Strategy — where it appears
Named by 55 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 an assertion 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
The rung below proposed the reverse of the rule that had just failed — discount a component by its answer's temperature rather than promoting it — and predicted, 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 this ladder has found 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.
Named alongside it
The objects these essays reach for when they reach for this one.
Exhaustive searchEnumerationCounterexampleHeuristicTemperatureNormal playCertificateDomineeringImpartialInvariantMean valueDisjunctive sum