Adapted from Lectures by Index Compression Adapted from Lectures by Prabhakar Raghavan (Yahoo and Stanford) and Christopher Manning (Stanford) DONE SO FAR ======== Token -> Term => Postings Lists (Doc Ids ::REFINED TO:: DocIds to Hit Positions) (Also maintain: cumulative freq in dataset, doc occurrence count, term freq in each doc) ----------- Large Dictionary structure for efficient search : B-Trees Tolerant IR: Map queries to sets of terms TO postings lists (robust wrt spelling errors (caused by typing error to sound-based verbalization) + flexible expression of (alternative) multiple terms) Prasad L07IndexCompression

Last lecture – index construction Key step in indexing – sort This sort was implemented by exploiting disk-based sorting Fewer disk seeks Hierarchical merge of blocks Distributed indexing using MapReduce Running example of document collection: RCV1 Prasad L07IndexCompression

Today Collection statistics in more detail (RCV1) Dictionary compression Postings compression heap’s law: vocabulary size as a function of collection size : impacts size of dictionary : zipf’s law : distribution of collection frequency of terms : impacts size of postings : enables gap analysis Prasad L07IndexCompression

Recall Reuters RCV1 symbol statistic value N documents 800,000 L avg. # tokens per doc 200 M terms (= word types) ~400,000 avg. # bytes per token 6 (incl. spaces/punct.) avg. # bytes per token 4.5 (without spaces/punct.) avg. # bytes per term 7.5 non-positional postings 100,000,000 stopwords are normally really short Total tokens = 160M Terms = 0.4 M Non-positional postings = 100M Postings per document = Terms per document = 125 (= 100 M / 0.8M) Prasad L07IndexCompression

Why compression? Keep more stuff in memory (increases speed, caching effect) enables deployment on smaller/mobile devices Increase data transfer from disk to memory [read compressed data and decompress] is faster than [read uncompressed data] Premise: Decompression algorithms are fast True of the decompression algorithms we use Assuming that one keeps compressed form in memory DMA : during disk to memory transfer the cpu is free, so it can concurrently decompress. Prasad L07IndexCompression

Compression in inverted indexes First, we will consider space for dictionary Make it small enough to keep in main memory Then the postings Reduce disk space needed, decrease time to read from disk Large search engines keep a significant part of postings in memory (Each postings entry is a docID) larger dictionary faster access to postings list Prasad L07IndexCompression

Index parameters vs. what we index (details Table 5.1 p80) size of word types (terms) non-positional postings positional postings dictionary non-positional index positional index Size (K) ∆% cumul % ∆ % Unfiltered 484 109,971 197,879 No numbers 474 -2 100,680 -8 179,158 -9 Case folding 392 -17 -19 96,969 -3 -12 30 stopwords 391 -0 83,390 -14 -24 121,858 -31 -38 150 stopwords 67,002 -30 -39 94,517 -47 -52 stemming 322 -33 63,812 -4 -42 Effectiveness: Case folding 0: Lower casing does not alter the number of occurrences, hence positional index invariant. Non-positional index does change because there are documents containing different case-variants of a word. Stopwords 0: Because we are talking about 30 and 150 stop words, dictionary size does not change much. Both indices affected substantially because lots of entries removed and gaps reduced? Stemming 0: Same reason as “Case folding”. Exercise: give intuitions for all the ‘0’ entries. Why do some zero entries correspond to big deltas in other columns?

Lossless vs. lossy compression Lossless compression: All information is preserved. What we mostly do in IR. Lossy compression: Discard some information Several of the preprocessing steps can be viewed as lossy compression: case folding, stop words, stemming, number elimination. Chap/Lecture 7: Prune postings entries that are unlikely to turn up in the top k list for any query. Almost no loss of quality for top k list. Latent Semantic Indexing and Dimension Reduction techniques are lossy. Prasad L07IndexCompression

Vocabulary vs. collection size Heaps’ Law: M = kTb M is the size of the vocabulary, T is the number of tokens in the collection. Typical values: 30 ≤ k ≤ 100 and b ≈ 0.5. In a log-log plot of vocabulary vs. T, Heaps’ law is a line. Check it out on test document collections in the assignment (What about the variation in individual documents?) Implications: As the document dataset grows, the vocabulary grows monotonically – dictionary will eventually overflow memory. the vocabulary saturates. Prasad L07IndexCompression

Heaps’ Law For RCV1, the dashed line Fig 5.1 p81 For RCV1, the dashed line log10M = 0.49 log10T + 1.64 is the best least squares fit. Thus, M = 101.64T0.49 so k = 101.64 ≈ 44 and b = 0.49. What is the effect of including spelling errors, vs. automatically correcting spelling errors on Heaps’ law? b increases with errors and decreases with corrections Prasad L07IndexCompression

Zipf’s law We also study the relative frequencies of terms. In natural language, there are a few very frequent terms and very many rare terms. Zipf’s law: The ith most frequent term has frequency proportional to 1/i . cfi ∝ 1/i = c/i where c is a normalizing constant cfi is collection frequency: the number of occurrences of the term ti in the collection. Heaps’ Law gives the vocabulary size in collections. Zipf’s law: 2nd most frequent term occurs half as often as the most frequent term. (Is the distribution similar in both the document and the collection in practice?) Prasad L07IndexCompression

Zipf consequences If the most frequent term (the) occurs cf1 times then the second most frequent term (of) occurs cf1/2 times the third most frequent term (and) occurs cf1/3 times … Equivalent: cfi = c/i where c is a normalizing factor, so log cfi = log c - log i Linear relationship between log cfi and log i Use to do gap analysis to estimate the index size (see old “Index Compression” lecture) most frequent : 60 occurrences, 2nd : 30, 3rd : 20, …, 60th : 1 etc So 60 >= rank >= 1 will appear in all documents (GAP = 1), … 120 >= rank >= 61 will appear in alternate documents (GAP = 2), … Prasad L07IndexCompression

Compression First, we will consider space for dictionary and postings Basic Boolean index only No study of positional indexes, etc. We will devise compression schemes Prasad L07IndexCompression

Dictionary Compression Prasad L07IndexCompression

Why compress the dictionary Must keep in memory Search begins with the dictionary Memory footprint competition Embedded/mobile devices Prasad L07IndexCompression

Dictionary storage - first cut Array of fixed-width entries ~400,000 terms; 28 bytes/term = 11.2 MB. Addr = WORD size (+ M/c word boundary alignment issues) Dictionary search structure 20 bytes 4 bytes each L07IndexCompression

Fixed-width terms are wasteful Most of the bytes in the Term column are wasted – we allot 20 bytes for 1 letter terms. And we still can’t handle supercalifragilisticexpialidocious. Written English averages ~4.5 characters/word. Ave. dictionary word in English: ~8 characters How do we use ~8 characters per dictionary term? Short words dominate token counts but not token type (term) average. Exercise: Why is/isn’t this (4.5 char / word) the number to use for estimating the dictionary size? Stop words, which tend to be short, are eliminated from the dictionary. Prasad L07IndexCompression

Compressing the term list: Dictionary-as-a-String Store dictionary as a (long) string of characters: Pointer to next word shows end of current word Hope to save up to 60% of dictionary space. ….systilesyzygeticsyzygialsyzygyszaibelyiteszczecinszomo…. Total string length = 400K x 8B = 3.2MB How do we map words to pointers to do binary search of pointers? We can still do binary search as usual as long as the strings are in sorted order. Only difference is that it requires an additional step of dereferencing from term ptr to string. -------------------------------------------------------------------------------------------------------------------------------- Average term = 8 bytes Total number of terms = 400,000 To address 3.2 MB requires 22 bits or approx. 3 bytes So total storage = 4 + 4 + 3 = 11 bytes (40% of 28 bytes) Table size = 400K * 11 B = 4.4MB Total compressed dictionary size = 7.6 MB Total saving = 1- 7.6/11 ~ 30% Pointers resolve 3.2M positions: log23.2M = 22bits = 3bytes Prasad L07IndexCompression

Space for dictionary as a string 4 bytes per term for Freq. 4 bytes per term for pointer to Postings. 3 bytes per term pointer Avg. 8 bytes per term in term string 400K terms x 19  7.6 MB (against 11.2MB for fixed width)  Now avg. 11  bytes/term,  not 20. 11 bytes per array entry + (Avg.) 8 bytes per string table entry = 19 bytes per term (as opposed to 28) Total savings = 33% (1 – 7.6/11.2) Prasad L07IndexCompression

Blocking Store pointers to every kth term string. Example below: k=4. Need to store term lengths (1 extra byte) ….7systile9syzygetic8syzygial6syzygy11szaibelyite8szczecin9szomo…. For every four terms : allocate 4 bytes extra for string length but save 3 * 3 bytes in term pointer Net saving per term = (+ 4 – 9)/4 = -5 / 4 So average storage per term = 19 (4+4+3+8) - 1.25 = 17.75 bytes per term  Save 9 bytes  on 3  pointers. Lose 4 bytes on term lengths.

Net Shaved another ~0.5MB; can save more with larger k. Where we used 3 bytes/pointer without blocking 3 x 4 = 12 bytes for k=4 pointers, now we use 3+4=7 bytes for 4 pointers. Shaved another ~0.5MB; can save more with larger k. 5 bytes saved for every 4 terms Overall saving : 5 / 4 * 400K terms = 0.5 MB Space-time Tradeoff : Need to search within a block at run time (For large k, run-time overhead can be prohibitive – non-responsive search) Why not go with larger k? Prasad L07IndexCompression

Exercise Estimate the space usage (and savings compared to 7.6 MB) with blocking, for block sizes of k = 4, 8 and 16. For k = 8. For every block of 8, need to store extra 8 bytes for length For every block of 8, can save 7 * 3 bytes for term pointer Saving (+8 – 21)/8 * 400K = 0.65 MB For k = 16. For every block of 16, need to store extra 16 bytes for length For every block of 16, can save 15 * 3 bytes for term pointer Saving (+16 – 45)/16 * 400K = 0.725 MB ----------------------------------------------------------------------------- Search trade-off explained in the next overhead Prasad L07IndexCompression

Dictionary search without blocking Assuming each dictionary term equally likely in query (not really so in practice!), average number of comparisons = (1+2∙2+4∙3+4)/8 ~2.6 Prasad L07IndexCompression

Dictionary search with blocking Binary search down to 4-term block; Then linear search through terms in block. Blocks of 4 (binary tree), avg. = (1+2∙2+2∙3+2∙4+5)/8 = 3 compares Prasad L07IndexCompression

Exercise Estimate the impact on search performance (and slowdown compared to k=1) with blocking, for block sizes of k = 4, 8 and 16. logarithmic search time to get to get to (n/k) “leaves” linear time proportional to k/2 for subsequent search through the “leaves” closed-form solution not obvious Prasad L07IndexCompression

Front coding Front-coding: Sorted words commonly have long common prefix – store differences only (for last k-1 in a block of k) 8automata8automate9automatic10automation 8automat*a1e2ic3ion Extra length beyond automat. Encodes automat Begins to resemble general string compression.

RCV1 dictionary compression Technique Size in MB Fixed width 11.2 String with pointers to every term 7.6 Blocking k=4 7.1 Blocking + front coding 5.9 Prasad L07IndexCompression

Postings compression Prasad L07IndexCompression

Postings compression The postings file is much larger than the dictionary, by a factor of at least 10. Key desideratum: store each posting compactly. A posting for our purposes is a docID. For Reuters (800,000 documents), we would use 32 bits per docID when using 4-byte integers. Alternatively, we can use log2 800,000 ≈ 20 bits per docID. Our goal: use a lot less than 20 bits per docID. Prasad L07IndexCompression

Postings: two conflicting forces A term like arachnocentric occurs in maybe one doc out of a million – we would like to store this posting using log2 1M ~ 20 bits. A term like the occurs in virtually every doc, so 20 bits/posting is too expensive. Prefer 0/1 bitmap vector in this case Arachnocentric hits seem to be just IR lectures! Arachnids are a class (Arachnida) of joint-legged invertebrate animals, including spiders, scorpions, harvestmen, ticks, and mites. Represent doc id via numbers or bit-vectors. Prasad L07IndexCompression

Postings file entry We store the list of docs containing a term in increasing order of docID. computer: 33,47,154,159,202 … Consequence: it suffices to store gaps. 33,14,107,5,43 … Hope: most gaps can be encoded/stored with far fewer than 20 bits. Prasad L07IndexCompression

Three postings entries Prasad L07IndexCompression

Variable length encoding Aim: For arachnocentric, we will use ~20 bits/gap entry. For the, we will use ~1 bit/gap entry. If the average gap for a term is G, we want to use ~log2G bits/gap entry. Key challenge: encode every integer (gap) with ~ as few bits as needed for that integer. Variable length codes achieve this by using short codes for small numbers Even though we store variable length codes on the disk we can always expand it out after moving to main memory for efficient handling (merge, word-boundary access, etc) Prasad L07IndexCompression

Variable Byte (VB) codes For a gap value G, use close to the fewest bytes needed to hold log2 G bits Begin with one byte to store G and dedicate 1 bit in it to be a continuation bit c If G ≤127, binary-encode it in the 7 available bits and set c =1 Else encode G’s lower-order 7 bits and then use additional bytes to encode the higher order bits using the same algorithm At the end set the continuation bit of the last byte to 1 (c =1) and of the other bytes to 0 (c =0).

Example Postings stored as the byte concatenation docIDs 824 829 215406 gaps 5 214577 VB code 00000110 10111000 10000101 00001101 00001100 10110001 For a small gap (5), VB uses a whole byte. Postings stored as the byte concatenation 00000110 10111000 10000101 00001101 00001100 10110001 Key property: VB-encoded postings are uniquely prefix-decodable. 2^7 = 128 decimal(824) = binary(110,0111000) = 6*128 + 56 decimal(214577) = binary(1101,0001100,0110001) = 13 * 128 * 128 + 12 * 128 + 49 Read first bit. If 0, use remaining 7 bits as part of MSB or its follower byte but not LSB. O/w, that is the LSB. (Do not include continuation bit for determining the value.) Prasad

Other variable codes Instead of bytes, we can also use a different “unit of alignment”: 32 bits (words), 16 bits, 4 bits (nibbles) etc. Variable byte alignment wastes space if you have many small gaps – nibbles do better in such cases. Prasad L07IndexCompression

Gamma codes Can compress better with bit-level codes The Gamma code is the best known of these. Represent a gap G as a pair length and offset offset is G in binary, with the leading bit cut off For example 13 → 1101 → 101 length is the length of offset For 13 (offset 101), this is 3. Encode length in unary code: 1110. Gamma code of 13 is the concatenation of length and offset: 1110101 Gap = 5 length = 110 offset = 5 – 4 Gamma-code = (110, 01) Gap = 4 length = 110 offset = 4 – 4 Gamma-code = (110, 00) Gap = 9 length = 1110 offset = 9 – 8 Gamma-code = (1110, 001) Gap = 2 length = 10 offset = 2 – 2 Gamma-code = (10, 0) Exercise: what is the g code for 1? Exercise: does zero have a g code? Gap = 1 length = 0 offset = 1 – 1 Gamma-code = (0, -) Zero does not have a gamma-code. Prasad L07IndexCompression

Gamma code examples number length offset g-code none 1 2 10 10,0 3 none 1 2 10 10,0 3 10,1 4 110 00 110,00 9 1110 001 1110,001 13 101 1110,101 24 11110 1000 11110,1000 511 111111110 11111111 111111110,11111111 1025 11111111110 0000000001 11111111110,0000000001 gamma-code(9) = transform(1001) = chop leftmost bit length -> 1110, offset -> 001 gamma-code(9) = 1110,001 decode(1110101) = decode(1110,101) = 2^3 + 101 = 8 + 5 = 13 decode(111101000) = decode(11110,1000) = 2^4 + 1000 = 16 + 8 = 24 Prasad L07IndexCompression

Exercise Given the following sequence of g-coded gaps, reconstruct the postings sequence: 1110001110101011111101101111011 Gamma Code : UNARY-LENGTH 0 BINARY-OFFSET (length of UNARY-LENGTH same as length of BINARY-OFFSET) (zero centered) (Binary value of the number is “1 BINARY-OFFSET”) Decoded Gaps: 9, 6, 3, 59 (=32+16+8+3), 7(4+3) Postings: 9, 15, 18, 77, 84 From these g-codes -- decode and reconstruct gaps, then full postings.

Gamma code properties Uniquely prefix-decodable, like VB All gamma codes have an odd number of bits G is encoded using 2 log G +1 bits Almost within a factor of 2 of best possible Offset ~ log G bits length ~ log G bits middle bit ~ 1 Prasad L07IndexCompression

Gamma seldom used in practice Machines have word boundaries – 8, 16, 32 bits Compressing and manipulating at individual bit-granularity will slow down query processing Variable byte alignment is potentially more efficient Regardless of efficiency, variable byte is conceptually simpler at little additional space cost Prasad L07IndexCompression

RCV1 compression Data structure Size in MB dictionary, fixed-width 11.2 dictionary, term pointers into string 7.6 with blocking, k = 4 7.1 with blocking & front coding 5.9 collection (text, xml markup etc) 3,600.0 collection (text) 960.0 Term-doc incidence matrix 40,000.0 postings, uncompressed (32-bit words) 400.0 postings, uncompressed (20 bits) 250.0 postings, variable byte encoded 116.0 postings, g-encoded 101.0 Prasad L07IndexCompression

Index compression summary We can now create an index for highly efficient Boolean retrieval that is very space efficient Only 4% of the total size of the collection Only 10-15% of the total size of the text in the collection However, we’ve ignored positional information Hence, space savings are less for indexes used in practice But techniques substantially the same. Prasad L07IndexCompression

Models of Natural Language Necessary for analysis : Size estimation

Text properties/model How are different words distributed inside each document? Zipf’s Law: The frequency of ith most frequent word is 1/i times that of the most frequent word. 50% of the words are stopwords. How are words distributed across the documents in the collection? Fraction of documents containing a word k times follows binomial distribution.

Probability of occurrence of a symbol depends on previous symbol Probability of occurrence of a symbol depends on previous symbol. (Finite-Context or Markovian Model) The number of distinct words in a document (vocabulary) grows as the square root of the size of the document. (Heap’s Law) The average length of non-stop words is 6 to 7 letters.

Similarity Hamming Distance between a pair of strings of same length is the number of positions that have different characters. Levenshtein (Edit) Distance is the minimum number of character insertions, deletions, and substituitions needed to make two strings the same. (Extensions include transposition, weighted operations, etc) UNIX diff utility uses Longest Common Subsequence, obtained by deletion, to align strings/words/lines.

Index Size Estimation Using Zipf’s Law SKIP : from old slides

Corpus size for estimates Consider N = 1M documents, each with about L=1K terms. Avg 6 bytes/term incl. spaces/punctuation 6GB of data. Say there are m = 500K distinct terms among these.

Recall: Don’t build the matrix A 500K x 1M matrix has half-a-trillion 0’s and 1’s (500 billion). But it has no more than one billion 1’s. matrix is extremely sparse. So we devised the inverted index Devised query processing for it Now let us estimate the size of index

Rough analysis based on Zipf The i th most frequent term has frequency proportional to 1/i Let this frequency be c/i. Then The k th Harmonic number is Thus c = 1/Hm , which is ~ 1/ln m = 1/ln(500k) ~ 1/13. So the i th most frequent term has frequency roughly 1/13i. Frequency is (unitless) fraction of words, not count. Graph ln x and area under the curve approximates area of rectangles = summation of 1/I : Application of integration (area under the curve) Rationale: 1 unit is divided as 1/13 + 1/26 + 1/39+ … 500,000 terms In a document with 1000 terms: most frequent term appears 76 times, 2nd most frequent term appears 38 times, …

Postings analysis (contd.) Expected number of occurrences of the i th most frequent term in a doc of length L is: Lc/i ≈ L/13i ≈ 76/i for L=1000. Let J = Lc ~ 76. Then the J most frequent terms are likely to occur in every document. Now imagine the term-document incidence matrix with rows sorted in decreasing order of term frequency: Then the most frequent term is likely to occur 76 times in the document with 1K terms 76th most frequent term is likely to occur once in every document So Gap size = 1 in both these cases.

Informal Observations Most frequent term appears approx 76 times in each document. 2nd most frequent term appears approx 38 times in each document. … 76th most frequent term appears approx once in each document. First 76 terms appear at least once in each document. Next 76 terms appear at least once in every two documents. … Prasad L07IndexCompression

Rows by decreasing frequency N docs J most frequent terms. N gaps of ‘1’ each. J next most frequent terms. N/2 gaps of ‘2’ each. m terms J next most frequent terms. N/3 gaps of ‘3’ each. etc.

J-row blocks In the i th of these J-row blocks, we have J rows each with N/i gaps of i each. Encoding a gap of i using Gamma codes takes 2log2 i +1 bits. So such a row uses space ~ (2N log2 i )/i bits. For the entire block, (2N J log2 i )/i bits, which in our case is ~ 1.5 x 108 (log2 i )/i bits. Sum this over i from 1 up to m/J = 500K/76 ≈ 6500. (Since there are m/J blocks.) 2 * N * J = 2 * 1 M * 76 ** Using Gamma code for encoding gaps ** Sigma_1 to 6500 log i / i = integrate 1 to 6500 ( ln i / i di ) (1 / ln 2) = [ln i]^2 / 2 ln 2 = (( ln 6500 ) ^ 2 ) / 2 ln 2 = = (8.78)^2 / 2 * 0.69 = 38 / 0.69 = 55

Exercise Work out the above sum and show it adds up to about 55 x 150 Mbits, which is about 1GByte. So we’ve taken 6GB of text and produced from it a 1GB index that can handle Boolean queries! Neat! (16.7%) Make sure you understand all the approximations in our probabilistic calculation. Sum adds up to 55 x 150 Mbits as given in the original lecture