State — where it appears
Named by 6 essays across 2 fields — each of them below, with the objects they name alongside it.
What a component has to carry
Three impartial games on this site break the sum, and they break it for the same reason: a component cannot say what its own legal moves are. Measured with one instrument — one number per part, exclusive-ored — the failure rate runs from a quarter to nearly half, against a control where the same recipe is a theorem and is never wrong.
What restores the theorem
Fibonacci Nim breaks the recipe every impartial game is supposed to obey: one number per heap, exclusive-ored, gets a quarter of two-heap sums wrong. Index the recursion on the pair of heap size and cap instead and the recipe is exact on every pair and every triple — and the number a heap of nine carries turns out to be five rather than one.
The family the Fibonacci numbers belong to
Fibonacci Nim lets a player take at most twice what the last one took, and the heaps the opener loses are the Fibonacci numbers. Two is an arbitrary number. At one the losing heaps are the powers of two, at three and four and five they are four more sequences, each with a linear recurrence whose lag is twice one less than the factor — until the factor is six, where the pattern stops.
What the numerals knew
Every factor in the Fibonacci Nim family gives a numeral system, and the rung below predicted its separation condition would be the lag of the recurrence the losing heaps satisfy. It is not. The gap is the factor — one at c = 1, Zeckendorf's two at c = 2, and c at every factor to eight — while the lag goes 1, 2, 4, 6, 8, 11, 14, 17 and leaves its own pattern at six. The numerals then solve every one of 194,480 states, cap and all.
One half multiplies, the other adds
The rung below priced the two halves of a substitution licence on sums of two Cram boards and predicted that the first half's saving would grow with the number of components while the second's would not. It is right, and both halves have closed forms: the component licence saves s^(k−1)/k and the subposition licence k·s over a shape count that never moves.
Two clauses and a third question
A component can carry its own rule when two things hold: its moves are a function of what it carries, and a move in it leaves every other component alone. Two rulesets built to fail one clause each are both caught on a named witness. The four real games sort exactly — every one the recipe gets right fails no clause, every one it gets wrong fails one — and the two clauses still miss something, because Fibonacci Nim and a held pass fail the same clause and only one of them can be repaired.
Named alongside it
The objects these essays reach for when they reach for this one.
ImpartialDecompositionEnumerationExhaustive searchInvariantComponentCounterexampleFibonacci nimSprague–GrundyZeckendorf representationContextFibonacci