Concept

Cost model — where it appears

The rule saying what a substitution and a gap cost, which is an input to an alignment rather than a property of it, and without which a distance is not a quantity. Two answers computed under different models are different quantities, so a distance printed without its model is a number rather than a measurement.

Named by 61 essays across 11 fields — each of them below, with the objects they name alongside it.

10010³10³10⁴10⁵10⁶10⁷Vcounted workBreadth-firstDijkstra, binary heapDijkstra, all V queuedBellman–Ford, all passesV from 64 to 2048, sparse, fixed average degreework = scans + visits + relaxations + queue comparisons

Counting on a graph

An instrumented array counts comparisons, swaps, reads and writes, and none of those is what a graph algorithm spends its time on. Three new primitives are needed — an adjacency scanned, a vertex first reached, an edge relaxed — and once they exist, breadth-first and depth-first search turn out to be the same algorithm by every count kept on arrays.

graphs · Graph
peak slots held at once, logarithmic18645124096Insertion sort11Selection sort11Bubble sort11Heapsort11Shellsort11Quicksort, median of three22log nQuicksort, random pivot28log nQuicksort, first-element29log nMerge sort with a cutoff4,106nMerge sort4,110nn = 4,096, random inputone slot = one array element or one stack frame

Measuring what an algorithm keeps

Four counters measure what an algorithm does and none of them measures what it holds. An in-place sort and an out-of-place one with identical comparison counts are different algorithms, and until this phase the site had no way to say so. Two primitives close the gap, and the second of them counts something no array counter can ever see.

space · Space
block transfersIn order, 0 to n−11,0241.0× a scan · 64.0 elements per transferEvery B-th element (B = 64)65,53664.0× a scan · 1.0 elements per transferUniformly at random61,40760.0× a scan · 1.1 elements per transfera scan of this array is 1,024 transfersB = 64, M = 4,096 (M/B = 64)64× between the cheapest order and the dearest

One access, eight kilobytes

An operation count charges one for an array access. A machine charges for a block. The same 65,536 accesses cost 1,024 transfers in one order and 65,536 in another, with nothing about the algorithm's work changed — a factor of 64, which is exactly the number of elements in a block, and which no counter here could see until now.

applied · Transfer
048162432characters every key sharesoperations072,581145,162Character comparisonsRadix sort, characters readElement comparisonsone unit = one character comparisonelement comparisons constant at 3,955

The comparison that is not one comparison

Sorting 512 keys costs 3,955 comparisons whatever the keys are, and between 7,849 and 134,409 character examinations depending only on how much those keys have in common. The first number is the one every bound so far is stated in. The second is the one the machine pays, it grows without limit, and nothing here has ever counted it.

text · Symbol
—bcab—abca6388328188632571322513518753111111one unit = one invocation of the recurrence481 calls, 25 distinct subproblems

The cost is the number of subproblems

The edit-distance recurrence, written down literally, makes 29,737 calls on a six-letter word and a seven-letter word. Written down with a table beside it, it makes 56. Nothing about the arithmetic changed, and the class did.

tables · Table
10⁵10⁶10³10⁴10⁵comparisonscache misses (modelled)Insertion sortSelection sortBubble sortMerge sortHeapsortQuicksort, firstQuicksort, median-3Quicksort, randomShellsortMerge + cutofffully associative · 64 lines × 8 elements · LRUa modelled count, not a time

The count is not the time

An operation count is exact, machine-independent, and not a running time. The gap between them is mostly memory, and it is large enough to reorder the rankings. A second count is kept alongside it — modelled cache misses from the same runs — and the two are required to disagree, because if they agreed the second one would carry no information.

machine · Machine
110100257amortised 2.000256512append numbercost of that append (log scale)growth factor 2, cost = 1 write + a copy of the array when it resizes9 resizes in 512 appends

What amortised means

Appending to a dynamic array is O(1) amortised. It is also, on 512 appends, an operation that costs one unit 503 times and 257 units once. The amortised bound is a true statement about the sequence and a false one about any append in it, and the picture that shows why is a sawtooth nobody draws.

structures · Structure
10³10⁴10010³n (elements)block transfersM = 256 · fan-in 7M = 1024 · fan-in 31B = 32, M as labelled4 passes against 3

Sorting what will not fit

Merge sort's Θ(n log n) is a statement about comparisons and says nothing about a file larger than memory. Counted in transfers the answer is (n/B)·log_{M/B}(n/B), and the base of that logarithm is the number of blocks that fit in memory — so doubling the memory does not halve the work, it moves a staircase. The measured cost jumps by 32,768 transfers at one step and by nothing for the next four.

applied · Transfer
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
—sitting—kitten012345678910111213141516171819202122232425262728293031323334353637383940414243444546474849505152535455one unit = one subproblem given a value56 cells, filled in row order

The same table, filled two ways

Top-down and bottom-up compute identical cells and return identical answers. One of them asks the table half a million questions and recurses four hundred frames deep; the other asks none and recurses none — and on a knapsack it fills twenty-two times as many cells as anything can reach.

tables · Table
—abrocadabro—abracadabra012101221012210122112222123321233212332123321233212322one unit = one subproblem given a value54 of 144 cells, 90 skipped

A band as wide as the answer

If two strings are close, the optimal route stays near the diagonal and nine cells in ten cannot be on it. A band of three finds the right answer on a pair 300 characters long — and a band of thirty-two is needed before anything can prove it.

tables · Distance
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
1010010B (elements per block)block transfers per searchtuned for B = 64Sorted arrayB-tree tuned for B = 64van Emde Boas — told nothingM = 16,384, B as drawnone layout, 7 block sizes, no parameter

The layout that is told nothing

A B-tree is built around a block size somebody looked up. The van Emde Boas layout is given neither the block size nor the memory size, and across seven block sizes spanning a factor of 64 it tracks the best structure that was told them. An algorithm with no parameters making a claim at every level of the hierarchy at once is a strange thing to be able to measure, and this is what it costs.

applied · Transfer
acgtacgtseen ->0313303113033130a gap of k characters costs 2krows: expected · columns: seen · unit: bitslinear gaps

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.

tables · Cost
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

The count somebody chose

Six quantities can now be measured for every sort. Ranking the ten algorithms by each of them and comparing the orders, comparisons and peak space disagree about 91% of all pairs, and memory traffic and modelled misses disagree about 7%. There is no ranking of sorting algorithms; there are six, and choosing between them is a statement about the data rather than about the algorithms.

counting · Count
pale: total slots ever allocated · dark: peak held at onceone buffer, allocated once16,38416,400peak 100% of na buffer per merge229,37616,386peak 100% of nn = 16,384, random input208,687 comparisons each — identical in time

The space the model does not see

A slot is not a byte, a frame is not a slot, sixteen thousand allocations are not one allocation of the same size, and none of these numbers includes the input. The space counters are the newest instrument here and the honest account of what they miss is longer than the account of what they measure — including one bound this phase set out to demonstrate and could not.

space · Space
01234567891011120123456789101112123456789101112123456789101112345678910123456789123456781234567123456123451234123121one unit = one subproblem given a value91 cells, 364 transitions, 4.0 per cell

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.

tables · Table
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
—agcacacggatcag—ccagggagta09101112131415161718192021229110101212141416171819192122101021011121314151718191920221110113101212141516171920192112121012411131314151718202119131312111351214131416181921211414131312146131413151719202115141514131314714151316181921161614161514141571416141719191717171517161515168151615182018171818151816161617816171519one unit = one subproblem given a value495 cells across three tables, 980 transitions

A cell that has to know where it is

A gap of four characters is usually one event, not four. No recurrence over a single table can charge it that way, because the price of a gap character depends on how the cell above it was reached and a cell holding one number has thrown that away. The repair is three tables, and it costs exactly three times the cells.

tables · Cost
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
least recently usedthe offline optimumA loop over 7 blocks35 accesses35113.18×Random over 7 blocks35 accesses971.29×A straight sweep35 accesses35351.00×B = 8, M = 48 (M/B = 6)3.2× on the loop

The block that is not a block

This field's model has one memory, one block size, one processor, no prefetcher, no queue and no clock, and every number in it is a count of transfers rather than a duration. A real machine has five levels, reads ahead, issues a hundred requests at once, and charges four orders of magnitude more for one kind of transfer than another. What survives the difference is worth stating exactly, and so is what does not.

wrong · Transfer
—execution—intention87777776569888888765one unit = one subproblem given a value100 cells computed, 20 held at once

The table nobody has to keep

A million-cell table, computed cell for cell in the same order, holding two thousand cells at its peak instead of a million. The saving is exactly (n+1)/2, it costs nothing on any operation counter, and what it buys is paid for with the one thing the table was for.

space · Table
01234567891011120123456789101112122332223333122222332231222222223122222233122222221222222122222122221222122121one unit = one subproblem given a value91 cells, 156 transitions, 1.7 per cell

The argmin that cannot go backwards

The same triangular table, the same ninety-one cells, the same tree at the end of it — and 364 transitions one way against 156 the other. At 256 keys the ratio is 38. What removes the factor is not a property of the recurrence but a property of the numbers it is given, and the recurrence does not mention them.

tables · Table
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

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.

tables · Cost
pairs where the two disagreerestricted · unrestrictedab → bca3 against 2ac → cba3 against 2ba → acb3 against 2bc → cab3 against 2ca → abc3 against 2triples the restricted rule breaksab → bca costs 3, but by way of ba it costs 2ac → cba costs 3, but by way of ca it costs 2ba → acb costs 3, but by way of ab it costs 240 strings, 1,600 pairs, 64,000 triplesunrestricted: 0 triples broken

The edit that reaches back two rows

Swapping two adjacent characters is one keystroke and costs two edits. Adding it as a fourth transition is four lines, it is what nearly everything ships, and the function those four lines compute is not the one they are named after. Over 1,600 pairs of short strings the two definitions differ on twelve, and the shipped one breaks the triangle inequality on twelve triples where the other breaks it on none.

tables · Cost
—gactacgat—gattacagt0123456789101234567821012345673211123456432212345554332123456543321234765443222387655432339876554333one unit = one subproblem given a value100 cells, 100 held at once

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.

text · Distance
Sorted array (binary search)2 blocksLevel order4 blocksvan Emde Boas2 blocksmemory address, left to right · alternating outlines are blocksB = 8, M = 64 (M/B = 8)4 blocks against 2, for the same 6 comparisons

Two searches, one comparison count

Three arrangements of the same binary search tree over the same million keys, walking the same path, making the same twenty comparisons. One costs 15 block transfers, one costs 13, and one costs 3. Nothing about the algorithm differs between them — only where the nodes were put — and no counter this site had before this phase could tell them apart.

machine · Machine
—sitting—kitten111111101111110-1111110-1-111110-1-1-11110-1-1-10110-1-1-10-11one unit = one subproblem given a value-1, 0, 1 — 3 values, 2 bits each

A column computed in machine words

Adjacent cells of a distance table differ by at most one, so a whole column is two bits per cell — and thirty-two of them fit in one register. Fifteen word operations per character replace three cell evaluations per cell, and below a pattern of fifteen characters the trade is a loss.

machine · Machine
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
Count-Min8,192 bitsCount-Sketch8,192 bitstug-of-war, F22,560 bitsGreenwald–Khanna0 bitsexponential histogram0 bitscash register+strict turnstile±general turnstile± may go negativesliding window+ expiresdeclareddeclared93% under—declareddeclareddeclared—declareddeclareddeclared—declared——————declared40,000 updates · deletion rate 0.5 · every count exact1,758 of 1,895 under

The model a bound was quoted in

Every accuracy figure in this field's first phase was measured under four unstated assumptions. Remove them one at a time and one structure loses its guarantee on 91% of queries, another's error stops falling when it is given more state, and a third has nothing to do at all.

bounds · Bound
1010010³10⁴B — elements to the blockblock transfersthey cross near B = 4one element at a timesorted by destination, 2passesM = 4,096, B as drawnthe lower bound is the smaller of the two, and it is proved rather than measured

Permuting is the harder problem here

Rearranging 65,536 elements into a stated order costs 63,601 transfers one at a time and 4,096 by sorting them into place. In the model every other field on this site uses, the first is the cheap method and beats the second by a factor of eight. The two models disagree about which problem is easy, and they disagree by about the same factor in opposite directions.

applied · Transfer
4 bytes32 bytes128 bytes512 bytesrecord:QuicksortMerge sortShellsortHeapsortInsertion sortSelection sortBubble sortQuicksortShellsortMerge sortHeapsortSelection sortInsertion sortBubble sortQuicksortShellsortMerge sortSelection sortHeapsortInsertion sortBubble sortSelection sortQuicksortShellsortMerge sortHeapsortInsertion sortBubble sort1234567n = 512, random input, key 8 bytesa read or a write moves the record; a comparison touches the key

The exchange rate nobody wrote down

Three earlier essays have said in passing that the ranking would change if the elements were large records. None of them computed it. Computed, selection sort goes from second-worst of seven at four bytes a record to best of seven at five hundred and twelve — and the crossover against each rival is a division that takes one line.

counting · Count
stay 0.9, 400 pairsacgtacgt0525505225045240stay 0.5, 400 pairsacgtacgt0212202112022120cost of aligning the row letter with the column letterpseudocount 1

The matrix a corpus wrote

A substitution matrix is not a property of an alphabet. Fit one to four hundred pairs of sequences that rarely change and the dearest substitution costs five; fit the same model to four hundred pairs that often change and it costs two. Two hundred test pairs aligned under each matrix give different alignments in 115 cases — and a matrix fitted to eight pairs of the first kind moves 79 of them, from sampling alone.

tables · Cost
levels on the path (outline) · transfers actually paid (filled)B = 411 levels · 6.99 transfers · 16,384 blocks residentB = 166 levels · 3.23 transfers · 4,096 blocks residentB = 644 levels · 1.99 transfers · 1,024 blocks residentB = 2563 levels · 1.30 transfers · 256 blocks residentB = 10243 levels · 1.01 transfers · 64 blocks residentbinary search over the same 4,194,304 keys: 20 transfersB = 1024, M = 65,536 (M/B = 64)20× between binary search and the widest tree

A tree with nodes the size of a block

A B-tree is a binary search tree that has read the hardware manual. Its node holds as many keys as fit in one transfer, so the height falls from log₂ n to log_B n — and the measured cost falls further still, to 1.01 transfers over four million keys, because the top of the tree is small enough to stay in memory. The comparison count goes up.

structures · Transfer
11010010³10⁴1010010³rows matching the queryblock transfersn/B rows — where the arithmetic says they meetindex, rows scatteredindex, file in key orderread the whole fileB = 64, M = 4,096 (M/B = 64)plus 3 transfers to descend the index

The index that is not worth reading

An index turns a query over 65,536 rows from 1,024 transfers into four. At a thousand matching rows it costs 654 and still wins; at sixteen thousand it costs 1,027 and has lost. Where it turns is decided by the block size — a number the query does not contain, the schema does not mention, and nobody writing either has seen.

applied · Transfer
1×10×10³10⁴the key side, as a multiple of memoryblock transfersblock nested looppartitioned hash joinsort–merge joinB = 64, M = 512 (M/B = 8)the other relation: 16,384 rows

Two ways to join, and the ratio that decides

The same join costs 260 transfers one way and 1,040 the other; at eight times the memory the same two costs are 2,880 and 1,280, the other way round. Neither number is a property of how large the tables are. The quantity that decides is how the smaller of them compares to memory, and a rule of thumb phrased in rows is a rule about somebody's machine.

applied · Transfer
123456123456cost to open a gapcost to extend a gapintentionexecution571 settingsinte---ntion---execution3 settingsinte-ntion-execution1 setting-intentionexec-ution1 settingintention against execution, affine costseach colour is one optimal alignment

The parameter plane has few answers

Sweep the cost of opening a gap against the cost of extending one over five hundred and seventy-six settings, and the optimal alignment of intention against execution takes four values — one of them at 571 of the settings. Under a linear model the plane divides into three wedges through the origin, because doubling every cost changes nothing and only the ratio is a parameter. Tuning an aligner is choosing a region, and most of the plane is one.

tables · Cost
11010010³10⁴12510rows the query actually matchescost ÷ the better plan's costrho plus the descent, 4.01estimate ÷64estimate ÷8estimate ×8estimate ×64n = 65,536, B = 64, M = 4,096 (M/B = 64), scattered read ×4dotted: the estimate switches at 512 rows

The estimate a plan rests on

A planner chooses between an index and a scan on how many rows it thinks will match, and the number it has is wrong by a factor. Guess sixty-four times too many on a narrow query and the scan it picks costs 13.5 times the index. Guess sixty-four times too few on a wide one and the index costs at most 4.01 times the scan — a ceiling that can be named before any query runs.

applied · Transfer
10³10⁴capacity Wsubproblems given a valueBottom-up table · 1.00Top-down, reachable only · 0.14one unit = one subproblem given a valuesubproblems given a value, W from 200 to 3200

A table wider than its input

The knapsack table has (n+1)(W+1) cells and is called polynomial. Adding one character to the input doubles it — across four settings the table grows sixty-four times while the input it is written from grows by half.

wrong · Bound
rounded to whole bitsunroundeda resample, 400 near pairs0.860.9340 pairs at stay 0.90.820.928 pairs at stay 0.90.710.88400 pairs at stay 0.70.680.86400 pairs at stay 0.50.620.790.5: no prediction200 test pairs, 16 directionsdashed: a coin flip

The ties a rounded matrix makes

Measure how far each optimal alignment is from a tie — the smallest change to any one cost that makes another alignment win — and it predicts which alignments a refitted substitution matrix will move. A resample of the same corpus moves 30 of the 63 test alignments that sit on a tie and 3 of the other 137. A matrix fitted to a different divergence moves alignments far from a tie as well, and the prediction weakens to a chance of 0.62. And a third of the alignments were on a tie only because the matrix was rounded to whole bits — fitted without rounding, 15 of 200 are, and every prediction improves.

tables · Cost
alignments on a tiedistinct costs · predictionwhole bits633 of 6 · 0.86half bits793 of 6 · 0.91quarter bits414 of 6 · 0.93a grain of 0.2235 of 6 · 0.89eighth bits156 of 6 · 0.93a grain of 0.05205 of 6 · 0.95unrounded156 of 6 · 0.93200 test pairs, 16 directionsbar: alignments within 0.01 of a tie

The lattice that decides the ties

Rounding a fitted substitution matrix to whole bits puts 63 of 200 alignments on a tie where the exact fit puts 15. Rounding to half bits — a finer grain, and the obvious repair — puts 79. What tracks the ties is not how fine the lattice is but how many of the six fitted costs it keeps apart: whole and half bits both leave three, an eighth of a bit leaves all six, and matches the exact fit exactly.

tables · Cost
unit cost — triples broken0.02%unit cost — pivot bound failed0.02%log-odds on a keyboard walk — triples broken6.61%log-odds on a keyboard walk — pivot bound failed13.22%42,840 triples per model, enumerated0.02% shown where the true rate is zero

A distance that is not a distance

Under unit cost the edit distance obeys the triangle inequality and this site asserts that it does. Under a stated substitution matrix it need not, and on 42,840 enumerated triples it fails 2,832 times — taking with it every structure that prunes by distance, at a measured 13.22% of the bounds they rely on.

wrong · Cost
1.522.533.544.55fitted cost, bitsgenerator: 1.644generator: 4.644ag1.667 ± 0.103ct1.711 ± 0.084gt4.323 ± 0.248cg4.687 ± 0.239ac4.706 ± 0.234at4.879 ± 0.291one corpus, 400 pairs at stay 0.9bars: two bootstrap standard errors

The grain a corpus chooses for itself

A fitted substitution matrix can choose its own rounding lattice: the coarsest grain that keeps all six of its costs apart. On eight resamples of the same 400-pair corpus that grain comes out anywhere from an eighth of a bit to a forty-seventh, because it is set by whichever two costs happen to land closest, and those two differ by less than a tenth of their own standard error. Asked instead which costs it can actually tell apart, the corpus pools six into the generator's two and chooses an alignment the generator agrees with for every one of 200 test pairs on every resample.

tables · Cost
01020share of queries for words the vocabulary lacksthousands of cells a query0%10%20%30%discover the boundone try at the medianone try at twoone, then two, then discovertold the distanceevery attempt chargeddotted: told the answer's distance

A second try before any feature

A nearest-word search that learned one starting bound for a whole stream was proposed a bound for each query instead, read off features the search can compute before it walks. The proposed feature separates nothing: 58% of misspelt vocabulary words and 53% of words the vocabulary lacks begin with a real three-letter prefix. Length and unseen letter-triples do predict distance, and still buy about 1%, because the best start is two for almost every class. What closes half the gap to an oracle is a second cheap walk: one, then two, then discover.

tables · Table
1001,000aligned sequence pairs in the corpusresamples choosing the generator's classes0481216held-out pairsa fixed penalty a classa penalty that growstwo classes, 16 seeds a size203 partitions each

The classes held-out pairs cannot rule out

A substitution matrix can choose how many distinct costs it has as part of the fit: try all 203 ways of grouping its six pairs into classes and keep the one that best predicts pairs it was not fitted to. On corpora from a generator with two costs, that choice finds the two on about a third of resamples, and no more often at 3,200 pairs than at 50. A spurious split gains the same bit or two of fit at every size, and held-out scoring charges it a fixed price. A penalty that grows with the corpus finds the two every time, and finds a third class half a bit away from 400 pairs.

tables · Cost
12481632641282561executed queries the fit sawexpected regrettrusting the estimatefitted from the queriesthe best divisor, 1.5ρ = 4, σ = 1.5, 120 fits a pointlower is better

What a planner pays to find out what to pay

Insurance against a row estimate is set from the error's median and spread, and a running system knows neither — it has to fit them from executed queries. Fitted from one query the divisor costs 1.263 times the better plan against 1.215 for a planner that never insures at all, so learning is worse than not learning until about sixteen queries have run. The tail, though, is bought immediately: one observation already holds the worst case to 24.7 against 36.6.

applied · Transfer
010203040the cap on an answer's distance, in editsthousands of cells a query123456nonewords it lacksa fifth unfamiliarmisspelt vocabulary wordsevery attempt charged2,000 + 600 queries

A checker that stops at two edits

A nearest-word search that must always answer pays most of its price on the queries whose answer is useless: misspellings of words the vocabulary does not have, whose nearest vocabulary word can be twenty edits away. Capping the answer at two edits keeps every suggestion a misspelt vocabulary word could get and costs 42% of the uncapped search. What it does not do is stop the false suggestions. It declines 95% of unfamiliar words of eleven letters or more and 1% of those of three or four, because two edits is most of a short word, and 72% of what it still says to an unfamiliar word is wrong — against 77% with no cap at all.

tables · Table
25501002004008001,6003,200sequence pairs in the corpusclasses chosen123456six different costsmedian chosentwo terms: three parameterssixteen corpora a sizedot area: corpora choosing it

A cost built from two properties

A substitution matrix fitted as a partition gives each class of pairs one cost, and the best number of classes can be chosen from the data. When the true costs are a sum of two terms — a transversion penalty and a graded difference between the letters — every pair costs something different, and the chosen partition's number of classes becomes a statement about the corpus: two classes at 25 sequence pairs, four at 200, five at 1,600. A model with the two terms and three parameters predicts a fresh corpus better than any partition at every size, and gets 191 of 200 alignments right from 50 pairs where the chosen partition needs 3,200. Told the wrong property it is worse than the plain two-class matrix, and nothing in its own fit says so.

tables · Cost
32641282565121010010³10⁴elements sortedword operations a comparison must cost to break evenSelection sort, pairsSelection sort, closureBubble sort, pairsBubble sort, closureHeapsort, pairsHeapsort, closureQuicksort, median-of-three, pairsQuicksort, median-of-three, closuresolid: pairs asked · dashed: everything impliedrandom input · counted exactly

The price of remembering an answer

Give selection sort a table of the pairs it has already compared and it makes 16,805 comparisons on 256 elements instead of 32,640. The table pays for itself once a comparison costs more than 3.1 word operations, and that price stays near three at every size measured. Give it the full closure of everything its answers imply and it makes 13,079. The closure pays only past 425 operations a comparison at that size, and past 1,659 at twice the size. The cheap memory is worth having on long keys. The complete one is worth having almost nowhere.

counting · Count
1001000entries in the cacheshare of the repeats it catches0%25%50%75%100%256 entries, one a pair asked in a passSelection sortBubble sortdashed: collisions alonerule at 100%: the table ofevery pair256 random elements · direct-mapped, one pair a slotoccupancy, not distance

The record that forgets on purpose

A table of every pair selection sort might ask is 131,072 bits at 256 elements and catches every repeated question. A direct-mapped cache of 1,024 pairs is 16,384 bits and catches 91% of them, at a break-even of 3.53 word operations a comparison against the table's 3.12. What decides the cache is not how long the sort waits before asking again — 98% of its repeats come within one pass — but how many other questions it asks meanwhile, and a model with nothing in it but collisions predicts every measured share to within 2.6%.

counting · Count
160170180190200aligned sequence pairs in the corpus (log scale)test alignments optimal, of 200502008003,200the sum, told the propertypositions, fittedsix free classesthe chosen partition200 test pairs, eight seedsoptimal under the generator's costs

A line the corpus finds

Substitution costs built as a transversion penalty plus the distance between the two letters' values on a graded property beat every partition of the pairs, when the model is told the property, and fall below the plainest matrix when told the wrong one. Fitting each letter's position on a line instead was predicted to find the property's order from about four times the told model's data and then behave like it. It finds the order on every corpus of 100 sequence pairs or more but one, and every miss starts with the two letters whose values differ by a fifth of a bit. But knowing the order is not knowing the costs. Its alignments are no better than six free classes' until 800 pairs, sixteen times the data the told model needed for the same score, and on a generator with no graded property at all it finds a line anyway, a quarter as long as a real one at 200 pairs, and pays for it against the plain partition.

tables · Cost
1.201.401.601.80executed queries the fit sawexpected regret248163264128256fitted spread75% bound95% bound99% boundtrusting the estimatethe best divisor, 1.5ρ = 4, σ = 1.5, 120 fits a pointlower is better

A planner that knows how sure it is

A planner that fits its insurance divisor from a handful of executed queries under-insures, because a spread fitted to a small sample is biased low. The textbook repair is to use the upper end of a confidence interval on the spread instead of its centre. At a read ratio of four it makes the planner worse than trusting its estimate for longer — until sixty-four queries rather than sixteen — and it buys the worst case immediately, from 22.4 times the better plan to 6.2 after two queries. The bound is not a correction towards the truth. It is a dial between two different planners.

applied · Transfer
0.0010.010.11aligned pairs in the corpusexcess bits a pair (log scale)25501002004008001,6003,200line, length charged (c = 1)free lineline, gaps countedthe chosen partitionthe generator with nonesixteen seeds a size

A line with a prior toward the classes

Substitution costs fitted as the letters' positions on a free line learn the letters' order quickly, and on a generator with no graded property they invent a line anyway. Charging the line for its length, at a rate growing with the corpus as the partitions' charge does, was predicted to collapse the invented line from 50 pairs and to cost the real one little. It does neither. On the generator with no property the charged line still has a median length of 0.12 bits at 3,200 pairs. On the generator with a property it takes the letters' order away at every size up to 400 pairs where the free line had it on sixteen seeds of sixteen. Charging the line for the number of gaps it keeps, the partitions' own rule, collapses the invented line on every seed at every size and matches the chosen partition to the bit — and then needs 1,600 pairs before it believes a real line is there.

tables · Cost
1,00010010³10⁴10⁵10⁶keys sortedcache missesno recorda cache of 1,024 pairsthe table, written twicethe table, by ordered pairselection sort, 512 lines of 64 bytesexact LRU

The record measured where it would run

A table of every pair selection sort has compared saves half its comparisons at 1,024 keys, and priced in word operations it was the better record at every size. Replayed through a 32 KB cache it misses 477,824 times to save 258,290 comparisons — nearly two misses for every comparison it removes — while a cache of 1,024 pairs misses 192. The table's size is not the problem. Its layout is: it looks up and writes each question at [key][minimum], a new row every time. Indexed by the ordered pair instead, the same bits miss 3,079 times, and that repair works only for a sort whose smaller operand stays put.

counting · Count
4-bit2.97× · mean 2.426-bit2.23× · mean 1.828-bit1.50× · mean 1.2211-bit3.61× · mean 1.5916-bit7.73× · mean 3.23from the counts1.23× · mean 1.06the best for each caseworst over 4 inputs × 2 cachesmisses ÷ the best fixed width's

The digit a pass chooses for itself

A radix sort's digit width trades passes against the streams each pass scatters into, and the best width depends on the keys and the cache together: 8 bits for uniform keys through 32 KB, 11 through 256 KB, 16 for keys with few distinct values. A digit widened because the previous pass's counts were concentrated cannot find it, since in a least-significant-first sort one digit's counts say nothing about the next. One read that counts every byte at once, and a sample to size each field, can. Charged for both, the resulting plan is never worse than 1.23 times the best fixed width on any keys or cache measured. The best single fixed width reaches 1.50, and 11 bits reaches 3.61.

counting · Count
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
12481632640.1110line size B, array elementstransfers × cells, per n²heapsortShellsortquicksort, random pivotquicksort, first pivotthe read-only familyn = 8,192, memory 512 cells, LRUrandom input

A product paid for in blocks

The published floor under sorting puts a bound on time times space, and a read-only family that scans its input over and over meets it, while every in-place sort sits far above it. Counted in block transfers rather than reads, the floor itself moves to n²/B: the family meets it at every block size, and so does every sort once memory holds the whole array. With a memory of 512 cells, quicksort sits a fixed eight times above the family at every block size, because partitioning is itself a scan. Only heapsort's distance grows with the block, from ten times at a line of one to 285 times at 64. The prediction that the family's lead would widen towards B holds against heapsort and nothing else.

bounds · Bound
a second hash, unordered0.054 entries read on an absent key14.435.79 storedthe bits below the bucket, unordered0.059 entries read on an absent key14.435.79 storedthe bits below the bucket, sorted0.057 entries read on an absent key8.975.21 storedthe bucket's own bits, sorted5.743 entries read on an absent key11.275.80 storedtag comparisons an absent key's lookup makes8,192 slots · buckets of 8 · 8-bit tags38% fewer

A tag that answers more than yes

A byte of hash in front of each entry has one bit of output per slot: this tag matches or it does not. Cut it from an ordered field instead and a bucket's tags can be kept sorted, so a lookup stops at the first tag above its own — 8.97 comparisons for an absent key instead of 14.43, for 3.55 tag-byte shifts an insertion. The obvious ordered field is the trap. Take the tag from the bucket index's own bits and every key in a bucket shares them: 5.74 entry reads on an absent key instead of 0.054, a hundred and seven times more.

machine · Machine
2,0484,0968,1921010010³10⁴nleast fast memory, cellsheapsortShellsortquicksort, random pivotquicksort, median of threefamily, 16 cellsfamily, 64 cellsfamily, 256 cellswithin 8 × n²/Bdoubling grid of memories

The memory a product does not charge

Time times space charges a sort for the cells it writes and not for the fast memory it needs to write them cheaply. Counted three ways, on random keys in lines of eight, a read-only family stays within a factor of eight of the floor with 32 to 256 cells of fast memory at every length, while heapsort needs all of n and quicksort an eighth of it. Quicksort's need falls far faster than predicted as the factor loosens. And the product the floor leaves out, transfers times fast memory, separates the two kinds of sort completely: no read-only sort can push it far below n²/B, and quicksort reaches a twentieth of it.

bounds · Bound
1664256102440961210100fast memory M, cellstransfers × writable cells, per n²/Btwo: one read, one write of nheapsortShellsortquicksort, random pivotfamily, 16 cellsfamily, 64 cellsfamily, 256 cellsn = 8,192, write-backs chargedthe floor at one

The lines a sort writes back

A replay that charges a transfer for every line brought into fast memory and none for a line sent back flatters every sort that writes in place. Charged for its write-backs, quicksort's transfers rise by 1.9 at every memory short of n, as predicted, and heapsort's by 1.8, not the predicted three, because no line can go back more often than it came in. The cost that changes the account is the last one: at n itself every in-place sort pays two reads of its array, so none of them can come within a factor of two of the floor, while the read-only family reaches 1.07.

bounds · Bound
fast memory, cells (log scale)transfers × fast memory, in n²/B (log scale)0.5124816321632641282565121,0242,0484,0968,192sorted buffer, with write-backsheap buffer, with write-backssorted buffer, reads alonebuffer of 64logarithmic scales

A buffer that does not shift its tail

The read-only family of sorts scans its input once for each batch of the smallest values not yet output, keeping them in a buffer; charged for the dirty lines it sends back to slow memory, its least transfers × fast memory rose from 1.29 of n²/B to 1.87, because a sorted buffer takes each value by shifting every larger entry one place. Kept as a max-heap instead, the buffer was predicted to write back about a third of the lines it reads below its own size and to bring the product back towards 1.29. At a buffer of 64 cells it writes back 33% at its best memory, against 45%, and its least product falls to 1.71: right about the share, wrong about the product, since a third written back is still a third more transfers. At 256 cells the heap does more than predicted — 1.47 million accesses against 6.94 million, a third of the misses — and its least product, 1.62, is under any buffer's on the earlier charge. At 16 cells it does nothing.

bounds · Bound

Named alongside it

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

Honest limitMeasured countAlignmentDynamic programmingEdit distanceBlock transferLocalityComparison countSubproblemSubstitution matrixTrade offCache

All concepts