Permutation — where it appears
Named by 18 essays across 9 fields — each of them below, with the objects they name alongside it.
The transform that emits nothing
The Burrows–Wheeler transform outputs exactly the characters it was given, in a different order. Its zeroth-order entropy is therefore identical to its input's, to fifteen decimal places, and by that measure it has done nothing at all. A Huffman coder handed the result spends 1.935 bits per symbol where the same coder on the same data spends 4.209, and the difference is entirely in the order.
What a reordering costs to undo
Sorting the characters of a text clusters them perfectly: a move-to-front pass then leaves 289 bits where the text's own floor is 31,931. Naming which arrangement of those characters the text was costs 31,827 bits, and the two numbers add to the floor it started from. The Burrows–Wheeler transform clusters less and costs nothing to undo, which is the only reason it is the one that is used.
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.
The permutation that moves almost nothing
Two ways to scramble sixteen thousand elements. Shuffling them inside windows of five hundred and twelve puts two million pairs out of order and costs 3,095 block transfers to carry out. Swapping a thousand pairs across the whole array puts seven million out of order and costs 1,189. Inversions are the textbook measure of disorder, and on a disk they rank these two backwards.
A worst case ten positions wide
Sorted input costs first-element quicksort 2,096,128 comparisons on 2,048 elements, 82 times its average. Reshuffle about eleven of the 2,048 positions and the cost halves — and it takes about ten at 128 elements, and between ten and thirteen at every size between. Reversed input costs insertion sort twice its average, and reshuffling half the positions still leaves 71% of the work. A worst case is a place in the space of inputs, and the two famous ones are places of very different sizes.
The structure paid for before the first query
The two grids that make a phrase index's search proportional to its answer are 3,240 bits on an 8,892-bit index — twenty-seven per cent of the whole structure, answering nothing on their own, and 38% of them is rank directory rather than payload — the lower-order term of the published bound, measured.
A block, a class and an offset
Replacing each block of a bit vector by how many ones it holds and which arrangement it is takes 7.6% off the grid. On a permutation with no structure at all it takes 2.6%, so five of the seven points are the data and two of them are the encoding.
The order inside a tie
Sort the rotations of a text by their first four characters rather than by everything that follows, and the output clusters slightly better than the full Burrows–Wheeler transform: 1.780 bits a symbol against 1.802, from 63% of the character reads. It also costs nothing to undo. The prediction that a shorter context leaves ties for the inverse to pay for was wrong. The cost to undo comes from how a tie is ordered, not from how long the context is, and at k = 0 the wrong tie rule is exactly the sort.
A result the size of its own noise
Sorting rotations by their first four characters beat the full Burrows–Wheeler transform by 0.022 bits a symbol. On sixteen streams of the same source the mean difference is 0.004 with a standard deviation of 0.011, and the short context is ahead on ten of them. The 1.2% was one draw. What survives is sharper than the result it replaces: from four characters upwards, every run break either rule makes is at a word boundary, and nothing above four characters decides anything at all.
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.
A bit for every bit
A grid over 3,612 points is 43,344 bits of payload. The smallest any structure can be that distinguishes one permutation of 3,612 things from another is 37,485. There is 16% to play for, and the deferral that asked for a compressed grid assumed there was much more.
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.
The operations a candidate count leaves out
The grid examines 33 candidates where the scan examines 4,355 — a factor of 132. Counted in the operations each of them performs, the same query is 2,471 against 4,355, and the factor is 1.8.
The level where compression stops paying
Choosing the best coding for every level of the grid separately, rather than one for all twelve, saves 26 bits out of 47,668 — five hundredths of one per cent. The apparatus for choosing costs more than that to describe.
Half an index is three permutations
A phrase index stores a length, a source and a literal per phrase — and three orderings of its boundaries, at forty-five per cent of the structure. One of the three is the inverse of another, and nothing needs both at once.
A count that names every point
A grid that can report the points in a rectangle can be no smaller than log₂(z!) bits, because reporting them recovers the permutation. A grid that only counts them was left open as possibly smaller. Exactly, it cannot be: counts alone decode every point, and on the phrase index's 3,612 points 42,860 count questions recover the whole permutation, so an exact counter pays the same floor of 37,485 bits. Approximately it can — and the rectangles a phrase index asks are where an approximate count is worst. They hold 2.6 points on average and a third of them hold exactly one; the wavelet grid cut to eight of its twelve levels, 8% under the floor, counts 46% of them right.
The moves no stable sort can skip
Any sort must move every key that starts out of place, and with one spare cell it must spend one more move on each cycle of the permutation. Against that floor, a stable merge sort trimmed to the keys not already in place moves 1.46 times the least possible on records written in overlapping batches — the two thirds the prediction named, and for its reason. On a sorted file with a late tenth appended it was predicted within twice the floor and moves 15 times it, because the late tenth arrives in random order and sorting it among itself is two thirds of the work.
A test that is the top of the search
A phrase index asked whether a pattern occurs only needs to know, split by split, whether a rectangle of its grid holds a point. Stopping the grid's descent at six of its twelve levels was predicted to turn away nine in ten of the splits that come to nothing, for half the grid's bits. It turns away three in four, and the ones it lets through sit a median of five rows from a point. The test is never wrong, but the existence query had little left for it to save: patterns that occur are settled by their first split 94% of the time, and most patterns that do not occur never reach the grid.
Named alongside it
The objects these essays reach for when they reach for this one.
GridWavelet treeIndex sizeTrade offEntropyLower boundMeasurementBit vectorBurrows-wheeler transformCompressionInvertibilityMove to front