Closed form — where it appears
Named by 41 essays across 9 fields — each of them below, with the objects they name alongside it.
The theorem that needed none of the theory
Bouton solved Nim completely in 1901, with an argument that mentions no value, no sum of games and no Grundy number, because none of the three existed. The argument is two closure properties and it is airtight — and run on any other game it fails at the step that does the work.
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.
A chess problem that turned out to be an octal game
Dawson posed it in 1934 as a puzzle about pawns. It is the octal game ·137, its Grundy sequence is eventually periodic with period 34 from heap 52 — and the word doing the work in that sentence is eventually, because five values below the start disagree with their repeats and always will.
The sequence nobody has settled
Guy and Smith surveyed the octal games by hand in 1956 and conjectured that every finite one is eventually periodic. Seventy years and a great deal more arithmetic later, some of them have settled and some have not — and the evidence for the conjecture is entirely that nobody has found a counterexample they were looking for.
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.
The digits say which move wins
Wythoff's cold positions are usually given as a pair of golden-ratio formulas. Written in Fibonacci base they are a statement about digits instead — the smaller heap ends in an even number of zeros and the larger is the same numeral shifted up a place — and a rule about digits answers a question about a heap of a trillion.
The recursion this site cannot run
Remove the stopping condition from the construction and it reaches ω, its reciprocal, and one third — none of which this site's evaluator can represent, because it interns a position from a finite list of options. The figure draws what it computes and names what it cannot, which is where the boundary belongs.
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.
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.
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.
Nothing worth fighting over
Shove is a strip of coins beside a cliff, and both players have completely different moves. Every one of its 728 positions is worth a number, so nobody ever wants to move; the winner is the owner of the coin furthest from the cliff, in all 728; and the number the board is worth is not the sum of its coins — that reading is exact on 126 strips and wrong on 588 of the other 602.
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.
A tree is still a number
A Hackenbush string spells its own value in binary. Put a fork in it and the numeral has nothing to read — there is no leftmost anything. The value is still a number, in all 10,066 forests up to six edges; it is still computable, by the ordinal sum, in all 3,238 single-trunk trees; and the reading is right on 762 of them, of which 126 are the strings it was written for.
The other way to move a row
Shove has a cliff and Push has a wall, and that is the whole of the difference. Both games make every one of the 728 strips up to six squares a number, so neither ever has anything worth fighting over — and the two rules do not agree on the value of a single position. The obvious board-reading is exact on 446 strips under the wall and on 140 under the cliff, and 486 strips contain a coin its owner cannot move at all.
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.
A sequence with a rule and no period
The values of the subtraction game where Left takes one or two and Right takes one or three never repeat — thirty-one heaps, thirty-one different values. They are nonetheless completely described: three seeds and the rule v(k + 3) = {0 | v(k)} generate every one of them, which is what a pattern without a period looks like.
The reading that survives too much
Counting the empty squares in front of each coin gets a Push position right half the time, and the rung below said the failures were exactly the positions with two coins of opposite colour side by side. Sixty-six of the 1,072 failures have no such pair, the smallest is five squares long, and the condition that does decide it is not about the board at all — it is about every position the board can reach.
A code that climbs by three
Five hexadecimal codes were known to repeat with a constant added, and every one of the five constants was a power of two — either a fact about exclusive-or or a coincidence over five cases. Sweeping all 255 two-digit codes settles it: seventy-one climb, seventy of them by 1, 2, 4 or 16, and one by three. The exception is ·3f, whose values are 3⌊n/6⌋ + (n mod 3) on every heap to twelve hundred.
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.
One domino every three cells
The rung below gave the optimistic packing count as a formula in odd runs and asked for the other end of the interval, expecting a formula in the even ones. Parity is the wrong arithmetic: the smallest maximal packing is a sum of ⌈(len−1)/3⌉ over the runs, exact on all 1,042 shapes. That makes the whole interval readable off a drawing — and shows it can never reach the value, because regions with the same runs have different values.
The same number in two currencies
The rung below found bent-walled values falling strictly inside the translation bound and asked how far. The shortfall is the value's own hottest follow-up's temperature — exactly, on 400 of 408 pairs, and twice it on the other eight — which makes the whole error one expression. And it is the switches ladder's constant: a half there and a whole here, because a temperature is half a stop gap.
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 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 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.
Where the value stops mattering
Fourteen straight-walled pairs missed the bound on a translated stop and had only a threshold to explain them. Their stops move by exactly the addend's temperature — a formula with the value nowhere in it — which turns the threshold into the boundary between two lines and closes a census of 1,440 pairs that has been open for four rungs.
A set with three descriptions, and a function with none
Wythoff's cold positions can be written three ways that share no arithmetic — an irrational constant, a greedy rule, a condition on Fibonacci digits — and all three are exact. The same game's Grundy values have no closed form at all. Both facts are about one table, and the gap between them is the subject.
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.
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.
Where the numeral stops
A Hackenbush string is a numeral and a tree is a trunk with a forest on it, so the obvious next question is a graph with a cycle in it. Green Hackenbush answers that by fusing the cycle to a point. In blue and red the fusion is right on every three-edge cycle, on fewer than half of the six-edge ones, and the smallest thing it gets wrong has four edges.
The residues as a sequence
Four of the eight residue sequences never repeat, and a recurrence is not a description. There is a closed form and it is not for the game: the stops and the temperature of the n-th residue are periodic with period one, two or four on every sequence in the pool, while three of them produce a different game at every n.
Three bits of rule
An octal code is three bits a digit. The Grundy sequence it determines costs anywhere from one bit to a hundred and thirty-six — a factor of two hundred and seventy-two across rules that differ by a single digit — or it cannot be written down at all. Of four properties of the rule table tested against that, exactly one holds on every code that never settles: whether a move may leave two non-empty heaps. It is necessary, it is not sufficient, and nine codes carry it and produce answers smaller than their own rules.
What the arithmetic cost in 1956
The rung below ends by respecting a hand computation without pricing it. Priced in the operations a person actually performs, ·137's certificate is 7,919 of them — and the same sweep says ·47's is sixty-three times that, that a splitting move is what makes the cost quadratic, and that seventeen of sixty-four codes have no certificate at any price.
Three complete solutions in nine years
Bouton in 1901, Wythoff in 1907, Moore in 1910 — three airtight solutions of three games, all published before there was any theory of games at all. Asked about each other's games they all fail, and two of them fail by being wrong while one fails by having no form for the question. Only the last kind of failure decides anything.
A set with a short description
Bouton's argument is a closure argument about a set, and every impartial game has such a set — its own losing positions. So the method is complete and proves nothing. What made 1901 a theorem is that his set had a description shorter than the game, and swept over fifty-six subtraction games, exactly seven have one of his kind.
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.
The formula is a limit
Cap the take in Lasker's Nim at k counters and the game is a finite rule table, 4.33…3, whose Grundy sequence repeats with period k + 1 rounded up to even and follows Lasker's formula until the cap bites. The formula is what those periods converge to. And the same column of codes, with a free split in front, holds Kayles itself: the rule 4.4 on a heap of n + 1 is Kayles on a row of n.
Named alongside it
The objects these essays reach for when they reach for this one.
Exhaustive searchGrundy valueEnumerationPeriodicityMexOctal gamePartizanImpartialSubtraction gameEventual periodicityInvariantNim-sum