Algorithms & Data Structures
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The thread: An index has a size — page 3

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bits heldbalanced49,593huffman41,589the answer comes backbalanced: sorted · 26 operationshuffman: sorted · 24 operations32 documents · zipf lengths83.9% of the ordered tree What the libraries do

The smaller tree hands it back unsorted

A frequency-shaped tree is sixteen per cent smaller and returns its documents in code order. A balanced one is larger and returns them sorted. The trade is d log d comparisons against a saving, which is not close — until the answer is truncated.

 equal lengths5.00 bits100.0%lengths as one over rank4.15 bits83.4%a few long, many short4.31 bits86.7%one document holding most of the text1.69 bits39.5%log₂ 3232 documents · 8,192 characters100.0% to 39.5% What is taught wrongly

A saving quoted without its collection

A check asking whether a compressed document array is smaller than the plain one passes on the rounding whenever the document count is not a power of two. It would report a saving of nothing as sixteen per cent, on a collection that has no redundancy at all.

 02040255075100125one sampled position in every …of the index, per centplain marks: 18.5%Elias-Fano: 3.8%16,384 characters14.9% off at one in 32 The index that replaces the text

What the locating apparatus becomes

The two parts that answer "where" are half an index at a dense sampling and a fifth at a sparse one, and the fifth does not fall further. Represent the marks properly and it keeps falling, to under four per cent.

 a loop over the alphabet1,478,400the compound walk, inside the loop134,40011xone descent per node18,91678x6 patterns · 1 error · sigma 2155.8% of the extensions were dead The index that replaces the text

Three savings on one structure

A factor of eleven on an extension, a factor of seven on a branching search, and a sixth of the bits. Applied to one bidirectional index they do not give a factor of seventy-seven, and the reason is that two of the three are the same saving.

 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% Structures

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.

 both halves locate250,584100.0%the reverse half counts only226,50490.4%and its marks are Elias-Fano210,93484.2%the saving spent on sampling241,47596.4%16,384 characters · one in 3284.2% smaller, or the same size and faster The other axis

The ladder, and the rung that spends

Two hundred and fifty thousand bits, then two hundred and twenty-six, then two hundred and eleven. The fourth rung takes the whole saving and buys a four-times denser sampling with it, landing at ninety-six per cent of where it started and locating several times faster.

 01e+42e+43e+44e+450100one sampled position in every …bitsplain, one bit a rowcompressed blocksElias-Fano16,384 rowscrosses at one in 4 What is taught wrongly

Where the sparse representation loses

At every row marked, Elias–Fano costs twice the plain vector. The crossing is at one row in four, which is a sampling rate a real index uses — so the choice between the two is a choice, not an improvement.

 0102030100200one sampled position in every …of both halves saved, per centplain marks: 13.5%Elias-Fano: 1.8%16,384 characters17.6x against 2.3x What the libraries do

The floor was the marks

A saving reported as about a sixth of a bidirectional index, falling to an eighth and levelling off. Represent one array properly and it falls to a fiftieth instead — most of what was being dropped was a badly encoded bit vector.

 phrase lengths2,71222.6%phrase sources2,71222.6%phrase literals1,1309.4%boundary orders5,42445.3%three permutations of 226 elements: 1,808 bits each, and one is the inverse of another226 phrases · 4,096 characters45.3% in the orders The other axis

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.

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

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.

 0100200300400102030copies of the basebits held during one querythe visited set: 420the frontier: 2666-character patterns20x to 1.58x What is taught wrongly

A constant factor, not a term

The substitution was expected to remove a quantity proportional to the answer. Both traversals hold a quantity proportional to the answer, and what it removes is a factor of 1.6 that shrinks as the collection grows.

 phrases examined per probethe linear scan207.0sorted, with a running maximum10.9 — 19xthe running maximum costs 2,821 bits — 21.6% of the index40 of 40 positions checked, and the two agree at all of them238 phrases · 6,144 characters19x on the scan What the libraries do

The scan the order does not touch

Two hundred and seven phrases examined per probe, against eleven. A running maximum of the source regions' right ends lets a leftward walk stop for good, and it costs twenty-two per cent more bits.

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