Certificate — where it appears
Named by 35 essays across 6 fields — each of them below, with the objects they name alongside it.
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 theorem that names a winner and no move
Strategy stealing proves that the first player wins Hex and wins Chomp, on every board, in about four lines. It exhibits no move, contains nothing a move could be extracted from, and is not going to. The moves have to come from somewhere else, and where they come from runs out almost immediately.
A winning strategy that is a spanning tree
The Shannon switching game was sold in a box in 1960 and solved in 1964, and the solution is not an assertion that somebody wins. It is a property of the graph anybody can check, and the strategy falls straight out of it — whichever link the opponent cuts, take its partner in the other tree.
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.
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.
"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.
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.
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.
A rule with a guarantee
Evaluating a sum of a dozen fights is impossible; following a rule is not. Move where the stake is largest, and over 220 sums of three hot components the rule scores exactly what perfect play scores in 196 of them, is never more than one point behind, and never ends more than the largest single stake below the mean. The rule that is supposed to be different — answer the threat — chose differently in none of the 220.
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.
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 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.
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.
Wrong in one direction only
The rung below asked for the simplified comparison test the dead-ending hypothesis is supposed to license, and predicted it would agree with the quantifier on the dead-ending rulesets and not on Toads and Frogs. Written three ways and scored on 492 pairs, it agrees best on the ruleset that is not dead-ending — and never once refuses a comparison that holds, which makes it a sound filter and not a test.
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.
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.
The first move is a link that is not there
Lehman's criterion answers one question about a switching game — who wins when Short moves second. The other question has the same answer asked of a different graph: add one link from A to B, and Cut is forced to spend its first move deleting it. The trees of that larger graph then name Short's opening, and on every subgraph of seven graphs they name a winner.
Cut is Short on another graph
Everything proved about the switching game is proved from Short's side, and Cut appears only as the player whose moves get enumerated. On a graph drawn without crossings Cut does not need a theory of its own: deleting a link is securing the link that crosses it in the dual, so Cut's game is Short's game on a different graph. Bridg-It is the board that is its own dual — one link short of two trees at every size, which is why its first player wins.
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.
Where the needle has a sentence
Strategy stealing proves the first player wins every Chomp bar and names no square to take. On two families the square can be said in a sentence — a square bar and a bar two rows deep — and in both the sentence is a pairing that names every later move too. Three rows deep the needle wanders, and the observation that every bar has exactly one needle survives ninety-four rectangles and fails on the ninety-fifth.
A board one column wider
Strategy stealing proves the first player wins Hex, and it needs three things: no draws, an extra stone never hurting, and rules that treat the two players alike. Add one column to the board and the third goes. The player whose edges are now nearer together wins whoever moves first — and does it with a table of pairs that names every reply, checked against every line to a board of twenty cells.
Where a search may stop
A search deepened until two consecutive depths agree carries a proof of its verdict, and on 4 × 5 Domineering it stops before the longest line on 17,589 of 48,670 positions. It also costs three times what the search that simply finishes costs. The rule that pays is the other one. Search on wherever the two players' counts of placements are within one, and at depth 2 the wrong verdicts fall from 2,140 to 86 for about a quarter more work per search.
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.
A potential that names every move
Strategy stealing names no move, and the pairings that do name moves need a board with the right symmetry. The Erdős–Selfridge potential needs neither: Down, moving second in Hex, takes the empty cell through which Across's unfinished chains weigh most. Its guarantee reaches only boards two rows deep. It wins far past the guarantee — on every board of three rows that Down can win — and then, on a four-by-five board that a table of pairs wins for Down with certainty, it answers Across's first stone in a different cell and loses along the bottom edge.
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.
The winning reply is the fourth choice
The repair proposed for the potential was to weigh an edge chain more heavily. Fifty-five weightings later, none holds the four-by-five board, and an edge bonus costs Down four boards it was already holding. The reason is not the numbers: over 393,660 turns of the pairing that does hold that board, the potential would take the same cell 26.1% of the time, and the winning cell is its 3.7th choice on average and as low as its seventeenth.
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 move whose every reply is struck
Read two moves at a time, the Sylver Coinage shortlist is no better ordered than read one at a time: preferring the survivor that leaves the opponent the fewest surviving replies puts a winner first on 24.7% of 583 positions against 22.3% by chance, and places it deeper than chance does. Read as a proof, the same count does what no order could. On 57 positions a survivor leaves no surviving reply at all, and wins by a certificate a few lines long; searched deeper, the survivors prove every position by thirteen moves — at a price that is never below the search that simply finishes.
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.
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.
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.
Named alongside it
The objects these essays reach for when they reach for this one.
Exhaustive searchStrategyComplexityIntractableCounterexampleGrundy valuePeriodicityStrategy stealingImpartialNormal playOutcome classSearch cost