Counting argument — where it appears
Named by 9 essays across one field — each of them below, with the objects they name alongside it.
A floor on the bits
Answering membership for n keys with a false-positive rate of 1% and no false negatives requires at least 6.64 bits per key, whatever the structure. A Bloom filter uses 9.59. The gap is 44.27% at that rate and at every other rate, and it is the first bound on this site that a real structure comes close to.
A floor under a run count
A structure whose size is a function of the number of runs in a transform must give a different bit string to every text with that many runs, so it needs at least the logarithm of how many such texts there are. That count is walked rather than estimated — all four thousand and ninety-six of them — and the representation everybody uses turns out to have five bits of slack.
The floor under moving data
The information-theoretic floor for comparison sorting is log₂(n!) and it says nothing about a file on a disk. In the external model the floor is (n/B)·log_{M/B}(n/B), it is a bound on every algorithm rather than on merge sorts, and a measured external sort sits 2.40 to 2.97 times above it. Both numbers are computable, and the gap between them is what a real implementation costs.
What a reordering costs to undo
Sorting the characters of a text clusters them perfectly: a move-to-front pass then leaves 289 bits where the text's own floor is 31,931. Naming which arrangement of those characters the text was costs 31,827 bits, and the two numbers add to the floor it started from. The Burrows–Wheeler transform clusters less and costs nothing to undo, which is the only reason it is the one that is used.
The floor under a summary
An exact one-pass distinct-counter over a universe of u keys needs at least log2 of u-choose-u-over-2 bits of state — the same counting argument as the sorting floor, applied to memory states instead of outcomes. At u = 12 that is 9.85 bits, and an eight-bit candidate is shown to collide by running all 924 subsets.
The floor under a window
An exact count of the ones in the last W arrivals needs W bits, and the argument is a pigeonhole that can be performed rather than quoted — 1,024 windows, an eight-bit state, the colliding pair produced, and the two answers it cannot tell apart.
A floor one pass cannot get under
An exact one-pass selector must reach a different memory state for every prefix it might have read, and the pigeonhole that proves it is small enough to perform — nine hundred and twenty-four prefixes through a nine-bit state, the collision produced, the suffix that separates it, and two true medians it cannot both return. Ten bits collide on none, so the bound is exact — and a second pass walks under it by a factor of ninety.
Every pair must be asked
Ask whether a six-vertex graph is connected, one pair of vertices at a time, and the best possible algorithm needs all fifteen questions on its worst graph. The claim that this holds for every monotone property of graphs is a conjecture fifty years old. At four vertices it can be settled completely: all 2,046 properties that do not depend on vertex names need every pair. Name one vertex, and the count drops from ten to four.
The order inside a tie
Sort the rotations of a text by their first four characters rather than by everything that follows, and the output clusters slightly better than the full Burrows–Wheeler transform: 1.780 bits a symbol against 1.802, from 63% of the character reads. It also costs nothing to undo. The prediction that a shorter context leaves ties for the inverse to pay for was wrong. The cost to undo comes from how a tie is ordered, not from how long the context is, and at k = 0 the wrong tie rule is exactly the sort.
Named alongside it
The objects these essays reach for when they reach for this one.
Lower boundInformation-theoretic boundHonest limitPigeonholeSpace lower boundState bitsBits per symbolBurrows-wheeler transformEntropyExhaustive searchFalsificationGuarantee