Parity — where it appears
Named by 30 essays across 6 fields — each of them below, with the objects they name alongside it.
How long it lasts
Move in every component at once and the game ends the moment any one of them does. Grundy values say nothing about that game; what decides it is the remoteness, a second number computed from the same tree that measures how long a component can be made to last. Over 2,268 positions the rule is right every time, and the two numbers determine each other in neither direction.
The strip where every number is a whole one
Delete the hop from Toads and Frogs and the halves, quarters and ups vanish completely: over 9,801 strips, every value that is a number is an integer, without a single exception. The guess that the hopless game therefore has a formula reading the gaps is half right and exactly wrong — 1,460 strips of eight squares are switches, and three strips with the same counts of toads, frogs and empty squares are worth 1, {2 | 1} and 2.
The number nobody needs
The compound theory carries a third quantity — the suspense number — computed by the remoteness recursion with both preferences reversed, for the compound that stops as soon as any component stops. It governs that compound correctly. So does remoteness, so does the plain Grundy value, and the shortening does not change the winner on any of 1,176 positions.
A compound of two different games
Every rule the compound theory has survives mixing exactly — the minimum-remoteness rule is right on all 5,184 mixed pairs and all 7,560 triples — and the reason is not that the rules are strong. It is that each of them reads one number per component, and a number does not remember which ruleset produced it. The thing mixing damages is the shortcut a reader carries instead.
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 patch that generalised
Misère Nim takes a one-line patch: play the normal-play strategy until every heap holds a single counter, then invert. Moore's Nim, where a move may take from up to k heaps at once, takes exactly the same patch with exactly the same modulus — and the two rules disagree on six positions out of 923.
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 check in front of a search
The rung below found a pairing one move away on 288 of the 767 even first-player shapes, and asked what a solver that tested for one before recursing would save on a real game. On an even Cram board it saves nearly the whole search — a 4 × 5 board takes 17,348 node expansions without the check and one with it — and the depth profile shows why that number flatters: the check settles every winning position at the opening and at the last two moves, and about one in ten in between.
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 parities, in size order
The rung below settled four of six parity classes in bounded Moore's Nim and asked whether the sizes pick out the losing positions in the two it could not. They do — but only through the order they put the parities in. Sort the heaps largest first, read off their parities, and that five-bit word settles the whole game at every width of move, with the losing words forming a subspace.
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 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.
The family with two witnesses
Six predictions about seven heaps were written down and deliberately not run. Five of them held. The one that broke is the condition that had been checked against two cases when it was proposed — the fewest of the four — and at seven heaps it does not merely give the wrong answer, it asks a question the parity word has stopped being able to answer.
The dual was the value table
A coin-turning game's losing rows form a linear code, and a code has a dual that nothing in the game appeared to read. It reads it constantly: the dual is spanned by the bit-planes of the Grundy values — the parity checks are the value table stood on end — and on Mock Turtles over eight coins the losing rows are exactly the span of the table that decides them.
The pairing the formula hides
Welter's closed form sums a function over every pair of coins and needs an extra term when the count is odd, which looked like a surprise. Read as a matching it is not: an odd number of coins cannot be paired, the left-over coin contributes its own square, and some matching gives the value on every position measured.
Three distances too many
The junction descriptor records how far a crossing sits from four ends, and the rung below asked what the value does when one crossing slides along its run. It reads one bit — the offset's parity — and only when the run has odd length. The other three distances reach the value not at all.
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.
Two ways to count a finished board
Territory scoring and area scoring are both in daily use and they are not variants of one rule. Over seventy-nine small endgames they agree exactly on the thirty-nine with an even number of neutral points and on none of the forty with an odd number — sixteen name a different winner, and twelve are played differently, which is not something a convention is supposed to do.
A key is a code, and two squares come free
The families of positions a Zobrist key confuses are the words of a binary linear code — the sets of squares whose words cancel — so choosing a key is choosing a code. On 4 × 5 Domineering the textbook choice, a code with the largest minimum distance, confuses more stored positions than a random key at fourteen, sixteen and nineteen bits. The choice that reads the board confuses none at eighteen: two squares of one shade, in rows of different parity, are decided by the other eighteen, and no seventeen-bit key is exact.
Search on in pairs of moves
Deepening until two depths agree gives a proved verdict, and searching on where the counts are close gives a better one; put together the obvious way, they stop on a wrong verdict at 3,231 positions of 4 × 5 Domineering. A guess one move past the cut has the other player to move and flatters the wrong side. Searching on two moves at a time keeps the proof, and the window that suits it is one-sided — but however it is widened, the certificate gets cheaper only by turning into the search that finishes, and on four boards it never gets below it.
A thousand positions and no exception
The parity law was fitted to constructed bags of chains and loops inside a string budget. A board's positions are a different population — the sizes are what the geometry allows, the components come correlated, and a six-box board holds exactly one position that is a loop of six. Tested on all 1,032 of them and all 160 of the four-box board's, the law is right every time, against a verdict computed from the strings by a walk that has never heard of a component.
Cancelling is not pairing
The coin rows that cancel against their own negatives looked like the rows whose coins pair off as nested equal pairs, and on rows of four they are exactly those. From six coins the description fails in both directions — twenty rows pair off perfectly and do not cancel, and one coin set has a hundred and thirty-six that cancel with no pairing at all — and a row of five coins cancels, though an odd row can never pair off. What does hold, on every row swept, is that the first player in a row plus its negative never finishes behind.
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 rare class exclusive-or respects
The rare values of ·007 were measured by rank — everything outside the twenty-two commonest — and found getting commoner. A sparse-space argument needs a different split: a rare class closed under nothing and a common class closed under exclusive-or, which is to say a parity of bits. Over every parity of every bit, ·007 and ·6 have no class that thins. Across the 79 growing octal codes of up to three digits, one does: ·354, whose odd values fall from 288 in its first two thousand heaps to seven in its last ten thousand.
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.
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.
The odd values keep coming
·354 is the one growing octal code of up to three digits whose values have a class that thins: 288 odd values in the first two thousand heaps, seven between ten and twenty thousand, the last of those at 18,908. A sparse-space argument needs that class to stop. Taken to a hundred thousand heaps it does not: 48 more odd values arrive after twenty thousand, every block of ten thousand holds some, and the last in range is at 99,600. They keep no schedule a remainder can see, their gaps are spread like arrivals, and most of them are one number, 37 — each one a heap where every split that could have reached 37 misses it.
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.
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.
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.
Named alongside it
The objects these essays reach for when they reach for this one.
Exhaustive searchCounterexampleImpartialEnumerationGrundy valueStrategyInvariantNormal playNimNim-sumP-positionProof