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

Expected case — where it appears

The mean cost over an algorithm's own randomness, which holds on every input and is the guarantee randomisation actually buys. It holds on every input because the randomness belongs to the algorithm, which is the difference between it and an average over inputs.

Named by 6 essays across 5 fields — each of them below, with the objects they name alongside it.

level 1level 2level 3level 4level 5key 21search: 8 comparisons, 3 hops26 keys, p = 0.5, seed 2026081057 coin flips decided the shape

A structure made of coin flips

Insert the same 512 keys into a skip list twice, once sorted and once shuffled, from the same seed, and the two structures are identical — the same 11 levels, the same height for every key, the same silhouette. Nothing about the data reached the layout. The 1,064 coin flips did all of it.

randomness · Randomness first-element pivot130,816 on sortedmedian of three5088 meanrandom pivot4937 meanblue bar: range over 400 random inputs · marker: the already-sorted inputn = 512a bad case that cannot be chosen is a different kind of bad case

What randomising the pivot buys

Quicksort taking the first element as its pivot costs 130,816 comparisons on an already sorted array of 512 — 27 times its cost on random data, and exactly the quadratic behaviour the algorithm exists to avoid. Randomising the pivot costs 5,490 on the same input. Randomisation does not make the bad case impossible; it makes it unchoosable.

wrong · Distribution quicksort on sorted input — comparisonsfirst-element pivot2,096,128random pivot25,318a search tree built from sorted keys — heightbinary search tree2,047treap26a hash table on keys chosen to collide — worst bucketthe low bits of the key2,048multiply–shift, a at random10n = 2,048, bars on a logarithmic scaledeterministic above, randomised below

Expected is not average

Quicksort on 2,048 sorted keys costs 2,096,128 comparisons with a first-element pivot and 25,318 with a random one. A binary search tree on the same keys is 2,047 deep; a treap is 26. A hash table on keys computed against its hash puts all 2,048 in one bucket; one drawn from a family puts at most 10 there. Three problems, one distinction.

bounds · Distribution 0%25%50%75%100%11.522.533.545what a matching character is worthreported as the shared region, of the longer sequencea random pair scores zero hereunit cost, alphabet acgtthe region found: 13 characters at the left, 34 at the right

The zero that moves the answer out of the corner

One extra term in the recurrence — a floor at zero — and the answer stops being in the last cell. It becomes a maximum over all 1,040 of them, the traceback's starting point is a search, and the whole mode is meaningless unless a randomly matched pair of characters scores negative on average. That last condition is on the scoring scheme, not on the sequences.

tables · Cost 024681,0244,09616,38465,536262,144keys, into as many bucketskeys in the busiest bucketone hashtwo hashes, take the emptierthree hasheslog n / log log nthe average load is 1 at every pointa lookup examines every choice, so two hashes is two probes

The second choice

Two hundred and sixty thousand keys into as many buckets. Under one hash the busiest bucket holds eight; under two, with each key going to whichever of its two is emptier, it holds four. The mean is exactly one in both. Nothing is rearranged afterwards, no key is ever moved, and the whole of the improvement is in a decision taken once, at the moment the key arrives.

randomness · Randomness trusts the estimateinsured ×2insured ×8always the index11.5235expected regret, logarithmic · label: worst within three standard deviationsrho 14.42.51.11.0rho 215.17.82.22.0rho 413.57.54.04.0rho 1616.016.016.016.0σ = 1.5, median error e^-1, 65,536 rowsexact over the error distribution

What insurance against an estimate costs

A planner that trusts its row estimate expects to pay 1.057 times the better plan and risks 7.76. One that insures itself by halving its estimate before it decides expects 1.057 and risks 4.10 — the insurance is free. At a read ratio of sixteen the same insurance costs six per cent in expectation and makes the worst case worse. Whether a conservative planner is paying a sensible premium depends on two numbers the planner can measure and usually does not — its device's read ratio and the direction its own errors run.

applied · Transfer

Named alongside it

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

DistributionAdversarial inputAverage caseComparison countGuaranteePivotPreconditionQuicksortRandomised algorithmRandomised data structureSkip listTrade off

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