Proof — where it appears
Named by 27 essays across 8 fields — each of them below, with the objects they name alongside it.
Two ways to end with no bound
Sylver Coinage and the hydra are both guaranteed to finish and neither will say when. The difference is that one of them carries its own bound: every move in Sylver removes at least one gap, the gaps can be counted in a moment, and over ten openings the longest play uses every single one. The hydra has no decreasing quantity a solver can hold — three hydras of five nodes each take seven chops, twenty-one, and a number past two hundred and seventy-nine that this machine never reaches.
The short side is not in the lemma
The closed form for a two-sided Maundy Cake rested on one unproved statement: that no divisor beats the largest prime. Written out, that statement never mentions the short side — it is an inequality between a multiset of primes and a term count — and once it is stated that way it has a two-line proof, term by term. The ladder ends in a theorem rather than a grid.
The case that was supposed to be hard
The mex rule for the mirror construction was to be proved by induction, and the step flagged as needing care was the one where an option is incomparable with the nimber. There is no induction: the argument is four lines, and incomparability is what makes two thirds of the cases go through — because a fuzzy sum is a first-player win and the first player is the opponent.
Which top is the top
The crossover law's proof rests on the walls above the crossover being governed by the top two options, and the check was never run. Run on 23,586 heights it holds exactly — but only when the options are ranked by mean value. Ranked by the temperatures the law is stated in, it fails on a fifth of them.
The obvious cut is the wrong one
Maundy Cake's rule was proved by restating its lemma so the short side vanished. Cutcake's collapses the same way — into a binary length instead of a multiset of primes — but the cut the argument needs is not the one the ladder predicted. Halving is wrong on a third of all cakes, and the smallest counterexample is six squares by two.
The square that cannot be halved
Every number in hopless Toads and Frogs is a whole number, which the rung below measured on seven thousand strips and could not explain. The reason is that every empty square is either one player's alone or split evenly between them — except one, and that one is where the numbers stop.
Which games end at which level
Between a game that ends within a computable bound and one that ends with no bound at all there are levels, each corresponding to a strength of induction. This site's games sit at three of them, and which level a game is at is decided by exhibiting its termination measure and checking that every move lowers it.
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 proof is sixteen cells
Lasker's Nim has a four-clause formula that was checked on two thousand heaps and never proved. The proof fits in a four-by-four table: the last two bits of a split's value are fixed by the last two bits of its parts, so no split can land in its own heap's class — except at 3 mod 4, where it lands exactly on the one value the takes leave missing and pushes the answer up by one.
The follower does the reversing
The gift-horse theorem for the ordinal sum needs two cases, and the second — the added option is reversible — was counted and not described. Recorded move by move, the reversing answer is always Right's move inside the follower: on all 410 escapes under five followers, and on every one of the 2,628 gift horses under every follower that gives Right a move at all. The case split is by follower, not by horse.
The split slips one day deeper
The reversal case of the gift-horse theorem was described in one line — the follower's own move reverses every gift horse, whenever the follower has one — and tested only where it was found. In the mirror it holds exactly, with 1 and −1 trading places. One day deeper it fails: under ↑ and ½, three gift horses on built day-four bases are not reversed by the follower's move. All three are dominated, so the theorem stands; the clean split by follower does not.
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.
A different question at every depth
A Sylver Coinage proof that deepens over survivors was never cheaper than the search that skips struck moves and finishes. Keep everything it proves and refutes and it gets cheaper — 41 per cent at twenty gaps, and more when the opponent's positions are kept too — and it is still never cheaper, on any of 1,766 positions, and the gap widens with the gaps. Told the right depth in advance it wins on seven. The finishing search asks one question of a position; a proof asks a different one at every depth.
The pairs are read from the bottom bits
Some matching of Welter's coins reads the position's value, and usually only one does — but which? The rule is short: pair the two coins whose squares agree in the most low-order bits, and repeat. It reads the value on every one of 3,696 positions of two to eight coins, where no other natural rule manages half. And it has a proof two lines long, because x − 1 flips a number's trailing bits, so a pair's animating value depends on where its squares part on the binary trie read from the bottom. What it is not is a strategy: a move from a lost position can be answered inside its pair only two times in five.
A proof inside a budget
A certificate — two depths of search that agree — was always dearer than the search that finishes, but that compared proofs with answers. Its whole purpose is the board where nothing finishes, so the fair test is a budget: cap the positions a search may expand and count the proved verdicts each method returns inside the cap. On a thousand deep positions of 4 × 5 Domineering, the search that finishes proves more at every budget from sixteen to eight thousand. The best certificate, with a one-sided two-move window, proves at most three positions finishing cannot, at any budget, and plain deepening proves none. Where deepening earns its keep is the guess it holds while it works, not the proof it ends with.
The seven were the order
A Sylver Coinage proof told its depth beat the search that finishes on seven positions of 1,766, and on all seven the winning move came fifth when moves were tried smallest first. Run both searches under five orders and the seven vanish under every other one; try the largest move first and there are fifty-five new ones instead. Give both searches the winning move first, free, and the proof is cheaper on none. The exceptions were never positions where proving is easy. They were positions where one order made the finishing search stumble — and what is left when the order is perfect is the price of asking a proof's question.
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.
No number bounds it
Sylver Coinage cannot be given a termination measure in the natural numbers: from {2} the game can last one move, two, or any number the second player chooses. The measure proposed for it — the gap count beside the number of distinct primes the named numbers share — rises on {4} + 6. Count the primes with their repeats and it falls on every one of 42,519 moves checked, and it is exact: the game tree is ω squared tall, and the height of a position is ω times the prime factors of its gcd plus the gaps that are left.
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.
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.
Two takes, one line
A subtraction game with two legal takes, a and b, has a Grundy sequence that fits in one sentence: cut the heap's remainder modulo a + b into blocks of a, alternate nought and one, and turn every even block at or past b into a two. The proof is three cases long and holds for every pair. Add a third take and the sentence has no successor — but it does not vanish. The period of a three-take game is a sum of two of its takes six times in seven, which sum is decided by the third take's remainder modulo the first two's period, and the long irregular periods cluster on the one arrangement where the third take is the other two added together.
The third take is the other two
A subtraction game whose third take is its first two added together was where the long irregular periods of three-take games clustered, and the question was whether that one arrangement has a formula or is where the formulas stop. It has one. Divide out any common factor and look at the heap one short of the middle take: if it would be a loss when only the smallest take is allowed, the period is the smallest take times twice the middle take plus a remainder, otherwise it is short and the whole sequence is two bands of blocks. The formula gives the exact period of every one of the 3,160 games to eighty.
An odd cycle is the only way to have none
A cycle of two positions has two stable sets and a cycle of three has none. Moses Richardson proved in 1953 that this is the whole story of absence: a graph with no odd cycle always has a stable set. Run over all 9,608 games on five positions, the theorem holds without exception, sits inside a chain of five nested families, and leaves open the larger question its converse invites, since 7,452 of those games have an odd cycle and a stable set anyway.
The list that cannot run out
Give each cell of an n × n square its own list of n colours, and ask for a colouring from the lists with no repeat in any row or column. Dinitz asked in 1979 whether that is always possible, and reading the cells in order fails about four times in ten. Galvin's proof of 1994 needs one page. Use an ordinary Latin square to turn each row and column into a preference, so that every cell has exactly n − 1 moves, and peel off a stable set for one colour at a time. A list of n cannot run out against n − 1 moves.
A rule that let the game run for ever
In 1929 German chess drew a game when the same sequence of moves came three times running, and the rule was meant to make chess finite. Max Euwe, six years from the world title and trained as a mathematician, showed that it does not, using a word that had already been found twice for other reasons. Two round trips are enough. The board returns to one position thousands of times, no block of play ever comes three times running, and the rule has nothing to say.
How strict a rule can be
A repetition rule can be tightened one notch at a time, and each notch either leaves unending games alive or kills them outright. With two moves a block may not come twice, and every game is over by the fourth letter; relax that by a single letter of overlap and unending games appear. Three moves stop dead at 38 letters under a rule that a whisker's relaxation frees. Four stop at 121. Past each boundary the words go on for ever, and the boundary sits at a fraction nobody would have guessed.
Named alongside it
The objects these essays reach for when they reach for this one.
Exhaustive searchInductionEnumerationCounterexampleCertificateClosed formGrundy valueStrategySylver CoinageMexParityBinary