Dynamic programming — where it appears
Named by 28 essays across 7 fields — each of them below, with the objects they name alongside it.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
The alignment that fits in one line
Compute the table twice and hold three rows of it. The factor of two is a geometric series and is predicted exactly; measured, it comes down from 2.269 to 2.052 as the strings grow, and the peak is 3(m+1) cells on the nose.
The filter that feeds the table
A self-index answers exact queries and nothing else. Approximate matching needs a table with twenty thousand columns in it. The pigeonhole joins them — cut the pattern into k+1 pieces and at least one occurs exactly, and the index that cannot answer the question decides where to ask it.
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.
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.
The row that starts at zero
The same 1,413 cells, filled by the same recurrence in the same order, answer 148 and 0. One line of initialisation decides which question the table was asked, and only one of the two answers is about whether the pattern is there.
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.
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.
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.
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.
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.
The branch that cannot reach an answer
Seventy-two rank operations over the pattern remove 27,906 of the 39,957 interval extensions a bounded-error index walk performs — 70% of the tree, at a budget of three. The share grows with the budget, which is what a pruning has to do to be worth its cost.
The columns the candidates share
Three thousand tables against one query, and most of them begin the same way. Stored as a trie, the 2,424-word vocabulary has 7,710 distinct prefixes holding 17,239 letters, and a search that computes one column per prefix reads 61,449 cells against 156,714 — before it applies any bound at all. Apply the bound at a prefix instead of at a word and it reads 16,958, beating a list search that was told the answer in advance.
The bound a block can and cannot have
Knuth's condition turns an interval table's cubic fill into a quadratic one by bounding each cell's best split between its two neighbours'. A blocked fill cannot use it a cell at a time, and the two cells that bound a block lie outside the block — one to its left, one below it. The schedule has finished both for ten per cent of the blocks, the bound then removes eleven per cent of the splits, and it removes half a per cent of the cache misses, because the splits it skips are the ones already in the cache.
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.
Eight cells at once
The anti-diagonal fill order exists because its cells do not depend on one another, and every table filled here has been walked one cell at a time anyway. Computed eight at a time, a step touches 5.71 cache lines on the layout that stores the table by diagonals and 10.87 on the one that stores it by rows — and per cell the first keeps falling to 0.42 while the second stops at 1.27. The prediction that a diagonal step would touch three or four lines was wrong, and line-aligning each diagonal only takes it to 4.94.
Named alongside it
The objects these essays reach for when they reach for this one.
SubproblemEdit distanceCost modelMeasured countTrade offAlignmentEvaluation orderRecurrencePruningHonest limitTracebackApproximate matching