Induction — where it appears
Named by 23 essays across 8 fields — each of them below, with the objects they name alongside it.
The condition the recursion rests on
Not that the moves run out, and not that the options are few. Poker Nim's heaps can grow without bound and it ends; the game called `on` has one option and never does. What every game value needs is that no infinite run of moves exists — and there are three separate ways to fail it.
Four values, and the sequence is settled for ever
The Grundy values of a subtraction game repeat with period 7, and proving it needs a window of exactly four of them — one for each size of move the game allows. Everything past the window follows by induction. A finite computation has settled a claim about every heap there will ever be.
It ends, and nothing says when
The recursion this site runs needs every line of play to reach a position with no moves, and the condition is usually met by an obvious decreasing quantity. The hydra meets it with no such quantity anywhere: the tree grows at nearly every step and the fight ends regardless, because the only thing that decreases is an ordinal. A four-node hydra dies in twenty chops; one level deeper and 279 chops reach forty thousand nodes with no end in sight.
The heap is not the position
Fibonacci Nim bounds a move by twice the previous move, which puts the state outside the board: a heap of six with a cap of two and a heap of six with a cap of five are different games. So there is nothing to add and no Grundy value to compute — and the game is completely solved anyway. The opener loses on exactly the nine Fibonacci numbers up to 120, and the smallest term of the Zeckendorf numeral is a winning move in all 110 winnable heaps.
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 birthday of a sum
Two values born by days m and n have a sum born by day m + n at the latest, which is the bound that stops a board made of many small parts from being unboundedly complicated. Over 231 pairs of day-two values the bound holds every time and is exact 163 times — and every pair it misses by three days or more has a sum that is a number or a nimber, so the slack is not noise but a measure of how much cancelled.
The short side only says how many
The rung below settled which cut to make in a Maundy Cake and left the value open. With the cut settled the recursion is a walk, the walk unrolls, and what it unrolls into is the running products of the long side's prime factors, largest first. The short side never enters the products at all — it decides how many of them there are and nothing else, so sixty-two different short sides give one value.
The premises an induction would need
The rung below settled by a grouping test that a position's crossover depends on its own temperature and its answer's and on nothing below them, and asked for the induction. The four paragraphs are not written here; the checking they would rest on is. The law holds at five levels, survives translation, heating and cooling — and none of that is the step.
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.
One proof, and one wrong lemma
Two measured identities were left for a proof: the move rule by induction, the gap condition from the reply bound. The induction is exact on 31,731 heaps at eight factors. The reply bound holds at c = 2 and on one index pair in twenty-seven at c = 3 — and the inequality that does the work is a third one nobody proposed.
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.
A hypothesis has to hold all the way down
Milnor's bound is proved by induction over the play, so the condition it needs has to hold at every position the play can reach. Checked on the row instead, ninety-two pairs pass the test and twenty-four of them break the bound. Checked at every subposition, twenty-eight pairs pass and none breaks it.
The paper was about how long
Zermelo's 1913 paper is remembered for a theorem it proves in passing. The question it actually asks is how many moves a forced win takes, the answer it can prove is the size of the whole position graph, and the round counter in the procedure is the real answer — a quantity nobody named for another forty years.
The gap between two answers
A draw is usually described as what the backward labelling never reached, which makes it sound like a shortfall of the algorithm. Written as one predicate the winning condition is an equation, the equation is monotone, and it has a least solution and a greatest one — and the set the two disagree about is exactly the drawn set, on every game checked.
Every play ends and no round settles
Take the finiteness hypothesis away carefully — not by adding a cycle, which has already been priced twice, but by adding infinitely many positions to a game every play of which still ends. Nothing is drawn, every line finishes, and the round the opening settles in grows with every cut: two, four, six, eight, twelve, sixteen, and no number in the column is the answer.
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 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.
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.
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.
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.
Named alongside it
The objects these essays reach for when they reach for this one.
Exhaustive searchProofClosed formTerminationEnumerationGrundy valueBoundMexOrdinalRecursionBackward inductionClosure