1. 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

2. 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
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L07IndexCompression

3. 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

4. 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

5. 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

6. 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

7. 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?

8. 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

9. 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

10. Heaps’ Law For RCV1, the dashed line
Fig 5.1 p81
For RCV1, the dashed line
log10M = 0.49 log10T is the best least squares fit.
Thus, M = T0.49 so k = ≈ 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

11. 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

12. 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), …
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L07IndexCompression

13. 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
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L07IndexCompression

14. Dictionary Compression
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L07IndexCompression

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

16. 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

17. 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

18. 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 = = 11 bytes (40% of 28 bytes) Table size = 400K * 11 B = 4.4MB
Total compressed dictionary size = 7.6 MB
Total saving = /11 ~ 30%
Pointers resolve 3.2M
positions: log23.2M =
22bits = 3bytes
Prasad
L07IndexCompression

19. 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

20. 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 ( ) = bytes per term
 Save 9 bytes
 on 3
 pointers.
Lose 4 bytes on
term lengths.

21. 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

22. 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 = MB
Search trade-off explained in the next overhead
Prasad
L07IndexCompression

23. 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
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L07IndexCompression

24. 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

25. 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

26. 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.

27. 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
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L07IndexCompression

28. Postings compression
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L07IndexCompression

29. 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

30. 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

31. 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.
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L07IndexCompression

32. Three postings entries
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L07IndexCompression

33. 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

34. 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).

35. Example Postings stored as the byte concatenation
docIDs
824
829
215406
gaps 5
214577
VB code
For a small gap (5), VB
uses a whole byte.
Postings stored as the byte concatenation
Key property: VB-encoded postings are
uniquely prefix-decodable.
2^7 = 128
decimal(824) = binary(110, ) = 6*
decimal(214577) = binary(1101, , ) = 13 * 128 * *
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

36. 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

37. 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:
Gap = 5 length = 110 offset = 5 – Gamma-code = (110, 01)
Gap = 4 length = 110 offset = 4 – Gamma-code = (110, 00)
Gap = 9 length = offset = 9 – Gamma-code = (1110, 001)
Gap = 2 length = 10 offset = 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 – Gamma-code = (0, -)
Zero does not have a gamma-code.
Prasad
L07IndexCompression

38. 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 ,
1025 ,
gamma-code(9) = transform(1001) =
chop leftmost bit
length -> 1110, offset -> 001
gamma-code(9) = 1110,001
decode( ) = decode(1110,101) =
2^ = = 13
decode( ) = decode(11110,1000) =
2^ = = 24
Prasad
L07IndexCompression

39. Exercise
Given the following sequence of g-coded gaps, reconstruct the postings sequence:
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 (= ), 7(4+3)
Postings: , 15, 18, 77, 84
From these g-codes -- decode and
reconstruct gaps, then full postings.

40. 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

41. 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

42. 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

43. 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

44. Models of Natural Language
Necessary for analysis : Size estimation

45. 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.

46. 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.

47. 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.

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

49. 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.

50. 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

51. 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, …

52. 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.

53. 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

54. 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.

55. 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 ≈ (Since there are m/J blocks.)
2 * N * J = 2 * 1 M * 76
** Using Gamma code for encoding gaps **
Sigma_1 to log i / i = integrate 1 to ( 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

56. 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