Impartial — where it appears
Named by 95 essays across 7 fields — each of them below, with the objects they name alongside it.
How hard is it
Every theorem on this site stays true at any size. The answers stop being reachable long before the games get interesting — deciding the winner of a generalised board game is PSPACE-complete, and an exact evaluator gives out after a few dozen moves.
Nim, and the nim-sum
Three heaps of counters, take as many as you like from one of them, and the player who takes the last counter wins. The winning condition is not a search, not a table, and not a heuristic — it is the bitwise exclusive-or of the heap sizes, and it was found in 1901.
Hard, proved
A game is as hard as a logical formula when the formula can be drawn as the game. Here is the drawing — a quantified formula turned into a graph with a token on it — with every formula over three variables played both ways and required to agree.
Every impartial game is a Nim heap
Sprague and Grundy proved, independently and four years apart, that any position in any impartial game is equivalent to a single heap of counters. Not similar to one — equal to one, interchangeable with it inside any larger game.
The chains decide it before the boxes do
Under every game of Dots and Boxes there is an impartial game with no score in it, and it settles the question the scoring game keeps asking — who ends up having to open. The rule players learn as folklore falls out of it, and so do the exceptions nobody mentions.
Grundy sequences, and where they stop being predictable
Computing one Grundy value is a mex. Computing all of them produces a sequence, and the sequences do something nobody has fully explained — most of them eventually repeat, some of them take thousands of terms to start, and for a few nobody knows whether they ever do.
What survives misère play
Misère play destroys the value theory, and something much smaller grows back. Fix one game, look only at sums of its own positions, and the classes that behave alike form a monoid — computed here, and larger than the normal-play answer every time.
Outcomes do not add
Knowing who wins each part of a position tells almost nothing about who wins the whole. Counted over every sum of two values born by day two, six of the outcome table's ten entries are settled and four are not — and every settled one is settled by the order rather than by anything about outcomes. Two first-player wins reach all four classes between them.
Two people, four years apart, one theorem
Roland Sprague proved it in 1935 and Patrick Michael Grundy proved it in 1939, neither knowing of the other. That looks like coincidence until the alternatives are examined — and the rule they both reached turns out to be the only one that can work at all.
A golden ratio thirty years early
Wythoff described the losing positions of his game in 1907 with an argument about partitions of the integers, and no Grundy value anywhere in it. The theory that arrived thirty years later computes the same positions — and has never produced a closed form for the values, which the older argument had for the zeros from the start.
Sprouts, and the game that is not one
Two games played with dots and curves, invented in the same room, all but indistinguishable on paper. One is unsolved past forty spots. The other has no decisions in it at all — the winner is fixed before the first curve is drawn.
Which part to move in
The value of a sum is the sum of the values. The move in a sum is not the move in any part, and there is no rule that reads it off the values — in the smallest interesting example, the only winning move is in the component worth nothing.
The move that gives counters back
Poker Nim adds one rule to Nim — a player may put counters back onto a heap from a private reserve. It looks as though a losing player could stall for ever. The winner is decided by exactly the same nim-sum, and the reason is the single most useful idea in the whole reduction apparatus.
Three ways to add the same games
A move in exactly one component is a choice, not a law. Move in every component at once and the game is different; move in any set of them and it is different again. The same two positions, added three ways, give three different answers — and only one of the three has values that add.
The game that is a number system
In Sylver Coinage two players name integers and nobody may name a sum of what has already been named. Its positions are not boards — they are numerical semigroups, its termination is a theorem of Sylvester's from 1884, and the question of who wins after the opening move 16 has been worth a thousand dollars since 2017.
A row of coins is already a sum
Everywhere else on this site a sum is several positions side by side. In a coin-turning game it is one row — each coin showing heads is a game in its own right, and the row is worth the exclusive or of them. The decomposition is inside a single picture.
Squash every loop to a point
Colour every Hackenbush edge green and the game becomes impartial, so the whole picture is worth a single Nim heap. Two principles find which one without playing anything — fuse the cycles, then run one pass up the tree — and a nine-vertex lattice that costs 1,283 positions to solve costs twelve steps to read.
Misère play has no negatives
Put a position beside its own mirror image and answer every move with the mirror move. Under normal play the answerer wins and the sum is worth zero. Under misère the answerer still has every reply and loses because of it — so there is no zero, no subtraction, and no comparison, which is why the misère theory had to be rebuilt rather than adjusted.
Take one, three or four
A heap and a list of legal takes. It is the smallest interesting impartial game there is, and the only family in the subject where eventual periodicity is not observed, not conjectured, but guaranteed — with a bound on when it must appear.
Where the impartial theory stops
Sprague–Grundy gives every impartial position one number, and the number is complete. The moment the two players have different moves no number works at all — not a harder one to compute, none — and three positions here are compared against every nimber to show it.
A green edge on a blue one
Blue over green and green over blue are the same two edges in the other order. One is worth 1∗ and the other ↑∗ — a number with a star on it against something smaller than every positive number — so a stalk with all three colours in it stops being a numeral and starts being a position whose value depends on what is underneath.
The tartan theorem
The nimbers are a field, with a multiplication defined by a mex-style rule that looks like an algebraist's amusement. Lay two coin-turning games on a grid and the Grundy value of each square is the nimber product of its two coordinates — which is the point at which the multiplication stops being a curiosity and starts computing answers.
Two misère outcomes are not enough
Knowing who wins each part does not say who wins the sum. Over 676 sums built from a pool of twenty-six positions, nine of the sixteen pairs of outcome classes settle the answer under normal play and not one of the sixteen settles it under misère — and the nine that work are theorems about a value being zero, which is exactly the thing misère play does not have.
A move that must be answered
Every argument on this site about sums assumes the parts are independent: a move in one leaves the others alone, and the reply may go anywhere. Top Entails denies it — take the top coin of a heap and the opponent must answer in that heap. The nim-sum then misreads 9 of 36 two-heap positions, and two heaps of two coins are a first-player win, which no impartial game the theory covers can be.
The losing positions are a code
Turn over one, two or three coins, and the rows a player has already lost turn out to be closed under adding two of them together. That makes them a linear code — and on eight coins it is the extended Hamming code exactly, sixteen words with a weight enumerator of 1 + 14x⁴ + x⁸, produced by a move rule that knows nothing about codes.
The period is small and the proof does not say so
Every subtraction game repeats eventually — that is a theorem, and its proof gives a bound of sixteen thousand for a three-move set. Over 112 sets the longest period measured is twenty-two. The proof and the fact are four orders of magnitude apart, and the rule of thumb that closes the gap is broken by one set in the sweep.
Taking from several heaps at once
Moore's Nim lets a move take from as many as k heaps at a time, and the losing positions are still read off the binary columns — divisible by k + 1 rather than by two. The rule agrees with exhaustive search over 54,264 positions and never disagrees, and it decides every outcome while supplying no value at all: reading the same columns as a base-3 number gets the Grundy value right on 42 of 330 positions.
What a tame heap may be replaced by
Calling a heap tame is only worth anything because a tame heap can be swapped for a Nim position with the same genus in any misère sum. The swap is not always a single heap: Kayles' heap of eight is worth ∗ under normal play and carries the genus of 2 + 3, and substituting ∗ instead gets three of the twenty-eight Kayles pairs wrong.
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.
A token on a graph
Geography is an impartial game whose position is a vertex and a history, so a ten-vertex graph has ten thousand states rather than ten. Take the arrows off and the same game is decided by a maximum matching — first player wins exactly when every maximum matching covers the start, verified on 41 vertices across eight graphs. One word in the rules separates a polynomial criterion from a PSPACE-complete problem.
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.
A pass is not a move
Put a single pass token on a Nim board and one clause decides everything. If it may be taken at any time — including as the move that ends the game — the value of the whole is the nim-sum with a one added, in all 120 positions swept: the pass is a heap of one. Forbid it as the final move and the value stops being a function of the nim-sum at all, and 3 and 1 + 2 come apart.
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.
When the nested sum only sees the value
The ordinal sum reads the form and not the value: three positions all worth zero, placed under a star, give three different answers. On impartial games it reads the value after all — 72 substitutions of an equal-valued heap from a different game, and every ordinal sum comes back unchanged. That difference is the whole reason a green Hackenbush tree can be collapsed one branch at a time.
Taking from the ends
End-Nim is Nim's board with a player at each end, and it takes one sentence to state. Not one of its 5,460 small positions is worth a non-zero number — the game is all-small, so zero is the only number any of them can reach — and there are 2,693 distinct values between them. The outcome says a great deal more: 4,738 of those positions are won by the same player whoever moves, and on two heaps the rule is that the larger end wins.
No two heaps alike
Welter's game is Nim with one extra clause — no two heaps may be the same size — and the clause is fatal to the nim-sum, which gives the right answer in none of the 120 three-coin positions. What replaces it is a function of pairs: ⟨a | b⟩ = (a ⊕ b) − 1, exact on all 55 two-coin positions, and nim-added over every pair it is exact on the whole board provided the number of coins is even.
The values that keep arriving
A Grundy sequence that repeats uses finitely many values and stops needing new ones. Six thousand heaps into ·007 the count of distinct values is 187 and still climbing, and the share of heaps carrying something outside the twenty-two commonest rises from 32% in the first thousand to 85% in the sixth. The rare values a periodicity argument needs to thin out are getting commoner.
Splitting is a move
Add to Nim a move that removes nothing — break a heap in two — and the Grundy sequence gets simpler, not harder. Lasker's Nim has a closed form with one clause per residue modulo four, exact on all 2,001 heaps checked: the identity with every fourth pair transposed. Kayles is the same kind of game with the taking bounded instead of the splitting, and it has no closed form at all, settling into a period of twelve only from heap 71 with fourteen values outside it for ever.
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 players, two lists
Give each player their own list of how many counters they may take and the impartial theory stops applying. What survives is the outcome: it settles into a repeat, for every pair of lists, and that is a theorem. What does not survive is the value — on four of six pairs swept it has no repeat inside sixty heaps, and the birthdays are still climbing at the edge of the window.
A period with a constant added
An octal code says what a player may do when removing k counters, in three bits; a hexadecimal code adds a fourth — leave three heaps — and the digits run to fifteen. Over twenty-two codes swept to six hundred heaps, five hexadecimal ones repeat with a fixed amount added each time round and no octal one does. Their values climb for ever and never repeat, so a search that looks only for repetition reports them unsettled.
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.
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.
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.
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.
Where the nimbers run out
A single End-Nim heap is a Nim heap and every palindromic row is worth a nimber, so the impartial theory looks as though it might get a long way into a partizan game. It gets one row in thirteen. Five nimbers occur in five and a half thousand rows, the palindromes account for two fifths of them, and the rows worth something else run to 2,693 distinct values.
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.
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 rule the symbols follow
Two genus symbols make a third by three lines and no lookup table: the base exclusive-ors, the sum is fickle only when every component is, and the symbol follows. Checked on 252 pairs across nine games it is right on 238 — and the fourteen failures are exactly the fourteen pairs with a wild heap in them, which is the boundary the genus is defined up to arriving as a measurement.
The rows that are their own mirror
Four hundred and ten End-Nim rows are worth nimbers and 168 of them are palindromes, so a condition covering the other 242 was outstanding. It is that the row is equal to its own negative — and on this game that condition is not merely sufficient but exact, which is more than the group law promises and is a fact about End-Nim rather than about games.
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.
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.
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.
The rule a smaller move breaks
Moore's Nim lets a player take from at most k heaps, and its winning condition is the binary columns summed modulo k + 1. Cap the amount as well and the obvious repair — reduce each heap modulo the cap plus one, then read the columns — is exact at every cap when k is one and wrong at every cap when k is two or three. The reason is stronger than a broken rule: at k ≥ 2 the residues do not determine the outcome at all, so nothing of that shape can work.
The condition that survived the wider sweep
Which pairs of subtraction lists have a value sequence that repeats? Over the 49 pairs drawn from one, two and three, two conditions answer it identically — a translation and all-odd — and both are exactly right. Over the 961 pairs drawn from one to five, all 83 translations still repeat with no exception and four all-odd pairs do not, at heap ninety with a period as long as forty-two. Neither condition is necessary: 104 pairs repeat that satisfy neither.
The wider move is the easier game
An earlier essay ruled out every rule that reduces the heaps and reads the residues, and asked for a two-part statistic: the residues plus one more count. Four second parts are tested here and none of them decides. What turns up instead contradicts the premise the request was made under — a move that may reach three heaps is more predictable than one that may reach two, on every cap, every candidate rule, and after the change in the base rate is taken out.
The third digit
The rung below found 71 of the 255 two-digit hexadecimal codes repeating with a constant added rather than exactly, and asked whether the same share holds one digit wider. It rises. Of the 4,095 three-digit codes, 1,433 climb and 617 repeat exactly — seven in ten of the settled ones — so a saltus is the ordinary way a hexadecimal game settles and the exact repetition the octal survey was built to find is the special case.
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.
A function with no formula
The rung below's composition rule is exact on tame pairs and wrong on all fourteen wild ones, which looked like an exact boundary. Two heaps further it is wrong on 34 of 35 and right on one — Kayles' five and nine — so the boundary was a boundary of the pool. What survives is stronger and stranger: the pair of symbols still determines the sum on the wild side, and no rule of that shape describes it.
A product against a sum
A company closed under both addition and options licenses a solver to rewrite any subposition, and the rung below found that exactly the nimber groups have both closures. Priced on Cram boards, that licence is the difference between walking a product of position sets and walking their sum — four to twenty-five times on two components, twenty-four to a hundred and sixty-one on three — and it is available to impartial games because their class representative is a heap rather than a form.
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 only way to split into three
Nineteen three-digit hexadecimal codes climb by three, and the rung below asked whether they share a form and what digits they have in common. The digits are exact: on eighteen of them the only way to split a heap into three is by taking exactly three counters, and taking three counters can do nothing else. The form is not shared — the eighteen carry four distinct sequences, and exactly one of the four counts in base three.
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.
The wild side does not close
The rung below asked for the wild composition table and for two things about it: whether the wild genus symbols form a small closed set, and whether that set is a misère quotient in disguise. Building the table needed a wider sweep — nine counters a heap gives a diagonal rather than a table — and both answers are no. Not one of the twelve entries is a symbol any wild heap carries, and two wild heaps added together are tame two thirds of the time.
Half a licence is nearly all of it
The rung below priced the substitution licence a restricted universe gives a solver and asked what half of one is worth — the licence to rewrite components but not subpositions. It is worth nearly the whole saving. Rewriting components collapses a million and a half states to three thousand six hundred; rewriting subpositions collapses those to eight hundred and eighty-four, and splitting the pieces takes it to fourteen.
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.
Two counters, and one displaced term
The rung below found four Grundy sequences in the odd-saltus class and asked which term each displaces and whether the digits predict it. They do — but there are two base-three counters and not one, chosen by whether a heap of one can be taken away. And there are three sequences rather than four: the fourth is the third with three isolated values, and was counted separately because its period had not settled.
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 pattern that has not started yet
A pre-period was supposed to be rarer in this family than a defect. Two hexadecimal codes in five have one, 321 have a pre-period longer than their own period, and the code the rung below found slow takes fifty-four heaps to settle rather than two blocks — which is also the account of three defects the rung below recorded and could not explain.
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.
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 quantity that carried nothing
The rung below proposed predicting a pre-period's length from the saltus and the period. The saltus correlates with it at −0.03, which is nothing; the period correlates at 0.77 with a coefficient of one, so a pre-period is about one period long. The digit that predicts whether there is one predicts nothing at all about how long.
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.
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 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 the rung below called 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.
Closing the wild side
The twenty-two wild genus symbols are not closed under addition, and the rung below offered two answers: a monoid nobody had guessed, or no algebra at any size. Neither. Five of the six games with wild heaps close at three or four heaps, with closures of two to five symbols, and the sixth is still growing.
A stopper and how to find one
The class a value theory for loopy games would need is the ones with no infinite alternating run, and the qualifier does the work: seventy-nine of the two hundred and fifty-six two-node loopy games qualify and seventy-two of them have a cycle. Every one has a decided outcome, and under eight tests the seventy-nine collapse to six.
The parts are worth nothing and the sum is not
Every chain and every loop in Nimstring, taken alone, has Grundy value nought. So the Sprague–Grundy theorem predicts that every position built from them is worth nought — and ninety-six of the two hundred and seven positions checked here are not. The theorem is not being misapplied; it does not apply, because a capture keeps the turn. What replaces it is smaller and sharper: count the short chains, and one long component of any kind reverses the parity.
A parity with a first exception
Sort every Sylver Coinage position by how many numbers are still unnameable and the game very nearly falls to parity: odd rows are between a fifth and a half positions the mover loses, and the first three even rows hold none at all. The rule has a first counterexample at genus eight, where it is a single position out of sixty-seven, and eleven more at genus ten. It is a tendency wearing away from both ends rather than a law with exceptions.
Four hundred and seventy steps
The tartan theorem replaces a search with a multiplication. Measured on every grid a brute-force solve can reach, the two agree on all of them — and the ratio doubles with every square added. On the 8 × 8 grid the theorem is normally drawn at, the search would have to value eighteen quintillion arrangements; the theorem needs twenty-six different nimber products, and computing all of them by the rule that defines them looks at four hundred and seventy pairs.
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.
Every move closes the largest gap
A census of Sylver Coinage by genus finds a parity that nearly decides the game and asks whether any known property of a numerical semigroup predicts the outcome. One does, completely: a semigroup whose gaps pair off around the largest one is never lost for the player to move — none of 583 up to genus sixteen. The reason is strategy stealing, and it is the same reason the top-right square decides Chomp: every move from such a position closes the largest gap.
Using up the edges instead
Undirected geography is decided by a maximum matching when a move uses up the vertex it leaves. Use up the edge it crosses instead and the matching is exact on every tree — on a tree the two games are one game — and on nothing else. Over every connected graph on up to six vertices it names the winner at 480 of 745 starts once there is a cycle, it gets a four-cycle wrong from every start, and the more cycles a graph has, the more of its misses are wins that are really losses.
The pairing removes moves it cannot name
Symmetric positions were settled by an argument that names a winner and no move. Turned on the moves instead, the same one-pass test strikes off 27,215 of the 159,728 moves in the census and not one of the 21,234 winning ones — a quarter of a full search — and still names nothing. On 583 paired positions nine arithmetic descriptions of the winning gap reach at most 123, and 367 of those positions have exactly one winning move.
Two graphs a rule cannot tell apart
The repair proposed for the matching criterion was to read the cycle as well. Over every connected graph with exactly one cycle up to six vertices — 21 graphs, 114 starts — eleven such rules reach at most 91, and the winner is not a function of the matching, the cycle's length, the start's distance from it, its degree and the edge count together: six cells of that table hold both verdicts, the smallest a pair of five-edge graphs.
A shortlist with nothing at the top
The one-pass test leaves 9.58 moves of 12.27 and names none of them. Seven quantities a scan could compute about the survivors were turned into orders and scored on 583 positions: the best puts a winning move first on 24.9% against 22.3% by chance, and every one of the seven places the winner deeper in its order than chance does. Read as sieves instead, the gentlest keeps half the list and throws the only winner away on 264 positions.
The endgame theory arrives late
Every component the chain-and-loop theory names has coins of degree two, so a position it can read is one where every surviving coin holds exactly two strings. Over a six-box board that is 1,033 of the 28,028 positions with no free box on the table — 3.7 per cent — and more than half of them only after twelve of the board's seventeen strings have been cut.
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.
A coin with three strings is worth something
Every chain and every loop is worth nought on its own, whatever its size, and that is exactly what makes their nim-sum useless. A coin with three strings on it is worth nought, one, two or three depending on its arms — 31 of the 35 measured are not nought, and the four that are are the ones whose arms are all long. A coin with four strings is back to nought every time.
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.
One split is enough
A heap of n in Lasker's Nim offers ⌊n/2⌋ ways to split, and the values use at most one of them. Allow only the split that takes a single counter off and every heap to six hundred keeps its value; of all sixty-three sets of split sizes up to six, a set keeps the formula exactly when it contains 1 or 2. Equal halves alone give back plain Nim, because a split into equal parts is a move to nought.
Three heaps and a pass
Nim with a single pass that may not end the game is easy on one heap and on two: a heap swaps each odd size with the even one above it, and two heaps lose exactly at (2k − 1, 2k). On three heaps the losses are known only as a list. Fix the smallest heap and each slice of the list settles into a pattern after an irregular start — period 4, 8, 10, then 160 at a smallest heap of ten, and nothing visible from eleven.
Named alongside it
The objects these essays reach for when they reach for this one.
Exhaustive searchGrundy valueEnumerationNormal playCounterexampleNimInvariantNim-sumDisjunctive sumMexOutcome classSprague–Grundy