The collection

Every essay — page 10

Page 10 of 13, continuing through the fields in the same order.

What a bound is Counting The floors What the machine does Structures Two parameters The other axis When the algorithm flips a coin What the libraries do When it does not fit One pass, and no room The data that is not a number When the algorithm is a table The index that replaces the text What is taught wrongly Ladders Objects Search

Structures

Heaps, trees, hash tables and dynamic arrays — each with its advertised bound put through the same measurement as everything else here.

rank among the boundaries, sorted by the text before themsorted by the text after0 in the rectanglepattern "ss is un"split after 40 end with the left half0 begin with the rightEnglish-like · 2 copies of 512z = 156 · 0 crossing

A rectangle over a permutation

Two orderings of one set of boundaries are two permutations, so a phrase index's intersection is a rectangle over a permutation grid — the one point set a wavelet tree stores exactly, at one bit per point per level and no coordinates at all.

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50318223749516674879310811positionthe minimum12 values · depth 3answer 6 (1)

The shape a range question is about

A range minimum is a question about a tree, and the tree is determined by the array. Twelve values, eleven parent links, and the answer to every one of the seventy-eight ranges is a lowest common ancestor — with the values themselves no longer needed.

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024601234567891011level of the wavelet treemean runthe parse's grida uniform permutationbreak-even3,612 points · 12 levels1.23x chance

The runs a permutation does not leave

A run of length L costs 2⌊log₂ L⌋ + 1 bits and replaces L, so coding runs pays above a mean run of six. This grid's mean run is 2.34, chance gives 1.90, and coding its runs makes it 17.8% larger.

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05101520510revisions in the collectiondepthone revision alone: 8worst depthmean depthfourteen revisions of another9 generations added exactly one

One revision, one level

The cap ladder assumed depth is generations of copying. On a real file it is exactly that — one more revision, one more level, nine times running — plus eight levels the first revision already had before any history existed.

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a fixed-length code97,718 bits5.00 ranks · in orderthe best ordered tree77,890 bits3.98 ranks · in orderthe best tree of any shape76,789 bits3.92 ranks · unorderedσ 21 · the ordered tree is 1.43% above the unordered optimum16,384 characters of englishorder costs 1.43%

The tree the operation insists on

The compound walk means "everything that went left is smaller", which is true only if the leaves are in the alphabet's order. Huffman's tree is the smallest and its leaves are in frequency order, so the operation that makes a bidirectional search affordable costs the shape that makes an index small.

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range consideredminimum atC[at] < lonew document[476, 489)483 · document 4yesyes[476, 483)477 · document 1yesyes[476, 477)476 · document 0yesyes[478, 483)482 · document 2yesyes[478, 482)478 · document 0nono[484, 489)486 · document 3yesyes[484, 486)484 · document 0nono[487, 489)487 · document 2nono8 steps, 5 documents, and the two columns never differrows 476–489 · 8 documents8 steps

A document already in the answer

The same walk, with the chain's test replaced by a lookup in the answer so far. It reports the same documents at the same cost, and two exchanged lines make it lose nine of seventeen without failing.

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051015255075100125documentsdistinct characters the documents end with+7+52+162+293+4+-1+9+19prosecopies of one basethe label is the runsthe cut added128 documents0.15 against 2.31 runs a separator

The price of a boundary is what precedes it

A separator sorts before everything, so its rows sit at the top of the suffix array and hold the documents' last characters. What a document boundary costs the transform is the entropy of the character in front of it, and nothing else.

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range minimum and chain838,797255.0%range minimum, no chain285,46086.8%one descent over D119,97236.5%the document array, the chain and a range minimumthe document array, a range minimum and one bit per documentthe document array, in a wavelet tree64 documents · 16,447 characters14.3% of the published apparatus

The last array in the apparatus

The document-listing apparatus began as three arrays beside a suffix array. Two of them turned out to be machinery for reading the third, and both have gone. What is left is the document array, and it is the only one of the three that was information.

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extensions attempted9,206 dead7,594 livebit-vector ranksthe loop: 168,000the descent: 24,0868 patterns · 1 error · sigma 2054.8% dead · 6.98x

The branches that find nothing

An approximate search over a twenty-symbol alphabet attempts sixteen thousand eight hundred extensions and nine thousand two hundred of them produce an empty interval. That is a full rank walk whose entire result is the discovery that nothing was there.

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plain bit vector19,475compressed blocks4,55723.4%Elias–Fano3,90520.1%positions, priced as Elias–Fano3,59118.4%16,385 rows · one in 32 markedElias-Fano at 20.1%

A position split in two

Write each sorted position as a high part and a low part. Store the low parts packed and the high parts as a bit vector in which the k-th one sits at position (p >> w) + k. A select on that vector and a low read recover any position.

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01,024120 phrases · 1,024 characters60 drawn, every one pointing right

Every copy points right

A greedy self-referential parse chooses each phrase's source from text already produced, so an occurrence copied from another lies strictly to its right. That one fact removes a visited set from a propagation, exactly as a left-first walk removed an array one strand ago.

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020406080255075100125one sampled position in every …of the plain index, per centsize: 84.7%67.5 steps16,384 characters · 8-character patternssize falls, the walk lengthens

Bits and steps on one frame

The size falls from ninety-nine per cent to eighty-six as the sampling thins, and the walk to a sampled position rises from two and a half steps to sixty-four. Neither line is the answer; the answer is a point on the pair.

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0a1b2split3split4a5a6b7split8a9b10split11a12b13split14accept9 character tests · 5 splits · 1 accepting(a|b)*a(a|b)(a|b)(a|b)15 states

Two states per operator

Thompson's construction adds a bounded number of states per rule and no rule copies a sub-machine, so the machine is linear in the expression and is built in linear time. Every exponential in this field is somewhere else.

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When the algorithm is a table

A dynamic program's cost is settled before its input is touched, by how many distinct subproblems its recurrence has. The unit is the subproblem; the second number is how many must be held at once; and the methods worth knowing are the ones that compute more in order to keep less.

bcababca6388328188632571322513518753111111one 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.

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sittingkitten012345678910111213141516171819202122232425262728293031323334353637383940414243444546474849505152535455one 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.

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abrocadabroabracadabra012101221012210122112222123321233212332123321233212322one 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.

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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.

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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.

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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.

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agcacacggatcagccagggagta09101112131415161718192021229110101212141416171819192122101021011121314151718191920221110113101212141516171920192112121012411131314151718202119131312111351214131416181921211414131312146131413151719202115141514131314714151316181921161614161514141571416141719191717171517161515168151615182018171818151816161617816171519one 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.

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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.

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executionintention0136101521283645247111622293746555812172330384756649131824313948576572141925324049586673792026334150596774808527344251606875818690354352616976828791944453627077838892959754637178848993969899one unit = one subproblem given a value100 cells, filled in diagonal order

The order that has a depth

One hundred cells, filled in three orders, producing one table. Row order takes ninety-one steps and anti-diagonal order takes nineteen. Nineteen is not a property of the order — it is the longest chain of cells in the recurrence itself, no schedule can get under it, and every count taken until now was a total that could not see it.

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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.

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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.

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gactacgatgattacagt112223334244352461647one unit = one subproblem given a value21 matching pairs, 21% of the rectangle

The cells that were never worth having

Two three-hundred-character strings over twenty-six letters give a table of 90,601 cells, and 3,421 of them are pairs of positions whose characters agree. Only those can lengthen anything. A method that enumerates exactly those computes a twenty-sixth of the table — and on a two-letter alphabet it computes half of it and is worse than the table it replaced.

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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.

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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.

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01234567801234567891724303539424410182531364043111926323741122027333813212834142229152316one unit = one subproblem given a valueeach number is a storage offset, of 45 slots

A triangle stored in a square

An interval table has a cell for every range of keys and nothing below its diagonal, and it can be stored as a square array, as packed rows, or as packed diagonals — the last matching the order it is filled in. On sixty-four keys, with every read replayed through a small cache, the square misses 39.7% of its reads, packed rows 38.8%, and packed diagonals 78.4%. Storing a table in the order it is written is storing it in the order it is not read.

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cache misses per split point consideredsquare array, by length1.1063,128,465 missestwo copies, by rows0.212598,455 missessquare array, split scans0.095268,386 missesfully associative · 32 lines × 8 elements · LRU256 keys, 32,896 cells

The split scan cut into blocks

Every way of filling an interval table one cell at a time stops at about one cache miss per split point considered once the table outgrows the cache — 1.01 at 128 keys, whether the cells go by length, by rows, or in a recursive tiling. Cut each cell's scan into blocks instead, and apply a block of split points to a block of cells whose inputs are all in hand, recursively at every scale, and the same 357,760 split points cost 0.094 misses each. The fill is told nothing about the cache, blocks of one and of four do equally well, and it needs no extra memory, where storing the table twice gets to 0.151 by doubling it.

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10³10⁴10⁵words in the vocabularysubproblems given a valueevery table in full · 1.16best so far · 0.99answer known · 0.96one unit = one subproblem given a valuesubproblems given a value, n from 250 to 2424

The bound the search finds for itself

A spelling checker that computes the full edit-distance table against every word in a 2,424-word vocabulary fills 156,714 cells for each misspelt query. Bound each table by the best distance found so far, and abandon it the moment a whole row exceeds that bound, and the same search fills 40,273 and finds the same words. Meet the candidates nearest in length first and it fills 26,203, starting a table for exactly the words a search that knew the answer in advance would start. The last factor of 1.7 is the price of not knowing, and it is largest when the misspelling is smallest.

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