Concept

Partition — where it appears

Rearranging an array so that everything below a chosen value precedes everything above it, which is one linear pass and the heart of quicksort. It is one linear pass and the whole of quicksort's work, and its three-way form turns a stream of duplicates from the worst case into the best.

Named by 23 essays across 8 fields — each of them below, with the objects they name alongside it.

comparisons ÷ n log nMerge sort0.855Merge sort with a cutoff0.992Quicksort, random pivot1.018Quicksort, first-element1.082Quicksort, median of three1.117Shellsort1.246Heapsort1.649all of these fit n log n1.9× between best and worst

The constant the notation drops

Six algorithms on this site are Θ(n log n) and their comparison counts differ by a factor of two. The class is what they have in common; the constant is what distinguishes them. It is measurable to three digits, it is the number that actually decides between them, and it is precisely the number the classification is designed to discard.

bounds · Bound
313roundconc 0.14536hashedconc 1.00309blockedconc 0.14how the arrivals were partitionedworst error over the top keysone summary, k = 32one summary, k = 256the merge of 8Space-Saving · stationary Zipf · 40,000 arrivals8 shards

The partition the analysis did not mention

Space-Saving and Misra-Gries are the same structure under a stream, related by subtracting one number. Sharded eight ways and merged, one of them is wrong by 313 where the other is wrong by 927 — and swapping how the arrivals were assigned to machines reverses which is which.

streaming · Merge
10010³1010010³nstack framesa stack of 512 framespivot: first-elementpivot: median of threepivot: random pivotMerge sortalready sorted input, n from 64 to 4,096one frame charged as one slot

The stack nobody counts

Merge sort makes 8,192 calls to sort 4,096 elements and holds fourteen of them at once. Depth-first search on a grid holds twelve vertices, or sixty-six, or a hundred and forty-four, depending on which of three equally standard implementations is running. The stack is a resource, it is the one that fails hard rather than slowly, and nothing that watches the data can see it.

space · Space
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
comparisonsswapsInsertion sort63,071 / 0Selection sort130,816 / 504Bubble sort129,688 / 62,563Merge sort3,964 / 0Heapsort7,653 / 4,170Quicksort5,049 / 2,380n = 512, random inputcounted in the same run

One run, four counts, four answers

The question “how many operations” has no answer until the operation is named. Selection sort makes more comparisons than any other algorithm here and fewer writes than almost all of them; bubble sort matches its comparisons and does 124 times the swapping. The ranking depends entirely on which count is chosen, and the choice needs justifying.

counting · Count
dark: with galloping · pale: with the mode removednearly sorted33,95221,373 jumped−37.7%few distinct values57,91241,178 jumped−36.9%random95,7702 jumped+0.0%n = 8,192, MIN_GALLOP = 7comparisons, counted exactly

When galloping pays

Timsort's merge does not always take elements one at a time. When one run has won seven times in a row it switches to searching for how many to take at once, and switches back when that stops paying. The mode saves 22,104 comparisons on nearly sorted input, 33,270 on input with few distinct values, and costs exactly six on random input — which is the whole design in three numbers.

practice · Practice
peak slots held at once, logarithmic18645124096Insertion sort11Selection sort11Bubble sort11Heapsort11Shellsort11Quicksort, median of three14log nQuicksort, random pivot32log nQuicksort, first-element4,097nMerge sort with a cutoff4,106nMerge sort4,110nn = 4,096, already sorted inputone slot = one array element or one stack frame

In place is a claim, and it is usually wrong about quicksort

Heapsort holds one slot at its peak. Quicksort holds twenty-two at n = 4,096 on random input and 4,097 on a sorted one. Merge sort holds 4,110. All three are described with the same two words, one of the three descriptions is false, and the false one is the algorithm the phrase is most often attached to.

wrong · Space
algorithmspread of count ⁄ f(n) — 1.0 is exactTimsortPython, Java objects, Rust, Android1.176nIntrosortC++ std::sort1.104n log nPattern-defeating quicksortRust unstable sort, Boost, libstdc++ 141.155nDual-pivot quicksortJava Arrays.sort, primitives1.624no class grantedtolerance 1.6fitted n = 256–16,384comparisons, counted exactly

The pattern that defeats the pattern

Quicksort's bad cases are patterns — sorted input, organ-pipe input, an adversary's construction. Introsort's answer is to notice the damage and switch algorithms. pdqsort's answer is to notice the pattern and break it, deterministically, with four swaps. On input with eight distinct values that turns a quadratic disaster into a linear sort, and the whole difference is one extra partition scheme.

practice · Practice
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
10³10³10⁴10⁵10⁶10⁷ncomparisonsadversary, no depth limitadversary, as it shipsrandom inputn²/2McIlroy's adversary, answering as it goescomparisons, counted exactly

The depth limit that almost never fires

Introsort counts how deep its recursion has gone and calls heapsort if it passes twice the logarithm. On every input measured, the mechanism handles under a tenth of a per cent of the elements. Then an adversary that answers the comparisons rather than choosing the array drives it to exactly n²/2 with the limit removed, and to one heapsort call with it — a factor of forty-two at n = 8,192, and growing.

bounds · Bound
Dual-pivot116,836shipsIntrosort130,863shipspdqsort114,408shipsTimsort95,770shipsalgorithmcomparisonsrandom, n = 8,192comparisons, counted exactly

Two pivots and what they cost

Java changed its primitive sort in 2011 on the strength of an analysis showing dual-pivot quicksort does fewer comparisons than the classical one. It does. It also does nearly twice the swaps, and the analysis that decided the matter counted neither — it counted a weighted combination that had to be chosen before any conclusion could be drawn.

counting · Count
0231436845165232shards mergedkeys whose count movedthree orders:folded in one at a timecombined pairwise, in a treefolded in, last shard firstMisra-Gries, k = 32 · hashedworst gap 298 arrivals

What a fold charges per level

Thirty-two counter tables folded in a chain come out wrong by 323 where the same thirty-two combined pairwise are wrong by 148, and the quantile summaries prefer the chain by exactly as much in the other direction. What is being charged in each case is the depth of the fold, and the two families are charged on opposite ones.

bounds · Merge
shard 1 · 120 → 119shard 2 · 121 → 120shard 3 · 120 → 120shard 4 · 121 → 120shard 5 · 119 → 118shard 6 · 124 → 123shard 7 · 121 → 120shard 8 · 122 → 121floor, in arrivalsnaive: tail ÷ kfixed pointmeasuredstationary Zipf · round · k = 321.006× the measured floor

The floor a histogram already knows

A summary of thirty-two counters settles at a smallest counter of 119, and the number can be computed from the shard's key frequencies before a single counter is allocated. The obvious way to compute it is wrong by a factor of two, and the reason is that the heavy counters carry no error at all.

streaming · Merge
010020030040021φ 382φ 7203φ 17434φ 42885φ 1012486φ 242measured 409counts chargedlevel of the foldcharged at this levelrunning total64 shards · k = 32 · hashedcharged 409 · measured 409

The floor charged at every level

A key surviving a fold of sixty-four shards is charged 2, then 8, then 20, then 43, then 88, then 248 — the floor of whatever summary it was merged against, level by level. They sum to 409, and the damage read off the merged table is 409. The model that charged sixty-three copies of the leaf floor said 222.

floors · Floor
0.111010010³10³10⁴floor, in countsarrivals in the shard, nround-robin — n^1.02hashed — fit refusedresidual 2.7%slope 5.2 → 1.19k = 32 · 40,000 arrivalsthe table holds 1.33 of a hashed shard's keys and 0.01 of the stream's

A floor with two variables in it

Under round-robin a Space-Saving summary's floor is 0.0203·n^1.018 over a hundred-and-twenty-eight-fold range of shard size, worst residual 2.7%. Under hashing the same measurement has no exponent at all — the local slope runs from n^5.17 to n^1.19 — and a least-squares line through it reports n^1.73 at a 441% residual.

floors · Floor
1,00010,000248163264ε = 0.02, α = 0.58ε = 0.01, α = 0.56ε = 0.005, α = 0.54tuples keptshards mergedlog-normal, σ = 1.2 — a latency distribution · 20,000 valuesα 0.58 / 0.56 / 0.54 · worst residual 2.0%

The tuples a merge does not give back

A merge of thirty-two quantile summaries keeps seven times the tuples of one summary over the same values, and sixty-four keeps ten and a half. Fitted across the sweep the count goes as the shard number to the power 0.56, which answers what it converges to — it does not.

streaming · Rank
f = 6,3628/8 holdingf = 3,0748/8 holdingf = 1,9638/8 holdingf = 1,3738/8 holdingf = 1,0988/8 holdingf = 9353/8 holdingf = 7690/8 holdingf = 6560/8 holdingpredicted damage, in arrivals — every row totals 967Space-Saving's shareMisra-Gries's share8 shards · round · k = 32bill 967 arrivals

The bill a partition only divides

The two predicted damages for any key sum to the same number under every partition — 967 arrivals here, whatever the arrangement. Round-robin hands nearly all of it to Misra-Gries and hashing hands most of it to Space-Saving, and neither of them is paying more than the other in total.

structures · Merge
0.000.250.500.751.00stationarydepartingburstydriftingprediction ÷ measurement, as a factorshare of the top k that moves between halves8 shards · round · k = 327.1× out where the statistic reads 1.00

The histogram that cannot see the order

A prediction accurate to one per cent on three streams is seven times out on the fourth, and the input that fails is the one every capacity plan is built from. A statistic computed from the same input says in advance which case is in hand — and misses one of the two ways it can go wrong.

wrong · Merge
weighted path lengthΣ wᵢdᵢ — what Huffman minimisescuts takenΣ over the mergesdamageworst error leftchaintreesmallest-firstlargest-first543k147k123k738k309254259388323148183403Misra-Gries · 32 shards · hashedleast path smallest · least damage balanced

The fold that minimises the wrong thing

A fold charges per level and a survivor pays the cuts on its path, so the bill looks like a weighted external path length — and Huffman's construction minimises that quantity by proof. Built and measured on thirty-two uneven shards it does minimise it, 181,407 against a balanced tree's 200,000, and leaves more damage than the tree does.

structures · Merge
counter tablesworst error, ratio to bestquantile summariestuples kept, ratio to bestchain2.18× (323)1.10× (2,807)tree1.00× (148)1.26× (3,211)smallest-first1.24× (183)1.28× (3,278)largest-first2.72× (403)1.00× (2,556)32 shards · hashed · k = 32each column against its own best shape

The shape one structure will not fold

Folding thirty-two shards largest-pair-first keeps 2,556 quantile tuples against a balanced tree's 3,211 — a fifth of the space saved. The same fold on the counter tables beside them leaves 403 counts of error against the tree's 148. A deployment holding both cannot fold once and be right twice.

structures · Merge
01002003004005006007008009001000roundloads 1.0×blockedloads 1.0×hashedloads 17.6×worst error over the heaviest keyschaintreesmallest-firstlargest-first32 shards · k = 32 · 40,000 arrivalseven 1.00× · uneven 2.7×

A parameter that waits for another

Four merge fold shapes over thirty-two evenly loaded shards leave errors of 665, 667, 665 and 667 — a fifth of a per cent apart. Give the same four shapes shards whose loads span seventeen-fold and they leave 148, 183, 323 and 403. The parameter did nothing until a second parameter moved, and every measurement that fixed the second one saw nothing.

wrong · Merge
0.60.81.01.21.41.61.84φ share 0.908φ share 0.8516φ share 0.8032φ share 0.7564φ share 0.70predicted ÷ measuredshards, mlevel floors, uncorrectedleaf floorslevel floors, correctedk = 32 · hashed · 40,000 arrivalsworst 22% against 73% and 46%

The floor a merge does not settle at

Compute a fold's level floors from the shard histograms and the prediction over-shoots by 1.73. A merged summary's floor is not the floor a summary settles at on the same arrivals — it is 0.90 of it at four shards and 0.70 at sixty-four, straight in log₂ m at a 3% residual, because merging preserves the heavy counters and never runs their eviction cascade.

wrong · Merge
how far the top k movedthe order warninghow far the floors are from doublingthe regime warningstationary Zipf0.160.53 (54%)one key floods a stretch0.170.59 (54%)a heavy hitter that stops0.170.65 (52%)the popular keys drift0.9134.00 (0%)k = 32 · 40,000 arrivalsin brackets: the leaf model at sixty-four shards

The warning that is silent for the right reason

The statistic shipped to warn that a merge prediction is about to fail reads 0.160 on a stationary stream, 0.172 on a bursty one and 0.909 on a drifting one. It was asked to be looked at again because it does not catch a burst. It does not, and the reason is that on a burst there is nothing to catch.

wrong · Merge

Named alongside it

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

ShardGuaranteeSpace-savingHeavy hitterMerge treeMergeable summaryPivotHistogramMeasurementMisra–GriesQuicksortFixed point

All concepts