Strategy — where it appears
Named by 75 essays across 8 fields — each of them below, with the objects they name alongside it.
The first theorem, and the winner it declines to name
Zermelo proved in 1913 that a finite game with no chance and no hidden information is decided before anybody sits down — every position is a win for one side or a draw, and which one is settled already. The proof is a labelling procedure, and watching it run shows exactly how little it says.
A winning strategy that is a spanning tree
The Shannon switching game was sold in a box in 1960 and solved in 1964, and the solution is not a claim that somebody wins. It is a property of the graph anybody can check, and the strategy falls straight out of it — whichever link the opponent cuts, take its partner in the other tree.
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.
"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.
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.
When never ending is a win
Retrograde analysis labels a position a win when somebody can force the opponent to be stuck, and leaves everything else blank. Calling the blanks draws is a rule from outside the game — and two other rules are available. The labelling does not change under any of them; only the residue does, and on a three-cycle that residue is every position on the board.
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.
Three players and no answer
Every theorem here is about two players, and the reason is not convenience. With two players the game is zero-sum, so 'play well' needs no further explanation. Add a third and the winner of a Nim position becomes a fact about the convention: two reasonable ones disagree on 56 of the 71 positions swept. The one question no convention touches — can a player force a win against the other two together — is answered 'nobody' in 65 of the 71.
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.
A rule with a guarantee
Evaluating a sum of a dozen fights is impossible; following a rule is not. Move where the stake is largest, and over 220 sums of three hot components the rule scores exactly what perfect play scores in 196 of them, is never more than one point behind, and never ends more than the largest single stake below the mean. The rule that is supposed to be different — answer the threat — chose differently in none of the 220.
A rule with no promise at all
Playing in a hottest component comes with a bound: never more than the largest single temperature below the mean of the board. Over 220 sums the bound holds 220 times — and so does the bound for a rule with nothing behind it, which scores exactly what perfect play scores on 205 sums against the hottest rule's 196. The control that shows the bound is doing work is the rule that plays the coldest component, which breaks it 74 times and loses up to eleven points.
A pool built to punish greed
The rung below found a rule with no theorem behind it beating the rule with one, and predicted that a pool of deliberate traps would reverse the result. It does not. The traps miss, because the rule called greedy scores a move by the stop it leaves and a stop already contains the follow-up — and the rule the traps do catch, losing sixteen points where the guaranteed rule loses five, had to be written to make the point.
The strategy that is a symmetry
A pairing strategy is a symmetry of the board that turns one player's moves into the other's, and it wins without computing anything. Tested by playing it out rather than argued, it wins one of seven candidate symmetries across four games — exactly the Cram boards with both sides even, which is exactly where no domino is its own image.
Looking for the symmetry
Answering every move with its mirror image wins Cram on a board with both sides even, which is the argument everybody meets. Asked of every connected shape of at most eight squares instead of of thirteen rectangles, it wins twelve — and accounts for a sixth of the second-player wins there are, because 852 of the 1,042 shapes have no symmetry to answer with in the first place.
The symmetry one move away
A pairing argument proves the second player wins and names no move to do it with. Asked of every shape of up to eight squares it settles twelve boards. Asked one move later — can the first player reach a position a half-turn pairs? — it settles 288, and which boards those are is decided by parity before anything about their outline is looked at.
A schedule instead of a number
Playing in the hottest component loses at most the largest temperature on the board — the classical guarantee, stated against one number. Sorting the temperatures and reading the guarantee one step further down gives a promise 47 per cent smaller that is never breached over 1,734 sums. Two steps down it fails 120 times, so the schedule has exactly one step of slack in it.
When to leave the environment
A Go player's question is not which fight to take but when to stop taking the small stuff. Put two fights beside a stack of coupons and the orthodox answer — leave when the coupon falls to the hottest temperature on the board — is exact on sixty of eighty-one pairs. All twenty-one departures have a fight with a follow-up in them, and every pair of plain switches leaves on time.
How big the answer is
The rung below found every early departure from a coupon stack caused by a position with a follow-up, and could not say more: its follow-ups were all of a similar size, so the class it measured was one bit. A pool graded by follow-up size answers it. With the position's own temperature held at one, the departure runs from coupon 1 to coupon 3.5 as the follow-up's temperature runs from 1 to 4 — and over the whole grid the players leave at the larger of the two temperatures.
A subtraction, not a factor
The crossover factor was a half on fights whose answer starts another fight, measured on a pool with two three-deep positions in it. A pool built to be deep gives twenty, and the factor does not survive them: the crossover is the follow-up's temperature less a half on eighteen of the twenty, and a factor of a half agrees with that only where the temperature is one — which nearly every position in the earlier pool had.
Half of the smaller temperature
The correction to the sente crossover has been priced twice — first as a factor of a half, then as a subtraction of a half — each time on a pool whose answers were all about the same size. Over 128 fights with answers from a number up to a temperature of six, the correction is half the answer's temperature, saturating at half the fight's own. Both earlier readings are regions of that one law.
What a strategy has to remember
A value answers who wins and by how much, and it settles neither how many moves achieve it nor whether the best one is unique. Counted over every position reachable inside the catalogue of regions, the gap has a size: 4,269 positions carry 128 values between them, and a player who wants to win rather than to predict has to store 3,308 choices — twenty-six entries for every number the theory supplies.
The quantity that does not order a board
The rung below found the players leaving an environment at the larger of a position's two temperatures, and proposed that a board should therefore be played in the order of that quantity. Over 220 boards of three components it plays exactly on 124 against playing-in-the-hottest's 196, loses 85 of the 97 disagreements, breaks Hotstrat's guarantee on six boards, and costs nine points on its worst one.
Three rules and a tie-break
An exhaustive table of what a Domineering strategy has to remember is 3,308 lines. Three rules applied in order answer 94.5 per cent of it — leave the opponent fewest replies, then keep the region whole, then take whichever placement comes first — and the fourth and fifth rules answer not one more. The residue is 181 decisions in which every rule scores the candidates the same and one of them is worse.
Cut small unless you are behind
The rung below found the greedy rule — cut at the largest prime — wrong on 104 of 552 Maundy Cakes and asked for a description of them. On all 104 the best cut is at the smallest prime, the exact opposite. A middle divisor is never needed on any cake in a sixty by sixty grid, and which of the two extremes wins is decided by Ω alone: cut small when Ω(m) + 1 ≥ Ω(n), large otherwise, and that is exact on all 3,540.
A rule that beats the hottest
When promoting a component by its answer's temperature failed, the reverse was proposed — discount it instead — with a prediction, made before the sweep, that it would not beat playing in the hottest component. It does. It plays exactly on 201 of 220 three-component boards against 196, wins two thirds of the boards where the two disagree, keeps inside a guarantee proved for the other rule, and the gap widens as the board grows.
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 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.
Seventy-two of them were not silence
The rung below said its 181 unanswered decisions were all the rules falling silent and asked whether the position's value picks the placement once the geometry cannot. Seventy-two of the 181 are the rules speaking and being wrong, which is a different failure. On the 109 that really are silence, a rule chosen per value answers more than half — and the star class the rung below singled out is settled outright by leaving the younger position.
The worst value in its own interval
The rung below scored a component by its temperature less its hottest answer's and asked what rate the answer should really be charged at. Every weight strictly between nought and one scores the same and beats the rung below's choice of one at every board size — because a ranking rule's score is a step function of its own coefficient, and one is exactly where two components tie.
A pairing, and the pairing
The rung below repaired the half-turn check and asked whether a reflection would fire where it does not. It does — forty positions of 58,830 on the largest board — and it is sound, and it is worth one node in a thousand to a solver. It can never fire on an empty rectangle at all, which is why the ladder's whole subject is the half turn.
The price of taking the maximum
The seventy-two decisions where a Domineering strategy's rules name the wrong placement are never wrong by more than one reply, and a third of them are the second rule's fault rather than the mobility count's. The repair that follows — keep every placement within one reply of the best — retains a best placement every time and costs thirty decisions for every one it saves.
A pool built to have an answer
The coefficient in the rule score a component by t − λa scored identically for every λ in the unit interval, because the rule reads an ordering and that pool's orderings changed at three places. A pool designed to have twelve crossings turns the interval into thirteen different rules, and all three board sizes agree on one cell: between a quarter and a third.
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.
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.
A pairing that is not a symmetry
Every pairing strategy found so far for these boards is a rigid motion of the square, and the requirement mentions no geometry at all. Searching all 8.8 million fixed-point-free involutions instead of the eight maps more than doubles what a pairing explains — and the smallest new one turns out to be a reflection with its two fixed squares swapped.
The table that changes its mind
The advice that ten entries chosen by use serve nine lookups in ten was untested: it describes a table sorted after the fact rather than a solver that only ever held ten. A solver that only ever held ten gets 94.2 per cent — beating the best ten chosen with the whole run in view, because there is no best ten.
The easy case was not the reason
The rung below found the mobility rule reaching a failure rate of exactly nought near the endgame and named what a proof would need: that a decomposed board's comparable options are ordered by reply count. That statement is false on all five boards, at margins up to two — and split positions go exact two squares of depth before whole ones, so decomposition is the easy case rather than the cause.
Close calls nothing resolves
The same value panel that settles 118 of the 202 silent decisions settles four of the seventy-two the rules get wrong. Every rule that helps at all must replace connectivity rather than follow it, and the cheapest one breaks twenty-six decisions for every one it saves.
A second pool, designed differently
One designed pool put the rule's best coefficient between a quarter and a third, and all three board sizes agreed. A second pool, built by the identical greedy criterion from different material, has no cell that is best at every size — so the coefficient is a property of the pool and there is no number to find.
An environment instead of a stack
The guarantee behind playing the hottest component survives one step down a board's sorted temperatures and fails at two. Against a coupon environment as hot as the board it survives all of them — because the rule stops being approximate and starts being optimal, on every sum in the pool built to punish it.
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.
Cut is Short on another graph
Everything proved about the switching game is proved from Short's side, and Cut appears only as the player whose moves get enumerated. On a graph drawn without crossings Cut does not need a theory of its own: deleting a link is securing the link that crosses it in the dual, so Cut's game is Short's game on a different graph. Bridg-It is the board that is its own dual — one link short of two trees at every size, which is why its first player wins.
A board one column wider
Strategy stealing proves the first player wins Hex, and it needs three things: no draws, an extra stone never hurting, and rules that treat the two players alike. Add one column to the board and the third goes. The player whose edges are now nearer together wins whoever moves first — and does it with a table of pairs that names every reply, checked against every line to a board of twenty cells.
What the play keeps coming back to
A draw is what the backward labelling never reaches, and handing every never-ending play to one player turns the draws into wins wholesale. Judge an infinite play instead by what it keeps returning to, and every draw gets a winner of its own: over the 262,144 three-node games, 15,432 send some of their draws to one player and some to the other, which no wholesale rule can do. Finding those winners takes a fixed point inside a fixed point.
One bit of memory
Judge an infinite play by whether one position keeps recurring and every winner can play from a table of one move per position, with nothing remembered. Ask for two positions to keep recurring and that stops being true. At a hub with two spokes a player has to alternate, and a table cannot alternate: over every three-node game, 49,487 position-and-mover pairs are won with one bit of memory and lost without it.
A potential that names every move
Strategy stealing names no move, and the pairings that do name moves need a board with the right symmetry. The Erdős–Selfridge potential needs neither: Down, moving second in Hex, takes the empty cell through which Across's unfinished chains weigh most. Its guarantee reaches only boards two rows deep. It wins far past the guarantee — on every board of three rows that Down can win — and then, on a four-by-five board that a table of pairs wins for Down with certainty, it answers Across's first stone in a different cell and loses along the bottom edge.
The rule decides who has to remember
Whether a winner needs memory is a property of the winning condition and not of the board, and the property everybody reaches for is the wrong one. Of the 128 conditions on which of three positions a play keeps returning to, 32 demand memory and 19 of those are closed under union. What separates them is measured two independent ways and the two agree on all 128: a condition needs no memory exactly when it can be rewritten as a number on each position.
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.
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.
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.
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.
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 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.
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
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.
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.
What survives a coalition
Three-player Nim has no winner without a convention and no sum theory at all. Fix one player against the other two and both come back: every position has a winner, the winner is read off the number of heaps of one modulo three and the number of larger heaps, and forty-five positions fall into six interchangeable classes — fewer than Nim's sixteen. Fix a scoring rule instead and the winner comes back, identical to one of the conventions, and nothing else does: no two positions are interchangeable.
The plays do not say who chose them
Every way a game of edge geography can go is a maximal trail, and the parity of each trail says who moved last. So the list of their lengths ought to decide the game, and over every connected graph to six vertices it nearly does — 780 of 810 starts, 108 of 114 on one cycle, where no rule on the matching and the cycle reached 91. It still fails, and the smallest failure is two trees: the same two plays from each start, and a different player choosing between them. Keep that choice one move at a time and the ambiguity falls from eighteen profiles to seven, one and none. The list costs sixteen times the search.
A coalition counts to three
Three-player Nim with one player against two collapses to six classes, because every heap of two or more is a store of tempo either side can draw on. Put the same coalition on games whose heaps are not all stores and it does not collapse: subtraction {1, 2} keeps five kinds of heap, Kayles six, subtraction {1, 3} ten and {2, 3} twelve, each count settling as the range grows. Past a threshold the large heaps are interchangeable again — and in every one of the five, unlike Nim, a single large heap hands the game to the coalition. The classes neither refine the two-player values nor coarsen them: {1, 3} has two Grundy values and ten coalition classes.
Two rows still reward the move
A single coin row with an even number of coins always rewards the player to move, because the first player can take a whole parity class. Two such rows side by side break that argument, and the conclusion survives anyway: on every one of 16,380 sums of two rows the first player is never behind, and there is a proof — open the row worth more, then answer in whichever row the opponent moves in — that guarantees the difference of the two rows' margins, which is what two switches would give. The sum meets that floor on three pairs in four and beats it on the rest, once by both margins together. At three rows the switch arithmetic stops being a floor. The incentive does not.
One square for every coin
Nim's strategy has two halves: every heap has exactly one target, the nim-sum of the others, and the heaps carrying the top bit of the nim-sum are the ones that can reach it. Welter's game keeps the first half on every one of 18,470 coins tried — one square per coin makes the value nought, and the winning moves are the coins above theirs — and the square is found by walking up the binary trie from the value of the other coins. It loses the second half. And at four coins it loses nothing: the losses and the winning moves are Nim's exactly, by a two-line proof, while at six coins no Welter loss is ever a Nim loss, because the last bit of a losing position is a count of pairs.
Units modulo three, and one choice
Nim's rule for one player against two has a shape: count the heaps that always last one move, modulo three, allow one heap that offers a choice, and give any second choice to the coalition. Subtraction {1, 2} has exactly that table — ten classes, nine of them a count of ones against nothing, a two or a three — and so do Kayles, where a heap of three is a heap of two with a unit beside it, and Dawson's Kayles. Subtraction {1, 3} and {2, 3} do not: a board with two choices on it can still be the lone player's, and the short key fails. And {2, 3} and Dawson's Kayles give every heap to ten exactly the same menu of game lengths, so what a coalition reads is not how long a heap can last.
Start at four
Every order a program could compute had made the Sylver Coinage search that finishes dearer than trying the smallest move first, and the free oracle that puts a winning move first costs under a third of it. Two orders a game program would actually use do better: a history table of numbers that have won before wins back 27 per cent of that gap, and a fixed order learned on small positions the same on larger ones. Both are learning one thing. Trying the smallest move first, but from four, wins back 31 per cent and beats smallest first on 1,695 of 1,766 positions — because naming 2 leaves a two-generator semigroup, which is symmetric and never lost for the opponent, unless 3 is already reachable.
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.
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.
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.
Blocking is not enough
Everybody knows noughts and crosses is a draw, and the usual explanation is that the second player can always block. That explanation is false. Against an opponent who can only block, the first player wins, from the centre or any corner. The game is a draw only because the second player's blocks build threats of their own. On fifteen boards solved outright, it is the only one where those two things come apart: everywhere else a draw is a draw because blocking is enough.
The opponent makes it repeat
Euwe's unending game needs both players to keep to it. Give the letters to two players in turn, with one of them wanting the draw that a repetition brings, and the rule of 1929 fires within seven letters on a plan of one sentence: always play A. The word that defeats the rule exists and nobody can be held to it. Add a third round trip and the same player can no longer force anything in twenty-six letters. Existence and force turn out to be separate questions, with separate boundaries.
Named alongside it
The objects these essays reach for when they reach for this one.
Exhaustive searchCounterexampleEnumerationHeuristicCertificatePairing strategyTemperatureNormal playDisjunctive sumMove selectionImpartialMean value