30,202 when it tries losing moves first — the same answer at twenty-seven times the price. The ordering that already knows which moves win is not the cheapest. A win needs one move and a loss needs all of them, so the price of an order is paid one level down, in the replies it leaves.

 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 position Left always wins, and not always. Values grouped by the outcome class alternating play assigns them, with the range of probabilities the coin gives Left inside each class. A class that alternating play calls a win for Left every time holds no position the coin makes certain. Where it stops

Left always wins, and loses more often than not

Alternating play answers with one of four classes and the coin answers with a chance, and the two do not have to agree. Over the twenty-two values born by day two they never disagree and the margin is exactly nothing — the lowest chance on a position Left wins whoever moves is a half. Over the 1,474 born by day three, seven of them sit at seven sixteenths, and seven mirror them on the other side.

 The coin's move is often a blunder. Positions where Left has a choice and at least one option wins under alternating play, with how often the option maximising Left's chance under random turns is an option that loses the alternating game outright. Where it stops

The best chance is the wrong move

Maximising a probability and denying an opponent a reply are different objectives, and on 189 of the 904 day-three positions where Left has a choice and a winning move, the option the coin prefers is one that loses the alternating game outright. The smallest case is two options and one line of arithmetic: five eighths beats a half, and a half is the move that wins.

 Even rows always reward the move. For four coin sets and rows of one to seven coins, the number of rows in which the player to move does at least as well as when the opponent moves first. Every even column is full. Out in the world

Even rows always reward the move

Milnor's mean-value theory needs an incentive to move — the player to move must do at least as well as if the opponent moved first. On a coin row with an even number of coins that is not a hypothesis but a theorem: the first player can collect one whole parity class of coins, and one of the two classes holds at least half the total. So the condition excludes no even row whatever the coins, the class the earlier table called 'incentive at the top' was every row of four, and the hereditary condition is a condition on odd intervals alone.

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