Pandu Nayak and Prabhakar Raghavan
Lecture 5: Index Compression
Problem set 1 due Thursday
Programming exercise 1 will be handed out today
Collection statistics in more detail (with RCV1)
How big will the dictionary and postings be?
Dictionary compression
Postings compression
Dictionary
Make it small enough to keep in main memory
Make it so small that you can keep some postings lists in main memory too
Postings file(s)
Reduce disk space needed
Decrease time needed to read postings lists from disk
Large search engines keep a significant part of the postings in memory.
Compression lets you keep more in memory
We will devise various IRspecific compression schemes
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
nonpositional postings 100,000,000
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 quality for top k list.
How big is the term vocabulary?
That is, how many distinct words are there?
Can we assume an upper bound?
Not really: At least 70^{20 }= 10^{37 }different words of length 20
In practice, the vocabulary will keep growing with the collection size
Especially with Unicode :)
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 loglog plot of vocabulary size M vs. T, Heaps’ law predicts a line with slope about ½
It is the simplest possible relationship between the two in loglog space
An empirical finding (“empirical law”)
For RCV1, the dashed line log_{10}M = 0.49 log_{10}T + 1.64 is the best least squares fit. Thus, M = 10^{1.64}T^{0.49 }so k = 10^{1.64} ≈ 44 and b = 0.49. Good empirical fit for Reuters RCV1 ! For first 1,000,020 tokens, law predicts 38,323 terms; actually, 38,365 terms. 
Compute the vocabulary size M for this scenario:
Looking at a collection of web pages, you find that there are 3000 different terms in the first 10,000 tokens and 30,000 different terms in the first 1,000,000 tokens.
Assume a search engine indexes a total of 20,000,000,000 (2 × 1010) pages, containing 200 tokens on average
What is the size of the vocabulary of the indexed collection as predicted by Heaps’ law?
Heaps’ law gives the vocabulary size in collections.
We also study the relative frequencies of terms.
In natural language, there are a few very frequent terms and very many very rare terms.
Zipf’s law: The ith most frequent term has frequency proportional to 1/i .
cf_{i} ∝ 1/i = K/i where K is a normalizing constant
cf_{i} is collection frequency: the number of occurrences of the term t_{i} in the collection.
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 = K/i where K is a normalizing factor, so
log cfi = log K  log i
Linear relationship between log cfi and log i
Another power law relationship
Now, we will consider compressing the space for the dictionary and postings
Basic Boolean index only
No study of positional indexes, etc.
We will consider compression schemes
We want to keep it in memory
Memory footprint competition with other applications
Embedded/mobile devices may have very little memory
Even if the dictionary isn’t in memory, we want it to be small for a fast search startup time
So, compressing the dictionary is important
↓ 20 bytes 4 bytes each
Dictionary search structure
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 or hydrochlorofluorocarbons.
Written English averages ~4.5 characters/word.
Exercise: Why is/isn’t this the number to use for estimating the dictionary size?
Ave. dictionary word in English: ~8 characters
How do we use ~8 characters per dictionary term?
Short words dominate token counts but not type average.
Store dictionary as a (long) string of characters:
Pointer to next word shows end of current word
4 bytes per term for pointer to Postings. → Now avg. 11 bytes/term,
3 bytes per term pointer ...Not 20
Avg. 8 bytes per term in term string
400K terms x 19 ⇒7.6 MB (against 11.2MB for fixed width)
Store pointers to every kth term string.
Example below: k=4.
Need to store term lengths (1 extra byte)
Where we used 3 bytes/pointer without blocking
3 x 4 = 12 bytes,
now we use 3 + 4 = 7 bytes.
Shaved another ~0.5MB. This reduces the size of the dictionary from 7.6 MB to 7.1 MB.
We can save more with larger k.
Why not go with larger k?
Estimate the space usage (and savings compared to 7.6 MB) with blocking, for block sizes of k = 4, 8 and 16.

Büinary search down to 4term 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
Estimate the impact on search performance (and slowdown compared to k=1) with blocking, for block sizes of k = 4, 8 and 16.
Frontcoding:
Sorted words commonly have long common prefix – store differences only
(for last k1 in a block of k)
8automata8automate9automatic10automation
→ 8automat*a1♦e2♦ic3♦ion
↑ ↑
Encodes automat Extra length beyond automat.
Begins to resemble general string compression.










The postings file is much larger than the dictionary, 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 4byte integers.
Alternatively, we can use log2 800,000 ≈ 20 bits per docID.
Our goal: use far fewer than 20 bits per docID.
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
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.
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 about as few bits as needed for that integer.
This requires a variable length encoding
Variable length codes achieve this by using short codes for small numbers
For a gap value G, we want to 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, binaryencode it in the 7 available bits and set c =1
Else encode G’s lowerorder 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 for the other bytes c = 0.
Instead of bytes, we can also use a different “unit of alignment”: 32 bits (words), 16 bits, 4 bits (nibbles).
Variable byte alignment wastes space if you have many small gaps – nibbles do better in such cases.
Variable byte codes:
Used by many commercial/research systems
Good lowtech blend of variablelength coding and sensitivity to computer memory alignment matches (vs. bitlevel codes, which we look at next).
There is also recent work on wordaligned codes that pack a variable number of gaps into one word
Represent n as n 1s with a final 0.
Unary code for 3 is 1110.
Unary code for 40 is
11111111111111111111111111111111111111110 .
Unary code for 80 is:
111111111111111111111111111111111111111111111111111111111111111111111111111111110
This doesn’t look promising, but….
We can compress better with bitlevel 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.
We encode length with unary code: 1110.
Gamma code of 13 is the concatenation of length and offset: 1110101
G is encoded using 2 ⌊ log G⌋ + 1 bits
Length of offset is ⌊ log G ⌋ bits
Length of length is ⌊log G ⌋ + 1 bits
All gamma codes have an odd number of bits
Almost within a factor of 2 of best possible, log_{2} G
Gamma code is uniquely prefixdecodable, like VB
Gamma code can be used for any distribution
Gamma code is parameterfree
Machines have word boundaries – 8, 16, 32, 64 bits
Operations that cross word boundaries are slower
Compressing and manipulating at the granularity of bits can be slow
Variable byte encoding is aligned and thus potentially more efficient
Regardless of efficiency, variable byte is conceptually simpler at little additional space cost
Only 4% of the total size of the collection
Only 1015% 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.
IIR 5
MG 3.3, 3.4.
F. Scholer, H.E. Williams and J. Zobel. 2002. Compression of Inverted Indexes For Fast Query Evaluation. Proc. ACMSIGIR 2002.
Variable byte codes
V. N. Anh and A. Moffat. 2005. Inverted Index Compression Using WordAligned Binary Codes. Information Retrieval 8: 151–166.
Word aligned codes