1. Index construction: Compression of postings
Prof. Paolo Ferragina, Algoritmi per "Information Retrieval"
Index construction: Compression of postings
Paolo Ferragina
Dipartimento di Informatica
Università di Pisa

2. Sec. 3.1
Gap encoding …
Then you compress the resulting integers with
variable-length prefix-free codes, as follows…

3. Prof. Paolo Ferragina, Algoritmi per "Information Retrieval"
Variable-byte codes
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) =
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 !!
T-nibble: We could design this code over t-bits, not just t=8

4. Prof. Paolo Ferragina, Algoritmi per "Information Retrieval"
PForDelta coding 10 01 … 42 23 2 1
a block of 128 numbers = 256 bits = 32 bytes
Use b (e.g. 2) bits to encode 128 numbers (32 bytes) or exceptions
Translate data: [base, base + 2b-2]  [0,2b - 2]
Encode exceptions with value 2b-1
Choose b to encode 90% values, or trade-off:
b waste more bits, b more exceptions

5. Prof. Paolo Ferragina, Algoritmi per "Information Retrieval"
g-code
Binary Length-1 Binary length
x > 0 and Binary length = log2 x +1
e.g., 9 represented as
g-code for x takes 2 log2 x +1 bits
(ie. factor of 2 from optimal)
Optimal for Pr(x) = 1/2x2, and i.i.d integers

6. It is a prefix-free encoding…
Prof. Paolo Ferragina, Algoritmi per "Information Retrieval"
It is a prefix-free encoding…
Given the following sequence of g-coded integers, reconstruct the original sequence: 8 59 7 6 3

7. Elias-Fano
If w = log (m/n) and z = log n, where m = |B| and n = #1 then
L takes n w = n log (m/n) bits
H takes n 1s + n 0s = 2n bits
z = 3, w=2
(Select1 on H)
In unary
How to get the i-th number ? Take the i-th group of w bits in L and then represent the value (Select1(H,i) – i) in z bits

8. Rank and Select data structures
Prof. Paolo Ferragina, Algoritmi per "Information Retrieval"
Rank and Select data structures

9. A basic problem ! D D B (n log m) bits = 32 n bits.
Abaco, Battle, Car, Cold, Cod .... D
Array of n string pointers to strings of total length m
(n log m) bits = 32 n bits.
it depends on the number of strings
it is independent of string length
Abaco Battle Car Cold Cod .... D B
Spaces are introduced for simplicity

10. Rank/Select Wish to index the bit vector B (possibly compressed). B
Rank1(6) = 2
m = |B|
n = #1
Rankb(i) = number of b in B[1,i]
Selectb(i) = position of the i-th b in B
Two approaches:
Takes |B| + o(|B|) bits of space,
Aims at achieving n log(m/n) bits, by deplyoing Elias-Fano + point (1)

11. The Bit-Vector Index: |B| + o(|B|)
m = |B|
n = #1s
The Bit-Vector Index: |B| + o(|B|)
Goal. B is read-only, and the additional index takes o(m) bits.
Rank B Z 8 18
block
pos #1 z
(bucket-relative) Rank1
0000 1
....
...
1011 2
(absolute) Rank1
Setting Z = poly(log m) and z=(1/2) log m:
Extra space is + (m/Z) log m + (m/z) log Z + o(m)
+ O(m loglog m / log m) = o(m) bits
Rank time is O(1)
Term o(m) is crucial in practice, B is untouched (not compressed)
There exists a Bit-Vector Index
taking o(m) extra bits
and constant time for Rank/Select.
B is needed and read-only!

12. The Select operation B Extra space is + o(m), and B is not touched!
m = |B|
n = #1s
The Select operation B
size r is variable until the subarray includes k = (log m)2 1s
Sparse case: If r > k2 = (log m)4 , we store explicitly the position of the k = (log m)2 1s, because we have at most (m/r) blocks of this type, each taking (m/r) * k * log m bits = O(m / log m) = o(m) bits
Dense case: k ≤ r ≤ k2, recurse by repeating the argument with now k’ = (log log m)2. If r’ including k’ 1s > log m bits, then store the k’ positions explicitly using O(log log m) bits each, thus O(m/log log m) = o(m) bits in total. Otherwise r’ < log m, and thus a precomputed table is enough.
Extra space is + o(m), and B is not touched!
Select time is O(1)
There exists a Bit-Vector Index
taking o(m) extra bits
and constant time for Rank/Select.
B is needed and read-only!

13. Via Elias-Fano (B is not needed)
Recall that by setting w = log (m/n) and z = log n,
where m = |B| and n = #1 then
Space = n log (m/n) bits + 2n bits
(Build Select1 on H
so we need extra
|H| + o(|H|) bits
= 2n + o(n) bits )
z = 3, w=2
Select1(i) on B  uses L and (Select1(H,i) – i) in +o(n) space
Rank1(i) on B  Needs binary search over B

14. If you wish to play with Rank and Select
Prof. Paolo Ferragina, Algoritmi per "Information Retrieval"
If you wish to play with Rank and Select
m/10 + n log (m/n)
Rank in 0.4 msec, Select in < 1 msec
vs 32n bits of explicit pointers