Algorithms & Data Structures
the bound is a claim, and it is checkable
Concept

Derandomisation — where it appears

Replacing an algorithm's coin flips with a deterministic rule, which removes the bad-luck case and usually costs a constant factor. It removes the bad-luck case and usually costs a constant factor, which is a measurable price rather than a theoretical one.

Named by 4 essays across 3 fields — each of them below, with the objects they name alongside it.

10³10⁴10³10⁴10⁵nrandom bitsskip list, one build — ntreap, one build — nreservoir, Algorithm R — n log nn from 256 to 16,384bits charged including rejections

Counting the coin flips

A skip list spends 2.03 random bits per key and a treap spends exactly 32. Reservoir sampling spends 1,356,399 bits on a stream of 65,536 and a better version spends 9,380. None of those numbers appears in any complexity class any of these structures is described by, and none of the site's other three counters can see them.

counting · Count randomised pivot164 random bits2.96 ±23%median of mediansno random bits7.94 ±2%0.04.38.6comparisons per element — bar is the range over seeds, tick is the mean30 seeds, n = 4,001, median2.68× the mean, 14× the spread

What derandomising costs

Randomised selection finds the median of twenty thousand elements in 3.21 comparisons per element and median-of-medians takes 8.15 — two and a half times as many for the same answer, both linear. The number that decides between them is not either of those. It is that the first varies by 28% from seed to seed and the second by 1.6%.

bounds · Distribution independent hashestwo values, h₁ + i·h₂2 choices, load 2+60,41859,9942 choices, load 3+2,2832,3672 choices, load 4+123 choices, load 2+46,45646,2433 choices, load 3+138145262,144 keys and buckets, seededbar length is log(1 + count)

Choices that are not independent

The power of two choices is analysed for choices drawn independently, and computing four independent hashes per key costs four hash evaluations. Compute two and take the choices to be h₁, h₁ + h₂, h₁ + 2h₂ and h₁ + 3h₂, and the choices are about as far from independent as they could be. On a million keys the buckets holding two or more come to 147,536 against 147,367 for four independent hashes, and the busiest bucket holds three either way.

randomness · Randomness 3579111311.523510positions per key, krate ÷ the independent rateh₁ + i·h₂h₁ + i·h₂, step oddh₁ + i·h₂ + (i³ − i)/6optimal load m·ln 2 / k; one set of hash functions per schemedashed: the account

Two hash values and the keys they copy

A Bloom filter that makes its k bit positions from two hash values, as h₁ + i·h₂, answers yes to 1.6% of absent keys on a 64-bit filter where k independent hashes answer 0.69%. The penalty is not the one expected. With the step forced odd no key ever repeats a bit, while a quarter of independent keys do. What costs the filter is a query whose start and step reproduce a stored key's whole progression, which happens with probability 4n/m², measured to within a few per cent from 64 bits to 4,096. The penalty fades as the filter grows and returns as the hash count rises — 1.13 times at seven positions on 1,024 bits, 5.35 times at thirteen.

randomness · Randomness

Named alongside it

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

Random bitsHash familyHash functionIndependence assumptionk-wise independenceMeasured countSkip listBloom filterBucket loadCacheClosed formComparison count

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