Algorithms & Data Structures
the bound is a claim, and it is checkable
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The thread: Measured, not assumed — page 2

Page 2 of 9, continuing the same thread in the same order.
acgtacgtseen ->0313303113033130a gap of k characters costs 2krows: expected · columns: seen · unit: bitslinear gaps When the algorithm is a table

A cost that is not one

The same eighty-one cells, filled by the same recurrence, return 6, 10, 10 and 15 — in edits, in cost, in bits and in bits again. Only the first is a count of anything, two of them are equal by arithmetic coincidence, and the alignment each one chooses is different.

 01325leading-zero rank keptregister, 0 to 255estimate 7,107truth 7,368error -3.54%0 still empty256 registers × 5 bits = 1,280 bitspredicted ±6.50% One pass, and no room

A count read off the leading zeros

Hash every key and watch for the longest run of leading zeros. Seeing k of them is evidence of about two to the k distinct keys — an estimator with a variance so large it is worthless, and the two devices that fix it are the whole of what a cardinality sketch is.

 46810120.010.1bits per element (m / n)false-positive ratemeasured(1 − e^(−kn/m))^kfrom the bits set60,000 absent-key queries per point, seed 80800 false negatives at every size When the algorithm flips a coin

A filter that is allowed to be wrong

A Bloom filter holding four thousand keys in five thousand bytes answers membership in four memory probes and gets 1.14% of its negative answers wrong. It never gets a positive one wrong. That asymmetry is the whole design, and the rate it makes errors at is a third quantity beside the operation count and the space.

 10³10⁴10³10⁴10⁵ncomparisonsbottom-up (Floyd)repeated insertionn from 128 to 65,536, random input, seeded1.21× between the two at the right-hand edge Structures

Building a heap from the bottom

Bottom-up heap construction is Θ(n) and repeated insertion is Θ(n log n), and the second of those is a worst case quoted as a behaviour. On random input, repeated insertion measures linear too — 2.22 comparisons per element against 1.87 — and the famous logarithmic factor never appears. On ascending input it appears in full, and it is a factor of six.

 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 What is taught wrongly

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.

 with the edge — connected01234567all 8 edges presentwithout it — two components01234567one edge withheld, everything else identicalV = 8, E = 8Ω(E), by adversary rather than by counting The floors

The adversary who hides the edge

The floor under comparison sorting comes from counting outputs — n! of them, so log₂(n!) comparisons. Connectivity has two outputs, so the same argument gives a floor of one comparison, which is useless. A different kind of argument gives Ω(E), and having both on the site is the point: lower bounds are not one technique.

 10³10⁴11010010³npointer hops per findneitherpath compressionunion by rankboth3n random unions, seededα(n) ≤ 4 for every n on this axis Two parameters

The constant that is practically constant

Everywhere else on this site the class is honest and the constant is hiding something. Union–find is the exact inverse — its bound is formally not constant, its growth term reaches four at n = 2,048 and stays there for every input anyone will ever run, and the measured path length is flat at 0.92 pointer hops across a two-hundred-fold range of n.

 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 What the libraries do

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.

 481632641281010010³10⁴noperations (mean of 60 runs)Insertion trafficMerge trafficInsertion cmpMerge cmptraffic crossessolid: reads + writes · dashed: comparisonstraffic crosses between n = 12 and 16; comparisons never do What the machine does

Where insertion sort actually wins

Every production sorting routine falls back to insertion sort on small subarrays, and the usual explanation is that below some threshold it does fewer comparisons. Measured, it does not — not at sixteen elements, not at eight, not at four. The crossover is real and it is entirely in memory traffic, which is a distinction the usual telling loses.

 01234567891011120123456789101112123456789101112123456789101112345678910123456789123456781234567123456123451234123121one unit = one subproblem given a value91 cells, 364 transitions, 4.0 per cell When the algorithm is a table

The cells are not the cost

This field opened by pricing a dynamic program in subproblems — 29,737 calls became 56 cells and the class changed. That is right when a cell is cheap. A table over intervals has 8,385 cells and considers 349,504 transitions to fill them, and the cubic in its bound is inside the cell rather than in the table.

 10010³10³10⁴10⁵10⁶10⁷Vcounted workTopological order, one passDijkstra, binary heapBellman–Ford, all passesV from 64 to 2048, directed, acyclicwork = scans + visits + relaxations + queue comparisons Two parameters

The precondition that removes the queue

Dijkstra maintains a priority queue to discover which vertex is safe to finalise next, and on a directed acyclic graph 65% of its counted work goes into that queue. The order it is discovering is already known. Relaxing in topological order makes exactly one relaxation per arc — 1,536 arcs, 1,536 relaxations — with no queue at all, and negative weights are fine.

 adjacency scansrelaxationsqueue comparisonsvisitsKosaraju, two passes5,632Tarjan, one pass2,560V = 1024, E = 1,536, directed, components plantedevery segment counted exactly Two parameters

Two passes or one, and what the second one costs

Kosaraju's algorithm and Tarjan's find the same strongly connected components of the same graph, in the same class, and one of them examines three times as many arcs as the other. The extra pass everybody counts is not where the difference is — building the reversed graph is, and no statement of "two depth-first passes" mentions it.

 No estimate543 cells expanded · path 58Straight-line estimate325 cells expanded · path 58Estimate doubled71 cells expanded · path 64V = 900, E = 895, every edge costs onethe estimate is a function of the vertex, supplied by the caller Two parameters

The precondition on a function the caller writes

Dijkstra expands 1,582 cells to find a path of 98 across a fifty-square grid. The same loop, with the straight-line distance to the goal added to each key, expands 405 and finds the same 98. The estimate has to be a function the caller supplies, and the guarantee holds only while that function never overestimates — a condition on somebody else's code, not on the graph.

 average 8.001,0242,048bucket, 0 to 255keys in the bucketthe low bits · keys built for this hashworst bucket 2,048 against 8.0 When the algorithm flips a coin

A hash is a family, not a function

Two thousand and forty-eight keys into two hundred and fifty-six buckets. Under a hash that takes the low bits of the key, all 2,048 land in bucket zero and 255 buckets are empty. Under a multiplier drawn at random, the worst bucket holds 11. The keys are the same keys, and they are the multiples of the table size.

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

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.

 10⁵10⁶10⁴10⁵comparisonsmispredictions (modelled)Insertion 0%Selection 1%Bubble 29%Merge sort 52%Heapsort 27%Quicksort, first 24%Quicksort, median-3 39%Quicksort, random 27%Shellsort 51%Merge + cutoff 40%Timsort 41%Introsort 32%pdqsort 43%Dual-pivot 37%2-bit counters, no historysquares are the sorts that ship What the machine does

The branch the machine guesses

Insertion sort does 176 times as many comparisons as Timsort at n = 8,192 and mispredicts a sixth as many branches. Merge sort's inner test is a coin flip and misses 51.5% of the time; selection sort's misses 0.6%. A processor does not wait to learn the answer to a comparison — it guesses, and throws away the work when it guessed wrong — and this is the fifth quantity this site counts.

 exact-11.2%-3.3%0.0%3.3%11.2%rmse 3.61%worst 9.71%23 of 60outside the band5,120 bits · 60 seeds · relative error of one runpredicted ±3.25% One pass, and no room

The correction that makes it work

HyperLogLog and LogLog read the same registers and differ only in how they average them. The harmonic mean is worth 30% of the error for nothing, and below two and a half registers' worth of keys the estimator both are built on is 137% high and has to be abandoned.

 the floor that applieslog₂(n!)Merge sort2.3×1,682Shellsort2.3×1,714Merge sort + cutoff2.6×1,938Heapsort4.2×3,101Quicksort, first6.6×4,887Quicksort, random7.2×5,291Quicksort, median-37.8×5,728Insertion sort18.5×13,644Bubble sort43.1×31,820Selection sort44.2×32,6408 distinct values, n = 256, seededthe two floors are 2.28× apart The floors

The floor when the values repeat

log₂(n!) counts orderings of distinguishable things. Two hundred and fifty-six values drawn from eight distinct ones have 1,684 bits of permutation entropy and 739 bits of distinguishability, so the real floor is less than half the one every table quotes — and merge sort, which sits exactly on the quoted floor, is 2.3 times above the one that applies.

 46810120.010.1bits per element (m / n)false-positive ratemeasured(1 − e^(−kn/m))^kfrom the bits set60,000 absent-key queries per point, seed 80800 false negatives at every size What is taught wrongly

The formula everybody sizes filters with

Fill a Bloom filter with four thousand random keys and its measured false-positive rate is within 4% of the textbook formula. Fill the same filter with the integers 1 to 4,000 and the rate is 30% worse than the formula says — not because the hash is bad, but because it is too good on that input.

 root, priority 0.9651.00.0prioritykey24 keys, sorted insertion, seed 20260810height 8 against an ideal of 4 Structures

The priority nobody supplied

Insert 4,096 sorted keys into a binary search tree and it reaches height 4,095, costing 8,386,560 comparisons to build. Give every key a second, random key and keep the tree heap-ordered on that instead, and the same insertion reaches height 26 for 32,750 comparisons. Nothing detected the imbalance, and nothing rebalanced.

 8 patterns of 10 characters over four symbols02,0004,0006,0008,00010,00012,00014,00016,00018,000one cell = 10 positions · shade = fraction read61.6% of the text The data that is not a number

The shift a set of patterns allows

Aho-Corasick reads every character of the text exactly once, whatever the number of patterns. Commentz-Walter reads backwards inside a window and steps over what it can, and on eight patterns of ten characters it looks at sixty-two per cent of a twenty-thousand-character text. The rule that does the skipping is not the one everybody implements.

 cost per column of the alignmentab / ba1.00 over 2 · 0.67 over 3kitten / sitting0.43 over 7 · 0.43 over 7intention / execution0.56 over 9 · 0.50 over 10gattaca / gactata0.29 over 7 · 0.29 over 7abracadabra / abrocadabro0.18 over 11 · 0.18 over 11unit cost, in edits per columnupper bar: the optimum, divided · lower bar: the best rate When the algorithm is a table

A distance divided by a length is not a rate

Two substitutions turn "ab" into "ba", a distance of two over an alignment of two columns — a rate of 1.00. Deleting, matching and inserting also costs two, over three columns, for 0.67. Both are alignments of the same pair, the second has the better rate, and the optimal alignment is not the one that achieves it. Over every pair of strings up to three characters on three letters, 21% disagree.

 busiest vertexuniform pairsbusiest 16 · Σd² 43,104 · degeneracy 4attachmentbusiest 114 · Σd² 90,296 · degeneracy 4pairs of neighbours examineduniform pairs — every pair18,480uniform pairs — oriented4,090attachment — every pair42,076attachment — oriented3,339V = 1,024, E = 3,072 on both30 triangles and 249 — the answers, which also differ Two parameters

Two parameters are not enough either

Two graphs on 1,024 vertices with 3,072 edges each — identical in both numbers every bound in this field is written in. Enumerating every pair of neighbours of every vertex costs 18,480 examinations on one and 42,076 on the other. The quantity that separates them is a third parameter, it is computable in linear time, and it appears in no statement of the problem.

 0.010.10.11Jaccard index of the two setsrelative error of the intersectionas wrong as the answerinclusion–exclusionthe minimum valuesHyperLogLog p = 12, 20,480 bits · bottom-4,096, 131,072 bitstruth from a hash map over 200,000 keys One pass, and no room

The error of a difference

Three sketches, each within a per cent or two of its own answer, subtracted into an intersection. At a Jaccard index of 0.82 the answer is 1.3% out. At 0.005 it is 146% out — the same three sketches, the same accuracy, a different question. The error never grew: it stayed a fixed fraction of the union, and the union stopped being the thing being asked about.

 —gactacgat—gattacagt0123456789101234567821012345673211123456432212345554332123456543321234765443222387655432339876554333one unit = one subproblem given a value100 cells, 100 held at once The data that is not a number

A distance that is a path through a grid

How far apart two strings are is a shortest-path problem on a grid whose every edge is drawn by the recurrence — and finding a string in a text costs 4,988 character comparisons where measuring how far it is from one costs 96,000.

 10110100runs in the transform, rbitsthe structurelog₂ N(r)12 characters · 2 symbols · all 4,096 texts walkedgap 14.1x–92.0x The floors

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.

 1,00010,000110100modulus qverifications in one searchq = 257: 693measuredn / qchosen text: 6%one unit = one character comparison · base 2568 colliding blocks at q = 1,009, none at q = 1,000,003 When the algorithm flips a coin

A match decided by a number

Rabin–Karp replaces a comparison of two strings with a comparison of two integers, and pays for it only when the integers agree by accident. On 65,521 windows at a modulus of 1,009 it verified 79 times against a predicted 64.9. On a text of 1,024 characters chosen with the modulus in hand it verified at 64 windows out of 64 possible, each one costing fifteen of the pattern's sixteen characters — and the closed form is silent about that, because it assumes nobody chose the text.

 average 8.001,0242,048bucket, 0 to 255keys in the bucketmultiply–shift · keys built for this hashworst bucket 2,048 against 8.0 What is taught wrongly

The adversary who knows the seed

Every randomised figure on this site is drawn from a stated seed, so that the numbers in the captions are the numbers on the reader's screen. That is also the exact condition under which none of the guarantees those figures demonstrate applies. A published seed is a published function.

 10010³10³10⁴10⁵10⁶10⁷Vcounted workBreadth-firstDijkstra, binary heapBellman–FordBellman–Ford, all passesV from 64 to 2048, sparse, fixed average degreework = scans + visits + relaxations + queue comparisons Two parameters

The bound with a precondition

Bellman–Ford is O(V·E), and on a graph of 2,048 vertices it stops after seven passes of the 2,047 the bound allows — a factor of 289 between the bound and the run. Dijkstra is faster and returns a wrong answer on four vertices if one arc is negative. Both facts are about the same clause: the qualifier at the end of the sentence.

 10³10³10⁴10⁵10⁶10⁷ncomparisonsadversary, no depth limitadversary, as it shipsrandom inputn²/2McIlroy's adversary, answering as it goescomparisons, counted exactly What a bound is

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.

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