The collection

Every essay — page 11

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

How it was found

The theory looks inevitable in retrospect and the record says otherwise. The older arguments are run here rather than recounted — and one of them still answers a question nothing since has answered.

One substitution, thirty-four years. Bouton's criterion and the Sprague–Grundy theorem run side by side over a family of games. They differ in one quantity: the heap's size against the heap's Grundy value. The exclusive-or that combines them is the same operation in both, and it is the one Bouton published in 1901.

The step nobody took for thirty-four years

Bouton's criterion is that the heap sizes exclusive-or to nothing. The 1935 theorem is that the heap Grundy values do. The exclusive-or is the same operation in both and it is his, so the whole of the intervening thirty-four years is one substitution — and run over eight games and 672 positions, the substituted criterion is exact on every one while the original is exact on Nim and nowhere else.

6 figures · Bouton
Bouton's argument, indexed by a value. Bouton's two closure properties stated for every Grundy value rather than for nought alone: no move stays inside a value class, and every class above a value can reach it. Checked on each game and each value in range.

The picture Bouton's proof leaves behind

His argument is two closure properties of one set, and the Sprague–Grundy theorem is the same two sentences with nought replaced by a variable — checked here on five games and every value in range, with no move staying inside a class and no class failing to be reachable from above. What the argument also leaves behind is a picture in which the values descend, and that is false: 99 of 444 moves here raise a value, and none of them is in Nim.

6 figures · Bouton
Every region of three positions, counted. The 262,144 graphs on three positions reduced to the regions that are genuinely three positions with a cycle in them, and then split by whether the two-position vocabulary has a name for both of their sides.

Four thousand nine hundred regions with no name

Two positions give 256 regions and ten names cover every side of all of them. Three positions give 262,144 graphs, 110,934 genuine loopy regions — and 4,931 of those have a side that no name in the two-position vocabulary reproduces, with 3,990 of them named on one side and blank on the other. The count the earlier essay left open comes back in the affirmative.

6 figures · Notation
The guess, and what it covered. Two attempts to name the leftover sides out of the old vocabulary: every pair of the six stoppers, and every two-position region that is a stopper, each with small finite games added. Both cover nothing, and the count of distinct leftovers is what remains.

The names are not built out of the old ones

The guess was that a three-position region's missing names would be sums of two loopy ones — on plus over, and that family. Built and tried, every pair of the six stoppers covers none of the 4,931 regions that need one, and so does every two-position stopper there is, all seventy-nine of them with small games added. Thirteen names have to be invented, and forty-eight cover the whole census against ten at two positions.

6 figures · Notation
Four positions, sampled. Three samples of three thousand loopy regions on four positions, drawn with each possible move present at a chance of one half, about a third and a quarter. For each: how many regions have both sides named by the thirty-five names two-position regions use, by those together with the thirteen invented for three positions, and how many need a new name.

Four positions, sampled

Ten names write both sides of every loopy region of two positions, and forty-eight every region of three. Four positions are over four billion graphs and cannot be counted, but they can be drawn. Three thousand regions at each of three densities: the forty-eight names cover between 95.9 and 99.5 per cent, the thirteen names invented for three positions come back at four almost all of them, and the sparsest sample meets thirty-five sides nothing earlier reproduces — a floor of eighty-three names, and a curve that grows by accretion rather than collapse.

6 figures · Notation
Half the new sides are sums. The 50 sides of four-position loopy regions that no earlier name reproduces, tested against every sum of two games from the day-two values, the stoppers and the thirteen three-position names. 23 are such sums, 15 of them using a three-position name; 27 are not.

Each size is built from the last

At three positions the sides that needed new names were sums of nothing older: no pair of stoppers, and no two-position stopper with a small game beside it, wrote any of them. At four positions the thirteen names three positions forced are available as parts, and they change the answer. Of the fifty sides the samples met that nothing earlier names, twenty-three are sums of two earlier games — fifteen of them using a three-position name — and twenty-seven are not.

6 figures · Notation
Three parts close most of the gap. The fifty sides four-position regions needed new names for, split by the smallest sum of earlier games that writes them: 23 by a sum of two, 16 only by a sum of three, and 11 by neither, with how often each group turned up in the samples.

Composition at every size

Twenty-seven of the fifty new four-position sides were written by no sum of two earlier games. Sums of three write sixteen of them — seven plain finite games born after day two, nine built on the names three positions forced — and leave eleven, met seventeen times in nine thousand regions. Run the same test one size down and the story the counts told falls apart: eight of the thirteen names three positions were thought to invent are sums of up to three older parts, two of them just two finite values. Every size composes. What each invents is a residue — five, then eleven.

7 figures · Notation
Four rows of lamps and three winning cuts. A Nim board of four rows of up to seven lamps, lit 7, 5, 3 and 2, with each row's binary beside it. The columns add without carrying to 011, a nim-sum of 3; the three rows carrying the middle bit each have a winning cut, to 4, to 0 and to 1, and the lamps each cut would put out are shaded.

A machine that knew one theorem

In 1940 a machine of relays played Nim against the public at the New York World's Fair, on four rows of up to seven lamps, and is reported to have won about nine games in ten. Everything it knew was Bouton's theorem, which is three column parities and one test per row. The record is the interesting number, and it is a measurement of the visitors rather than of the machine: against a visitor choosing at random the machine wins 98.5 per cent, and nine in ten is what a visitor earns by playing perfectly once seven lamps or fewer are lit. On the board the game is usually set up on, a visitor handed the first move cannot win at all.

7 figures · Machines
A Hex board read as a circuit. A three-by-four Hex board with Down to move, one Down stone and two Across stones. Each empty cell carries a disc whose area is the current through it when both players' edges are held at a unit voltage; the busiest cell is the current rule's move. The three cells that actually win for Down are shaded, and the busiest cell is not among them.

A current through the board

In the early 1950s Shannon and Moore built a machine that played Hex by wiring the board as a network of resistors and reading its move off the electrical potential. There is no theorem behind that rule, and on every Hex board small enough to solve it can be scored against perfect play. On boards to twelve cells it takes a winning cell on more than 98 per cent of the positions where one exists, far above chance. And it loses: every board of three rows or more it has to defend, and four by four moving first, because an opponent needs only one of the two per cent and can steer the game to it.

7 figures · Machines
Twelve openings, four answers. A three-by-four Hex board. Each cell is labelled with the cell the current rule of Shannon and Moore's Hex machine takes for Down when Across has a single stone on it. The twelve answers are only four cells, shaded, each answering three openings.

A circuit asked once

The Hex machine's circuit rule loses every board past two rows that it has to defend, and the strategy that holds those boards is a pairing — a partner for every cell, fixed before the game begins. So make the circuit commit: consult it once per cell on an empty board and never again. Its table answers twelve openings with four cells, and loses every board the running circuit lost. Forced to choose a whole pairing, it picks a losing one under seven of eight ways of reading its scores; the eighth finds the winning pairing of three by four by a margin of rescaling and gives up two by four for it.

7 figures · Machines
One chain given up. Two pairings of the three-by-four Hex board. Left, the pairing the chain-sharing scores choose, with the one chain it misses shaded. Right, the diagonal pairing, which meets all 25 chains.

A pairing is checked, not measured

The Hex machine's circuit chose pairings near the winning ones and never at them, which left a finite question: does any score of a pair of cells, computed from the board alone, have a winning pairing as its best? The obvious scores count the chains a pair shares. All four hold two by three; on three by four the plain count cannot choose between a winning pairing and a losing one, and the corrected counts choose the loser; on four by five every one loses. The reason is arithmetic — a sum will give up one chain to cover two others twice. Told which chains it missed, the count finds the diagonal pairing at one rate in three, and by then it is checking, not measuring.

6 figures · Machines
One rate in twelve. The chain count reweighted by the chains it misses, at 12 rates from 1.1 to 3, on the Hex boards 2 × 3, 3 × 4 and 4 × 5. The small boards are held at every rate by the diagonal pairing; 4 × 5 is held only at × 1.5, round 46, and every other rate stalls with chains still missed.

The weights were not the argument

A chain count corrected by its own misses held the four-by-five Hex board with the diagonal pairing, the one pairing there with a sentence behind it, and the hope was that its weights were the sentence in numbers. They are not. One rate in twelve holds the board and one start in twenty; the weights it ends on survive five per cent of noise and not twenty; and weighting exactly the chains the argument relies on finds no winning pairing at all. Told which pairing to prefer, a measurement reaches any of the three in a dozen corrections. What chain weights can do is rule a pairing out, and on three by four they rule out one that wins.

7 figures · Machines
Every winner but one. A perceptron over chain weights taught to make each winning Hex pairing the heaviest, on boards 3 × 4 to 6 × 7. Every winner is reached in at most sixteen corrections except the cross pairing of 3 × 4, which another winning pairing covers on every chain.

The teacher's game

A machine that weighs Across's chains could be taught every winning Hex pairing on four by five, and on three by four it could never be taught the one pairing another winner covers chain by chain. Whether anything else could hide from it — a winner covered not by one rival but by a mixture of several — is a question about a game between a teacher choosing weights and a rival choosing pairings, and Jean Ville's 1938 theorem of the alternative, written for von Neumann's games, answers it in general. On five by six and six by seven every winner is taught in sixteen corrections or fewer, and nothing hides at all.

6 figures · Machines
Three rows that do not add. Four diagrams of Dawson's pawns valued whole and as the sum of their rows. 3 + 1 is worth ∗ against ∗3 by rows. 4 + 3 + 1, three rows, is worth ∗ against ∗3, a difference of ∗2, over 4,153,158 positions. 5 + 2 + 1 and 3 + 2 + 1 add.

The rows that cancel

Every one of Dawson's diagrams that failed to be the sum of its rows had an even number of rows, and twenty-one with an odd number all added. One file past the edge of that sweep, four files beside three beside one fails: three rows, worth ∗ against ∗3 by its rows. What the parity was tracking is two deletions a sum allows — a row worth nought, and two identical rows — which the compulsory board turns out to allow as well. Each of 44 diagrams is worth exactly what is left after them, and in the fifty-diagram sweep that remainder happened to have as many rows, odd or even, as the diagram it came from.

7 figures · Dawson
Three winners, none of them teachable. A graph of ten points on which Short, claiming points to join A to B while Cut deletes them, wins moving second. Three pairings of the contested points win, each holding one whole pair in every one of the six minimal separating sets; one, ce df gh ij, is drawn. No weighting of the separating sets makes any winning pairing outscore every losing one.

Two losers that average to a winner

On Hex boards to 6 × 7, a machine that weighs the opponent's winning sets could be taught every winning pairing except one, which another single winner covered. Ville's alternative allowed a second way to be out of reach — covered only by a mixture — and the switching game on points, where winners are many, is where it turns up. None in all graphs of seven points or in 4,000 of eight; 35 covers at nine and ten points. Every one is the even average of two losing pairings, and on six graphs Short wins and no weighting can teach the machine any winning pairing at all.

6 figures · Machines

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