Suffix Arrays in Linear Time.

1. Suffix Arrays in Linear Time

2. Index text, so substring
queries can be answered fast

3. The Text
C G A C G C T
Suffix Tree
A C G T
G T A C
A C

4. The Text
C G A C G C T
A C G T
G T A C
A C
Substring
C G C Query

5. Trees take too much space.
Are there smaller indices?

6. The Text
C G A C G C T
Suffix Tree
A C G T
G T A C
A C
Suffix Array
Sorted List of
Suffixes 3 1 4 6 2 5 7

7. The Text
C G A C G C T
Burrows-Wheeler
Index (an array)
Suffix Array
3 1 4 6 2 5 7

8. How can one compute the
Suffix Array in Linear Time?

9. Task
String of length n
with characters
in the range 1..n
Sort these
suffixes
lexicographically
Obtain two arrays, O(n log n)
f[i]: sorted order of ith comparisons
suffix, g[i]: which each taking up
suffix is ith highest to n time

10. Divide and Conquer
Separate odd and
even suffixes; sort
each recursively,
then combine

11. Sorting Even Suffixes
A1 A2
A3 A4
Sort these n/2
pairs and map
them to single
chars in the range
1..n/2
New text of half
the length; sort
suffixes
recursively

12. Sorting Odd Suffixes
O1 O2 O3 O4
A1,E1 A2,E2 A3,E3 A4,E4
Sort these n/2
pairs, E’s are the
even suffixes,
whose order we
know

13. Time Complexity
T(n) = O(n) + T(n/2) + Time for merging even and odd suffixes
O(n)

14. Merging
O E
A,E B,O
Do we have any info
to determine the
relative order of an
odd suffix and an even
one?

15. The Trick
Sanders, Karkkainnen
0 1 2
Split suffixes
into 3 groups
instead of 2, so
0 mod 3, 1 mod
3 and 2 mod 3

16. Sorting 0 and 1 Together
ABCDEFGHIJKL
Sort these 2n/3
triplets and map
them to single
chars
New text of
length 2n/3; sort
suffixes
recursively

17. Sorting Suffixes in 2
21 22 23 24
A1,01 A2,02 A3,03 A4,04
Sort these n/3
pairs, 0’s are the
mod 0 suffixes,
whose order we
know

18. Merging
1 2
AB,0 CD,1
We know the
order of all 0,1
suffixes!

19. Time Complexity
T(n) = O(n) + T(2n/3) + O(n)
O(n)

20. Generalization
Set D of indices mod v
v 2v 3v
Sorting suffixes of
this string gives the
This string has size Time taken to create sorted order of all
|D|n/v this string is O(n |D|) suffixes which begin
at indices j such that
j mod v is in D

21. Key Property of D
x<v
x<v
For any 2 indices i and j
i-j mod v is the distance between some two beads in D
D is a Difference Cover if
distances between beads in
D generate 0,1…,v-1

22. Size of D
sqrt(v)
sqrt(v)
There exists a Difference
Cover of size 1.5*sqrt(v)!

23. Time Complexity
T(n) = O(n|D|) + T(|D|n/v) + O(nv)
T(n) = O(n sqrt(v))+ T(n/srqt(v)) + O(nv)
For |D|=2.5 sqrt(v)