Concept

Information-theoretic bound — where it appears

A floor derived by counting how many distinct answers a procedure must be able to give, and therefore binding on every possible algorithm. It binds every algorithm and not every implementation, so a structure sitting far above one may be badly built or may be paying for something the count ignores.

Named by 9 essays across 3 fields — each of them below, with the objects they name alongside it.

bits per elementε = 0.13.3 → 4.8 (+1.5)ε = 0.035.1 → 7.3 (+2.2)ε = 0.016.6 → 9.6 (+2.9)ε = 0.0038.4 → 12.1 (+3.7)ε = 0.00110.0 → 14.4 (+4.4)ε = 1e-413.3 → 19.2 (+5.9)floor log₂(1/ε) filled; Bloom's log₂(1/ε)/ln 2 outlined44.3% above the floor at every rate

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.

floors · Floor
10⁴10010³n (elements)block transfersmeasured sortthe bound3.20×2.67×2.29×2.94×2.67×2.40×B = 32, M = 512 (M/B = 16)3.20× the floor at worst

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.

floors · Floor
thousands of bits · the floor is 31.9kthe text as it is34.0k streamto undo itnothinga shuffle agreed in advance34.4k streamto undo itnothingthe characters sorted0.3k streamto undo it31.8kthe Burrows–Wheeler transform14.8k streamto undo itnothingmodel: order 0 after move-to-front · Words from a fixed vocabularydashed: coding the text as it stands

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.

floors · Bits
bits per symbolhuffman3.9371.01× the floorarithmetic (adaptive)3.8981.00× the floorLZSS (window 4096)2.1130.54× the floorLZ783.1470.81× the floorH₀ = 3.89216,384 symbols · alphabet 21 · floor shown is H0 = 3.892 bits/symbolmodel: order 0 · Words from a fixed vocabularybest here: 2.113 bits/symbol

The bits a coder emits

A stream of 16,384 symbols with a zeroth-order entropy of 3.891 bits per symbol was coded by a Huffman coder into 3.937 and by an arithmetic coder into 3.898, and neither went under 3.891 because neither can. That floor is a third kind of limit, the first that is a property of a model rather than of a question, and the same stream has a different one under every model of it.

floors · Bits
011228101214161820universe size ubits of statea u-bit bitmaplog₂ C(u, u/2)⌈log₂(u+1)⌉, a counterat u = 12: 924 subsets, 8 bitstwo collide → answers 6 and 7floor computed exactly · collision found by exhaustion at u = 12floor 9.85 bits

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.

floors · Cardinality
1,00010³window length W, in arrivalsbits of state heldW bits, exactε = 0.2ε = 0.1ε = 0.05sliding-window model · 40,000 arrivals · state from the shape of the structure2,808 bits at W = 8,000

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.

floors · Window
0.010.11101001000QuicksortQuicksortQuicksortShellsortHeapsortMerge sortMerge sort with a cutoffInsertion sortBubble sortSelection sortn²time × space, as a multiple of n²n = 8,192 · light: auxiliary slots · dark: the input counted toonone under n²

A floor under a product

Every lower bound written here so far puts a floor under one quantity. The published time–space bound for sorting puts one under a product: T·S is Ω(n²), so buying time with space and space with time cannot both be done. Measured, heapsort's product is 0.0028 n² — four hundred times under a bound nothing can beat. The bound is not wrong. It is stated in a model where the input may not be written, and every sort here writes to the input, which is where its working memory actually is.

bounds · Bound
copy 1copy 8copy 16copy 24copy 32position within the copy32 copies × 2,048 charactersdivergence 0.0005

Sparse on average, dense where it gathers

The r-index keeps one suffix-array value for every run of its transform and prices the set as Elias–Fano: on 32 exact copies of 2,048 characters that is 666 values among 65,537 positions at 9.04 bits a value, as though they were spread evenly. They are not — 663 sit in one copy. Cut into chunks of 32 values, each held by whichever of a bitmap, its own Elias–Fano or a run is cheapest, the same set costs 4.48 bits a value, under the 8.05 that any representation of an arbitrary set of that size must spend on some set. At a divergence of 0.0005 the saving is a fifth, and from one character in five hundred it is nothing but the select directory, which the chunks replace with reading. On the scattered marks of an ordinary index the partition saves only that directory too.

space · Space
bitmap chunkElias–Fano chunka chunk of 32 beginscopy 1copy 8copy 16copy 24copy 32position within the copy32 copies × 2,048 charactersdivergence 0.0005

Cuts that follow the gathering

Cutting an r-index's run-start values into chunks of a fixed count left chunk edges inside the places the values gather. Choosing the cuts as a shortest path through the sorted values was predicted to save a further tenth on exact copies by cutting once at each edge of the gathering. It cuts there, and saves more than a quarter: two chunks, one of them a bitmap of 664 values that a select must scan from its start. Capped at the chunk's old size, the optimum keeps the reads and saves 9% on exact copies and 15% at a divergence of 0.0005. On moderately drifted copies the uncapped optimum is smaller and cheap to read at once.

space · Space

Named alongside it

The objects these essays reach for when they reach for this one.

Lower boundCounting argumentGuaranteeSpace lower boundBits per symbolCompressed bit vectorCost modelElias fanoEntropyHonest limitIndex sizePigeonhole

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