zone indexes paolo ferragina dipartimento di informatica università di pisa reading 6.1

Zone indexes

Paolo FerraginaDipartimento di Informatica

Università di Pisa

Reading 6.1

Parametric and zone indexes

Thus far, a doc has been a term sequence

But documents have multiple parts: Author Title Date of publication Language Format etc.

These are the metadata about a document

Sec. 6.1

Zone

A zone is a region of the doc that can contain an arbitrary amount of text e.g.,

Title Abstract References …

Build inverted indexes on fields AND zones to permit querying

E.g., “find docs with merchant in the title zone and matching the query gentle rain”

Sec. 6.1

Example zone indexes

Encode zones in dictionary vs. postings.

Sec. 6.1

Tiered indexes

Break postings up into a hierarchy of lists Most important … Least important

Inverted index thus broken up into tiers of decreasing importance

At query time use top tier unless it fails to yield K docs

If so drop to lower tiers

Sec. 7.2.1

Example tiered index

Sec. 7.2.1

Index construction:Compression of postings

Paolo FerraginaDipartimento di Informatica

Università di Pisa

Reading 5.3 and a paper

code for integer encoding

x > 0 and Length = log2 x +1

e.g., 9 represented as <000,1001>.

code for x takes 2 log2 x +1 bits (ie. factor of 2 from optimal)

0000...........0 x in binary Length-1

Optimal for Pr(x) = 1/2x2, and i.i.d integers

It is a prefix-free encoding…

Given the following sequence of coded integers, reconstruct the original sequence:

0001000001100110000011101100111

8 6 3 59 7

code for integer encoding

Use -coding to reduce the length of the first field

Useful for medium-sized integers

e.g., 19 represented as <00,101,10011>.

coding x takes about log2 x + 2 log2( log2 x ) + 2 bits.

(Length) x

Optimal for Pr(x) = 1/2x(log x)2, and i.i.d integers

Variable-bytecodes [10.2 bits per TREC12]

Wish to get very fast (de)compress byte-align

Given a binary representation of an integer Append 0s to front, to get a multiple-of-7 number of bits Form groups of 7-bits each Append to the last group the bit 0, and to the other

groups the bit 1 (tagging)

e.g., v=214+1 binary(v) = 10000000000000110000001 10000000 00000001

Note: We waste 1 bit per byte, and avg 4 for the first byte.

But it is a prefix code, and encodes also the value 0 !!

PForDelta coding

10 11 11 …01 01 11 11 01 42 2311 10

2 3 3 …1 1 3 3 23 13 42 2

a block of 128 numbers = 256 bits = 32 bytes

Use b (e.g. 2) bits to encode 128 numbers or create exceptions

Encode exceptions: ESC or pointers

Choose b to encode 90% values, or trade-off: b waste more bits, b more exceptions

Translate data: [base, base + 2b-1] [0,2b-1]

Index construction:Compression of documents

Paolo FerraginaDipartimento di Informatica

Università di Pisa

Reading Managing-Gigabytes: pg 21-36, 52-56, 74-79

Uniquely Decodable Codes

A variable length code assigns a bit string (codeword) of variable length to every symbol

e.g. a = 1, b = 01, c = 101, d = 011

What if you get the sequence 1011 ?

A uniquely decodable code can always be uniquely decomposed into their codewords.

Prefix Codes

A prefix code is a variable length code in which no codeword is a prefix of another one

e.g a = 0, b = 100, c = 101, d = 11Can be viewed as a binary trie

0 1

a

b c

d

0

0 1

1

Average Length

For a code C with codeword length L[s], the average length is defined as

p(A) = .7 [0], p(B) = p(C) = p(D) = .1 [1--]

La = .7 * 1 + .3 * 3 = 1.6 bit (Huffman achieves 1.5 bit)

We say that a prefix code C is optimal if for all prefix codes C’, La(C) La(C’)

Ss

a sLspCL ][)()(

Entropy (Shannon, 1948)

For a source S emitting symbols with probability p(s), the self information of s is:

bits

Lower probability higher information

Entropy is the weighted average of i(s)

Ss sp

spSH)(

1log)()( 2

)(

1log)( 2 sp

si

s s

s

occ

T

T

occTH

||log

||)( 20

0-th order empirical entropy of string T

i(s)

Performance: Compression ratio

Compression ratio =

#bits in output / #bits in input

Compression performance: We relate entropy against compression ratio.

p(A) = .7, p(B) = p(C) = p(D) = .1

H ≈ 1.36 bits

Huffman ≈ 1.5 bits per symb

||

|)(|)(0 T

TCvsTH

s

scspSH |)(|)()(Shannon In practiceAvg cw length

Empirical H vs Compression ratio

|)(|)(|| 0 TCvsTHT

Statistical Coding

How do we use probability p(s) to encode s?

Huffman codes

Arithmetic codes

Document Compression

Huffman coding

Huffman Codes

Invented by Huffman as a class assignment in ‘50.

Used in most compression algorithms gzip, bzip, jpeg (as option), fax compression,…

Properties: Generates optimal prefix codes Cheap to encode and decode La(Huff) = H if probabilities are powers of 2

Otherwise, La(Huff) < H +1 < +1 bit per symb on avg!!

Running Example

p(a) = .1, p(b) = .2, p(c ) = .2, p(d) = .5a(.1) b(.2) d(.5)c(.2)

a=000, b=001, c=01, d=1There are 2n-1 “equivalent” Huffman trees

(.3)

(.5)

(1)

What about ties (and thus, tree depth) ?

0

0

0

11

1

Encoding and Decoding

Encoding: Emit the root-to-leaf path leading to the symbol to be encoded.

Decoding: Start at root and take branch for each bit received. When at leaf, output its symbol and return to root.

a(.1) b(.2)

(.3) c(.2)

(.5) d(.5)0

0

0

1

1

1

abc... 00000101

101001... dcb

Problem with Huffman Coding

Take a two symbol alphabet = {a,b}.Whichever is their probability, Huffman uses 1

bit for each symbol and thus takes n bits to encode a message of n symbols

This is ok when the probabilities are almost the same, but what about p(a) = .999.

The optimal code for a is bits So optimal coding should use n *.0014 bits,

which is much less than the n bits taken by Huffman

00144.)999log(.

Document Compression

Arithmetic coding

Introduction

It uses “fractional” parts of bits!!

Gets nH(T) + 2 bits vs. nH(T)+n of Huffman

Used in JPEG/MPEG (as option), Bzip

More time costly than Huffman, but integer implementation is not too bad.

Ideal performance. In practice, it is 0.02 * n

Symbol interval

Assign each symbol to an interval range from 0 (inclusive) to 1 (exclusive).

e.g.

a = .2

c = .3

b = .5

cum[c] = p(a)+p(b) = .7

cum[b] = p(a) = .2

cum[a] = .0

The interval for a particular symbol will be calledthe symbol interval (e.g for b it is [.2,.7))

Sequence interval

Coding the message sequence: bac

The final sequence interval is [.27,.3)

a = .2

c = .3

b = .5

0.0

0.2

0.7

1.0

a

c

b

0.2

0.3

0.55

0.7

a

c

b

0.2

0.22

0.27

0.3(0.7-0.2)*0.3=0.15

(0.3-0.2)*0.5 = 0.05

(0.3-0.2)*0.3=0.03

(0.3-0.2)*0.2=0.02(0.7-0.2)*0.2=0.1

(0.7-0.2)*0.5 = 0.25

The algorithm

To code a sequence of symbols with probabilities

pi (i = 1..n) use the following algorithm:

0

1

0

0

l

s iiii

iii

Tcumsll

Tpss

*11

1 *

p(a) = .2

p(c) = .3

p(b) = .5

0.27

0.2

0.3

2.0

1.0

1

1

i

i

l

s

03.03.0*1.0 is

27.0)5.02.0(*1.02.0 il

The algorithm

Each message narrows the interval by a factor p[Ti]

Final interval size is

1

0

0

0

s

l

n

iin Tps

1

iii

iiii

Tpss

Tcumsll

*1

*11

Sequence interval[ ln , ln + sn )

Take a number inside

Decoding Example

Decoding the number .49, knowing the message is of length 3:

The message is bbc.

a = .2

c = .3

b = .5

0.0

0.2

0.7

1.0

a

c

b

0.2

0.3

0.55

0.7

a

c

b

0.3

0.35

0.475

0.55

0.490.49

0.49

How do we encode that number?

If x = v/2k (dyadic fraction) then the

encoding is equal to bin(v) over k digits (possibly padded with 0s in front)

0111.16/7

11.4/3

01.3/1

How do we encode that number?

Binary fractional representation:

FractionalEncode(x)1. x = 2 * x2. If x < 1 output 03. else {output 1; x = x - 1; }

.... 54321 bbbbb

...2222 44

33

22

11 bbbb

01.3/1

2 * (1/3) = 2/3 < 1, output 0

2 * (2/3) = 4/3 > 1, output 1 4/3 – 1 = 1/3Incremental Generation

Which number do we encode?

Truncate the encoding to the first d = log2 (2/sn) bits

Truncation gets a smaller number… how much smaller?

Truncation Compression

d

i

id

i

id

i

ididb

222212

11

)(

1

)(

2222

log2

log 22nss snn

ln + sn

ln

ln + sn/2

....... 32154321 dddd bbbbbbbbbx 0∞

Bound on code length

Theorem: For a text T of length n, the Arithmetic

encoder generates at most

log2 (2/sn) < 1 + log2 2/sn = 1 + (1 - log2 sn)

= 2 - log2 (∏ i=1,n p(Ti))

= 2 - log2 (∏ [p()occ()])

= 2 - ∑ occ() * log2 p()

≈ 2 + ∑ ( n*p() ) * log2 (1/p())

= 2 + n H(T) bits

T = acabc

sn = p(a) *p(c) *p(a) *p(b) *p(c)

= p(a)2 * p(b) * p(c)2

Document Compression

Dictionary-based compressors

LZ77

Algorithm’s step: Output <dist, len, next-char> Advance by len + 1

A buffer “window” has fixed length and moves

a a c a a c a b c a a a a a aDictionary

(all substrings starting here)

<6,3,a>

<3,4,c>a a c a a c a b c a a a a a a c

a c

a c

LZ77 Decoding

Decoder keeps same dictionary window as encoder. Finds substring and inserts a copy of it

What if l > d? (overlap with text to be compressed) E.g. seen = abcd, next codeword is (2,9,e) Simply copy starting at the cursor

for (i = 0; i < len; i++) out[cursor+i] = out[cursor-d+i]

Output is correct: abcdcdcdcdcdce

LZ77 Optimizations used by gzip

LZSS: Output one of the following formats(0, position, length) or (1,char)

Typically uses the second format if length < 3.

Special greedy: possibly use shorter match so that next match is better

Hash Table for speed-up searches on triplets

Triples are coded with Huffman’s code

You find this at: www.gzip.org/zlib/

Dictionary search

Exact string search

Paper on Cuckoo Hashing

Exact String Search

Given a dictionary D of K strings, of total

length N, store them in a way that we can

efficiently support searches for a pattern P

over them.

Hashing ?

Hashing with chaining

Key issue: a good hash function

Basic assumption: Uniform hashing

Avg #keys per slot = n * (1/m) = n/m = (load factor)

Search cost

m = (n)

In practice

A trivial hash function is:

prime

A “provably good” hash is

Each ai is selected at random in [0,m)

k0 k1 k2 kr

≈log2 m

r ≈ L / log2 m

a0 a1 a2 ar

K

a

prime

l = max string lenm = table size

not necessarily: (...mod p) mod m

Cuckoo Hashing

A B C

E D

2 hash tables, and 2 random choices where an item can be

stored

A B C

E D

F

A running example

A B FC

E D

A running example

A B FC

E D

G

A running example

E G B FC

A D

A running example

Cuckoo Hashing Examples

A B C

E D F

G

Random (bipartite) graph: node=cell, edge=key

Natural Extensions

More than 2 hashes (choices) per key.

Very different: hypergraphs instead of graphs. Higher memory utilization

3 choices : 90+% in experiments 4 choices : about 97%

2 hashes + bins of B-size.

Balanced allocation and tightly O(1)-size bins Insertion sees a tree of possible evict+ins paths

but more insert time(and random access)

more memory...but more local

Dictionary search

Prefix-string search

Reading 3.1 and 5.2

Prefix-string Search

Given a dictionary D of K strings, of total

length N, store them in a way that we can

efficiently support prefix searches for a

pattern P over them.

Trie: speeding-up searches

1

2 2

0

4

5

6

7

2 3

y

s

1z

stile zyg

5

etic

ialygy

aibelyite

czecin

omo

Pro: O(p) search time

Cons: edge + node labels and tree structure

Front-coding: squeezing strings

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2-level indexing

Disk

InternalMemory A disadvantage:

•Trade-off ≈ speed vs space (because of bucket size)

2 advantages:• Search ≈ typically 1 I/O

• Space ≈ Front-coding over buckets