Complexity — where it appears
Named by 42 essays across 6 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.
The game with the shortest rule is the hard one
Deciding a generalised board game is PSPACE-complete, which is a statement about families and encodings rather than about size. Nim in the same subject is settled by one pass over the input at any size, and green Hackenbush by one pass over the edges — while Domineering, whose rules take a single line, has no shortcut anybody has found.
A position reached eleven ways is one position
A 4×4 Domineering board has 5,700 positions in it and 6,257,129 routes through them. Three heaps of 7, 11 and 13 have 480 positions and 7.6 × 10¹⁶ routes. The gap between those two numbers is not an optimisation — it is the difference between a search that finishes and one that does not.
An outcome with no value behind it
Retrograde analysis labels positions in rounds, outward from the ones already lost. Whatever is still blank when nothing more can be deduced is a draw — and there is no separate test for a draw, because a draw is exactly the residue the method never reaches.
The board falls apart, and the arithmetic changes
A 4×5 Domineering board with a wall down the middle has 2,916 positions in it, and that number is exactly 54 × 54 — the product of its two halves. Solving the halves separately costs 108. Decomposition is the one saving in this subject that turns a product into a sum.
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.
What counts as the same position, and what that is worth
Folding a 4×4 Domineering board by its symmetries takes the table from 5,700 entries to 1,522 — a saving of 3.75, against a ceiling of exactly 4. An orbit cannot be larger than the group acting on it, so this is the one saving in the subject that can never change an exponent.
A puzzle asks once, a game asks alternately
Quantifier alternation is the whole difference between a puzzle and a game. One chooser is an existential and its answer is a witness somebody can check; two choosers taking turns is a prefix of alternating quantifiers, and the witness stops being an assignment and becomes a strategy.
Comparing two positions means playing a third
There is no way to look at two games and see which is better. The question "is G at least H?" is answered by building G − H and asking who wins it — so the most basic operation in the theory is a decision problem, and every canonical form is built out of them.
Nim is easy, in binary
Three heaps of a thousand counters take thirty bits to write down and three thousand counters to lay out. The nim-sum does three exclusive-ors either way. Whether that counts as fast depends entirely on which of the two numbers the work is compared against.
"Left wins" has no short proof
A complete solution of Nim on heaps of 7, 11 and 13 is 480 table entries. A winning strategy for the same position — one move of the winner's at each of their turns, and an answer to every reply — has 56,167,022 nodes in it. The answer is smaller than the proof by a factor of a hundred thousand.
A conjecture from hand play
Sprouts was invented over tea and its outcome pattern was guessed from games played with a pencil. Computers have checked it far past where a person could go, and this site's own solver gives out at three spots — so the honest figure states the frontier it reaches rather than the number somebody else published.
The class is named after memory, and that is not an accident
A 4×4 Domineering board has 6,257,129 routes through it, 5,700 distinct positions, and a deepest line eight moves long. Those three numbers are three different resources, and the smallest of them is the one that gives games their complexity class.
Three different claims are all called solved
Hex is solved in the sense that the first player provably wins, by an argument that names no move whatever. Nim is solved in the sense that a formula gives the right move from any position at any size. Between them sit strategies for one opening, and databases of a few billion positions. The word covers all four.
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.
A rule that is never right and cannot be far wrong
Playing the hottest component is not optimal — over 440 measured lines it costs something on 17 of them. What makes it worth having is that the cost is bounded by the temperature, provably, and that the same test run with the ordering reversed breaks the bound on 54.
The cost is in the closure, not in the positions
Under normal play, Dawson's chess needs four classes for every heap up to twelve, because its Grundy values stay at three or below there. Under misère play the same game needs six, then twelve, and the number rises with the universe rather than with the position — which is a different kind of expense entirely.
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.
Finding the parts
Decomposition turns a product into a sum and is the largest saving in the subject. Nobody labels the regions. The pass that finds them costs the same on every board of a size — including the boards where there is nothing to find — and what it buys ranges from four orders of magnitude to nothing at all.
Knowing who wins, and knowing what it is worth
Deciding a winner expands positions. Computing a canonical form expands pairs of positions, because a comparison unfolds as a recursion over one subposition of each and the reduction makes many comparisons. Measured on the same nine positions by an evaluator that starts empty every time, the second costs between 1.3 and 279 times the first, and the ratio grows with the tree.
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 question in the middle
Between knowing who wins each part and knowing what each part is worth sits the question a player actually has: who wins the board. Priced on sums of two it lands between the other two on seven of eight, cheaper than the values by up to eight times. On sums of three, with nothing repeated, it is dearer than the values on five of six — because a component multiplies a search and only adds to a value.
How often a board falls apart
A decomposition turns a product into a sum, so a solver wants to know how often one arrives. Over every position of a 4 × 4 Domineering board the answer is 47 per cent — nought for the first two moves, three fifths in the middle, and nought again at the end. What one decomposition is worth is the other half of the answer and it is a factor of 1.8.
When the catalogue starts paying
The rung below priced two questions — who wins one board, and what it is worth — and named the third: a program pays for a family of regions once and answers every board over them by addition. The crossover is between five boards and two hundred, depending on how far the catalogue reaches, and it falls as the board grows. The whole catalogue of every region to eight squares costs one part in seventy-six of one undecomposed five-by-five board.
Where to stop building
The rung below priced a catalogue of small regions against the search it replaces and found the crossover. What it could not say is how far to build, and the coverage answers that: going from four squares of reach to ten multiplies the catalogue by 860 and lifts the share of regions it answers from 54 per cent to 74. The price of a point of coverage runs from five shapes to five thousand.
A wall an amazon can walk through
An arrow burns a square for good, so an Amazons board that has fallen into pieces should stay in pieces. Over 127,583 positions it does not: fifty-one thousand moves put two regions back together. Every one of them is a single diagonal step, and what is wrong is not the game but the rule used to find the regions — which was borrowed from a game whose pieces lie along the board's own lines.
What it costs to notice a repetition
Folding a 4 × 4 Domineering board by its symmetries takes the table from 5,700 entries to 1,522. It also spends 559,424 square-mappings to work out where each entry goes — seventeen and a half times the entire cost of not folding. The saving has a ceiling of four and the price has no ceiling at all, and knowing which currency each is paid in is the difference between an optimisation and a habit.
What computing further has bought
Sprouts has been searched harder and longer than almost any game, and the period-six pattern has survived every extension. This site's own exhaustive search settles three spots; the published results reach forty-seven, and the gap is not a gap in hardware — the gentler of the two measured growth factors puts forty-seven spots at ten to the hundred and twenty-fifth positions. Beside it sits Brussels Sprouts, which has five million positions holding a choice and not one choice that changes who wins.
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.
A point with three neighbours
The switching game on links is settled by counting — enough links, arranged as two trees. Played on points instead, it is the game Hex belongs to, and the count is gone. The link game turns out to be the point game in which every contested point has exactly two neighbours; give one a third, and two graphs with the same points, the same links and the same number of separate routes can have opposite winners.
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.
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.
The cheap fights make the rule cheaper
A conjecture stands that playing the hottest part costs at most the coolest temperature times the number of parts sharing it — proposed on a range where that number never exceeds two. Swept to five-part boards over 10,410 lines it is false, and false the other way round: every line costing more than the coolest part has one or two parts at that temperature, and over the 3,230 lines with three or more, not one does.
Twelve turns, and three different prices
The earlier essay prices a universal quantifier at a doubling and leaves it there. Twelve turns with six of them the opponent's cost 6, 63 or 384 decisions to write down, depending on nothing but the order the turns come in — and the cheap arrangements are cheap for only one of the two players. What a claim costs is the number of times the choosing changes hands.
Eleven moves and one decision
A prefix has one quantifier a turn, so a game of eleven moves is eleven alternations. Counted on the boards themselves, a Toads and Frogs strip of eleven moves has twenty-six turns with exactly one move available and one turn anywhere at which the choice changes the answer; a Clobber board has a hundred and fourteen turns and none. Nim, the game everybody calls solved, decides at four turns in five.
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 opponent stops choosing
Replace one player by a rule with no search in it and the question has one chooser left, which is a puzzle rather than a game. Nim recovers five of its six lost positions that way, and six of seven on three heaps of five. Domineering recovers six of a hundred and twenty-two while the fixed rule throws away a winning move eighty-eight times, and one Clobber board recovers none at all — because on that board no rule can misplay.
A turn is not a bit
The prefix a game is read as gives each player one quantifier a turn, and a turn on a board is a choice among however many moves there are. Nim on heaps of 3, 4 and 5 lasts twelve moves and carries 23.6 bits of choice; a Toads and Frogs strip lasts eleven and carries two. Corrected for that, the model predicts a strategy 539 times too large on one board and 67 times too small on another, and the two failures have different causes.
A misère sum is searched, not added
Under normal play the outcome of a sum of heaps is a nim-sum of numbers already known: twenty stored values decide every sum of Dawson's chess with heaps up to nine, however many heaps it has. Under misère play each sum is a new position to search. One outcome costs six positions for a single heap, two hundred for four heaps and over five thousand for eight, and a table of every eight-heap outcome costs a hundred thousand. The misère quotient is the only thing that brings the price back down.
Two heaps of testing are enough
A misère quotient is computed by testing positions against positions, and the universe used to find twelve classes of Dawson's chess was every position of up to four heaps tested against every other — 511,225 outcomes. Varied one size at a time, the count stops growing at tests of two heaps and positions of three: 12,100 outcomes find the same twelve classes. The narrower universe the earlier essay drew did not merge anything; it held fewer positions. And the corner that is enough moves: for Kayles at heap twelve, two-heap tests miss a class.
Twelve classes, seven questions
Twelve misère classes of Dawson's chess were found by testing 715 positions against 715 others. Seven of those tests are enough to tell every class from every other — a greedy choice against a floor of four, since each test is one yes-or-no question. Kayles needs nine of 715 and Nim sixteen. The seven cost almost nothing to use and cannot be found without the whole closure, and they do not carry: the tests found with heaps up to seven tell apart only seven of the twelve classes with heaps up to nine.
A staircase, not a slope
With the misère closure cut forty-fold, Dawson's chess can be classified at heaps far beyond nine. The count of classes is a staircase: six from heap three to eight, twelve from nine to twelve, seventeen from thirteen to sixteen. Normal play steps once in that range, from four to eight at heap thirteen, where a Grundy value of four first appears. Misère play steps there too, and once more at heap nine, where normal play does not move at all — the first wild heap. Heaps eleven, fifteen and sixteen are also wild and move nothing.
Named alongside it
The objects these essays reach for when they reach for this one.
Exhaustive searchDomineeringIntractableCertificateNimMemoisationDecompositionDisjunctive sumPSPACEGrundy valueExact evaluationPosition graph