Break-even — where it appears
Named by 10 essays across 5 fields — each of them below, with the objects they name alongside it.
An estimate borrowed from an easier problem
On a grid where every step costs one, the straight-line distance to the goal cuts a search from 543 cells to 325. On terrain where steps cost between one and nine it cuts 1,572 to 1,550, because it still believes every step costs one. Four exact distance tables, computed once, cut the same search to 252 — and cost 6,328 reads to build, so they pay for themselves on the fifth query.
The index that is the text
A suffix array sorts all 4,097 suffixes of a text — 8.4 million characters of string, in total — and examines exactly zero characters doing it. It then answers a search in 91 characters where a scan costs 1,472, and the whole thing pays for itself at six queries. Both halves of that are worth the same amount of attention, and the first is the one that is usually skipped.
What a planner pays to find out what to pay
Insurance against a row estimate is set from the error's median and spread, and a running system knows neither — it has to fit them from executed queries. Fitted from one query the divisor costs 1.263 times the better plan against 1.215 for a planner that never insures at all, so learning is worse than not learning until about sixteen queries have run. The tail, though, is bought immediately: one observation already holds the worst case to 24.7 against 36.6.
Where the landmarks stand
Four tables of exact distances, each from a chosen cell, turn a straight-line estimate that barely helps on rough terrain into one that cuts a search by a factor of six. Averaged over 1,200 queries on eight maps, the same four tables expand 126 cells a query when their cells are the map's corners and 423 when they are near its centre. The standard choice, each landmark as far as possible from the ones before, expands 141 and loses to the corners on all eight maps. Moving four landmarks to the right places buys more than doubling their number.
What the queries know that the map does not
A greedy rule that chooses landmark cells by rerunning a sample of past queries needs two hundred of them to draw level with a rule that reads only the map — and what it finally chooses, on map after map, is the four corners. Give the queries a destination instead of scattering them, and twenty are enough to beat the corners by 29% on eight maps out of eight. A query log is worth reading exactly to the extent that it is not uniform.
The price of remembering an answer
Give selection sort a table of the pairs it has already compared and it makes 16,805 comparisons on 256 elements instead of 32,640. The table pays for itself once a comparison costs more than 3.1 word operations, and that price stays near three at every size measured. Give it the full closure of everything its answers imply and it makes 13,079. The closure pays only past 425 operations a comparison at that size, and past 1,659 at twice the size. The cheap memory is worth having on long keys. The complete one is worth having almost nowhere.
How long a reweighting stays true
Johnson's one Bellman–Ford run costs under one per cent of an all-pairs computation because it is divided over every source. Asked one query at a time it is divided over nothing, and it still repays itself after 2.7 queries — because the preparation is one Bellman–Ford and every query saves a third of another. What decides the trade is not the query count but whether the graph holds still: at half a per cent of arcs redrawn between queries the stored potential is worth exactly nothing, and its life is geometric at a per-arc failure rate of 6.6%.
The runs a permutation does not leave
A run of length L costs 2⌊log₂ L⌋ + 1 bits and replaces L, so coding runs pays above a mean run of six. This grid's mean run is 2.34, chance gives 1.90, and coding its runs makes it 17.8% larger.
A potential mended where it broke
A stored reweighting on a 256-vertex graph with negative arcs costs 10,045 relaxations to rebuild, and rebuilding it every time an update breaks it stops paying once half a per cent of arcs change between queries. Mending it from the arcs that broke costs 16 to 442 relaxations instead, and the stored potential stays at two thirds of the per-query cost at every rate of change. When the change is a vertex whose costs all move together, a repair reaches nearly every vertex. It still costs a third of a rebuild.
The level where compression stops paying
Choosing the best coding for every level of the grid separately, rather than one for all twelve, saves 26 bits out of 47,668 — five hundredths of one per cent. The apparatus for choosing costs more than that to describe.
Named alongside it
The objects these essays reach for when they reach for this one.
PreprocessingShortest pathPotential functionHeuristic searchLandmarkSpace time tradeTriangle inequalityAdmissibilityAmortisationCompressionControlCost model