Hex — where it appears
Named by 11 essays across 2 fields — each of them below, with the objects they name alongside it.
The theorem that names a winner and no move
Strategy stealing proves that the first player wins Hex and wins Chomp, on every board, in about four lines. It exhibits no move, contains nothing a move could be extracted from, and is not going to. The moves have to come from somewhere else, and where they come from runs out almost immediately.
The winning reply is the fourth choice
The repair proposed for the potential was to weigh an edge chain more heavily. Fifty-five weightings later, none holds the four-by-five board, and an edge bonus costs Down four boards it was already holding. The reason is not the numbers: over 393,660 turns of the pairing that does hold that board, the potential would take the same cell 26.1% of the time, and the winning cell is its 3.7th choice on average and as low as its seventeenth.
Three pairings and no order
The potential lost four-by-five Hex for Down, and fifty-five re-weightings lost it too, which left the question whether anything that orders the cells can hold the board. The plainest ordering — one fixed list, whatever the position — holds none of the boards the pairing is known for, and on four by five the best it can do is the left column. At the other end, every table of replies that wins a small board is a pairing, and four by five has exactly three pairings, each using all twenty cells.
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.
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.
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.
Two neighbours keep a pairing, three lose it
In the switching game on points, Short wins moving second by a pairing when every smallest set of points that cuts A from B holds both points of some pair. On every graph of seven points — 971,104 of them — every Short win has one, at every degree. In random link games, where each contested point has two neighbours, every win has one too, although the winning strategy follows trees that change as the game goes. The first wins with no pairing appear at nine points with four neighbours and at ten points with three: a cubic graph Short wins whoever moves, which no fixed set of answers explains.
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.
A pairing that changes once
The cubic ten-point graph is a win for Short that no fixed pairing explains, and the full strategy that does explain it is a tree of 68 decisions. One change is enough: after Cut's first deletion Short claims a stated point, and from then on a fixed pairing of what is left holds every cutting set. Four such plans, eight pairs between them, answer every opening. Of the 23 pairless wins in the samples, 21 need one change and two need a second, and in both the second is a forced claim. Unless the hard problems collapse, no fixed number of changes can be enough for every graph.
Two reflections and a corner
Hex boards of n rows and n + 1 columns are won by the player with the nearer edges, by a pairing, and the count of pairings that do it ran two, three, three. It is three again on five by six, against 1,225 chains, and three on six by seven, against 16,097. They are the same three every time, and they are not three strategies: one is the classical reflection, and the other two are a reflection across the board's other diagonal, which joins its two far corners, in two versions that disagree about six cells. Each is a rule a sentence long, and each holds on every board out to ten by eleven.
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.
Named alongside it
The objects these essays reach for when they reach for this one.
Exhaustive searchCounterexamplePairing strategyStrategyCertificateHeuristicMove selectionStrategy stealingIntractableThe Shannon switching gameSolved gameSymmetry