Search cost — where it appears
Named by 22 essays across 5 fields — each of them below, with the objects they name alongside it.
The table that changes its mind
The advice that ten entries chosen by use serve nine lookups in ten was untested: it describes a table sorted after the fact rather than a solver that only ever held ten. A solver that only ever held ten gets 94.2 per cent — beating the best ten chosen with the whole run in view, because there is no best ten.
The price of asking what the parts are
The third licence lets a solver look up a region rather than a position, and the rung below priced it by the entries it stores. Priced by the work it costs, it saves between a third and two thirds of the expansions and pays for them with a flood fill at every node — six times the total. A square would have to be ten times cheaper than a table probe before it broke even.
Two measures bounded, and one not
A sum is born no later than its parts' birthdays together, and it has no more options than they have between them — a bound nobody had checked, and it is attained. What runs away is the length of the written form: 27 pairs of 231 exceed it, the worst by 29 characters, on a sum with exactly as many options as it was entitled to.
A factor, and not an overhead
Deciding who wins a form searches the form's own tree. Reducing it to canonical form searches a difference game for every comparison, and a difference game is a sum. Over 256 forms the reduction expands 5.46 times as many positions — and the ratio runs from 0.58 at one option to 9.80 at eight.
Close calls nothing resolves
The same value panel that settles 118 of the 202 silent decisions settles four of the seventy-two the rules get wrong. Every rule that helps at all must replace connectivity rather than follow it, and the cheapest one breaks twenty-six decisions for every one it saves.
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.
A count that forgets
A Domineering solver with room for ten component values does better evicting whatever it used least recently than evicting whatever it used least often, and the explanation offered was that a use count never forgets. Halve every count at a fixed interval and the count overtakes recency at every table size — by less than half a point, and only with the right interval. The right interval grows with the table: a quarter of a game's worth of lookups at ten entries, five games' worth at forty.
One board, and recency still wins
A Domineering solver's table of component values did best evicting whatever it used least recently, and the explanation was that the run changed board size three times. Take the change away — play all 650 games on one board — and counting wins back its lead only on the smallest board. On 5 × 5, 6 × 6 and 7 × 7 recency still beats both counting and the best fixed table, by the most on the largest. The locality recency exploits is not between boards or between opening and endgame. It is inside a single move.
The order a solver tries the moves in
A memoised search asking who wins 4 × 5 Domineering expands 1,125 positions when it tries first the move that leaves the opponent fewest replies, and 30,202 when it tries losing moves first — the same answer at twenty-seven times the price. The ordering that already knows which moves win is not the cheapest. A win needs one move and a loss needs all of them, so the price of an order is paid one level down, in the replies it leaves.
A verdict that changes with the depth
A who-wins search of 4 × 5 Domineering cut at a fixed depth, guessing that the player with more placements wins where it stops, is right about 72.5 per cent of positions at depth 0 and about every one of them by depth 7. On the way, 4,697 positions are right at one depth and wrong at a deeper one. With a guess that knows nothing, going one move deeper makes the search worse — and its errors alternate in kind with the parity of the depth, so that half its verdicts are proofs.
A key shorter than the position
A who-wins table for 4 × 5 Domineering addressed by a 16-bit Zobrist key stores a wrong verdict in 59 runs of 60 and names the wrong winner of the empty board in 19. The pairs of positions sharing a key follow the birthday count exactly while addresses are scarce, and fall away to nothing once the key has more bits than the board has squares, because a Zobrist key is linear. Symmetry and value identify positions that really are the same; a short key identifies positions that differ, at a rate set by arithmetic.
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.
A check bit halves the average and not the key
Real transposition tables keep a few of a key's bits beside each verdict and trust an entry only when they match. On 4 × 5 Domineering each such check bit halves the average number of wrong verdicts, exactly as the birthday count says. It does not halve any one key's. A Zobrist key confuses positions in families — every pair that differs on one set of squares whose words cancel — and a bit removes a family whole or not at all, so from eighteen bits to nineteen thirty of fifty-eight keys lose every confusion and fourteen keep every one.
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.
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.
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.
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.
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.
Named alongside it
The objects these essays reach for when they reach for this one.
Exhaustive searchDomineeringMemoisationHeuristicCertificateNormal playTransposition tableComplexityDecompositionDisjunctive sumEnumerationDawson