CS5263 Bioinformatics

1
CS5263 Bioinformatics

• Lecture 9-10
• Exact String Matching Algorithms

2
Overview

• Pair-wise alignment
• Multiple alignment
• Commonality allowing errors when comparing
  strings
• Two sub-problems
• How to score an alignment with errors
• How to find an alignment with the best score
• Today exact string matching
• Do not allow any errors
• Efficiency becomes the sole consideration

3
Why exact string matching?

• The most fundamental string comparison problem
• Work processors
• Information retrieval
• DNA sequence retrieval
• Many many more
• Is it still an interesting research problem?
• Yes, if database is large
• Exact string matching is often the core of more
  complex string comparison algorithms
• E.g., BLAST
• Often repeatedly called by other methods
• Usually the most time consuming part
• Small improvement could improve overall
  efficiency considerably

4
Definitions

• Text a longer string T (length m)
• Pattern a shorter string P (length n)
• Exact matching find all occurrences of P in T

length m
T
length n
P
5
The naïve algorithm
6
Time complexity

• Worst case O(mn)
• Best case O(m)
• e.g. aaaaaaaaaaaaaa vs baaaaaaa
• Average case?
• Alphabet A, C, G, T
• Assume both P and T are random
• Equal probability
• In average how many chars do you need to compare
  before giving up?

7
Average case time complexity

• P(mismatch at 1st position) ¾
• P(mismatch at 2nd position) ¼ ¾
• P(mismatch at 3nd position) (¼)2 ¾
• P(mismatch at kth position) (¼)k-1 ¾
• Expected number of comparison per position
• p 1/4
• ?k (1-p) p(k-1) k (1-p) / p ?k pk k
• 1/(1-p)
• 4/3
• Average complexity 4m/3
• Not as bad as you thought it might be

8
Biological sequences are not random

• T aaaaaaaaaaaaaaaaaaaaaaaaa
• P aaaab
• Plus 4m/3 average case is still bad for long
  genomic sequences!
• Especially if this has to be done again and again
• Smarter algorithms
• O(m n) in worst case
• sub-linear in practice

9
How to speedup?

• Pre-processing T or P
• Why pre-processing can save us time?
• Uncovers the structure of T or P
• Determines when we can skip ahead without missing
  anything
• Determines when we can infer the result of
  character comparisons without doing them.

ACGTAXACXTAXACGXAX
ACGTACA
10
Cost for exact string matching

• Total cost cost (preprocessing)
• cost(comparison)
• cost(output)

Overhead
Minimize
Constant
Hope gain gt overhead
11
String matching scenarios

• One T and one P
• Search a word in a document
• One T and many P all at once
• Search a set of words in a document
• Spell checking (fixed P)
• One fixed T, many P
• Search a completed genome for short sequences
• Two (or many) Ts for common patterns
• Q Which one to pre-process?
• A Always pre-process the shorter seq, or the one
  that is repeatedly used

12
Pre-processing algs

• Pattern preprocessing
• Karp Rabin algorithm
• Small alphabet and short patterns
• Knuth-Morris-Pratt algorithm (KMP)
• Aho-Corasick algorithm
• Multiple patterns
• Boyer Moore algorithm
• The choice of most cases
• Typically sub-linear time
• Text preprocessing
• Suffix tree
• Very useful for many purposes

13
Karp Rabin Algorithm

• Lets say we are dealing with binary numbers
• Text 01010001011001010101001
• Pattern 101100
• Convert pattern to integer
• 101100 25 23 22 44

14
Karp Rabin algorithm

• Text 01010001011001010101001
• Pattern 101100 44 decimal
• 10111011001010101001
• 25 23 22 21 46
• 10111011001010101001
• 46 2 64 1 29
• 10111011001010101001
• 29 2 - 0 1 59
• 10111011001010101001
• 59 2 - 64 0 54
• 10111011001010101001
• 54 2 - 64 0 44

15
Karp Rabin algorithm

• What if the pattern is too long to fit into a
  single integer?
• Pattern 101100. But our machine only has 5 bits
• Basic idea hashing. 44 13 5
• 10111011001010101001
• 46 ( 13 7)
• 10111011001010101001
• 46 2 64 1 29 ( 13 3)
• 10111011001010101001
• 29 2 - 0 1 59 ( 13 7)
• 10111011001010101001
• 59 2 - 64 0 54 ( 13 2)
• 10111011001010101001
• 54 2 - 64 0 44 ( 13 5)

16
Algorithm KMP

• Not the fastest
• Best known
• Good for real-time matching
• i.e. text comes one char at a time
• No memory of previous chars
• Idea
• Left-to-right comparison
• Shift P more than one char whenever possible

17
Intuitive example 1
abcxabc
T
mismatch
P
abcxabcde
Naïve approach
abcxabc
T
?
abcxabcde

• Observation by reasoning on the pattern alone,
  we can determine that if a mismatch happened when
  comparing P8 with Ti, we can shift P by four
  chars, and compare P4 with Ti, without
  missing any possible matches.
• Number of comparisons saved 6

18
Intuitive example 2
abcxabc
T
mismatch
P
abcxabcde
Naïve approach
abcxabc
T
?
abcxabcde

• Observation by reasoning on the pattern alone,
  we can determine that if a mismatch happened
  between P7 and Tj, we can shift P by six
  chars and compare Tj with P1 without missing
  any possible matches
• Number of comparisons saved 7

19
KMP algorithm pre-processing

• Key the reasoning is done without even knowing
  what string T is.
• Only the location of mismatch in P must be known.
  

x
t
T
y
z
t
t
P
i
j
y
z
t
t
P
i
j
Pre-processing for any position i in P, find
P1..is longest proper suffix, t Pj..i,
such that t matches to a prefix of P, t, and the
next char of t is different from the next char of
t (i.e., y ? z) For each i, let sp(i) length(t)
20
KMP algorithm shift rule
x
t
T
y
z
t
t
P
i
j
y
z
t
P
t
i
j
sp(i)
1
Shift rule when a mismatch occurred between
Pi1 and Tk, shift P to the right by i
sp(i) chars and compare x with z. This shift
rule can be implicitly represented by creating a
failure link between y and z. Meaning when a
mismatch occurred between x on T and Pi1,
resume comparison between x and Psp(i)1.
21
Failure Link Example

• P aataac

If a char in T fails to match at pos 6,
re-compare it with the char at pos 3 ( 2 1)
a
a
t
a
a
c
sp(i) 0 1 0 0 2 0
aaat
aataac
22
Another example

• P abababc

If a char in T fails to match at pos 7,
re-compare it with the char at pos 5 ( 4 1)
a
b
a
b
a
b
c
Sp(i) 0 0 0 0 0 4 0
ababaababc
abab
abababab
23
KMP Example using Failure Link
a
a
t
a
a
c
T aacaataaaaataaccttacta
aataac

• Time complexity analysis
• Each char in T may be compared up to n times. A
  lousy analysis gives O(mn) time.
• More careful analysis number of comparisons can
  be broken to two phases
• Comparison phase the first time a char in T is
  compared to P. Total is exactly m.
• Shift phase. First comparisons made after a
  shift. Total is at most m.
• Time complexity O(2m)

aataac .
aataac
aataac ..
aataac .
24
KMP algorithm using DFA (Deterministic Finite
Automata)

• P aataac

If a char in T fails to match at pos 6,
re-compare it with the char at pos 3
Failure link
a
a
t
a
a
c
If the next char in T is t after matching 5
chars, go to state 3
a
t
t
a
a
c
a
a
1
2
3
4
5
0
6
DFA
a
a
All other inputs goes to state 0.
25
DFA Example
a
t
t
a
a
c
a
a
1
2
3
4
5
0
6
DFA
a
a
T aacaataataataaccttacta
1201234534534560001001
Each char in T will be examined exactly once.
Therefore, exactly m comparisons are made. But
it takes longer to do pre-processing, and needs
more space to store the FSA.
26
Difference between Failure Link and DFA

• Failure link
• Preprocessing time and space are O(n), regardless
  of alphabet size
• Comparison time is at most 2m (at least m)
• DFA
• Preprocessing time and space are O(n ?)
• May be a problem for very large alphabet size
• For example, each char is a big integer
• Chinese characters
• Comparison time is always m.

27
Boyer Moore algorithm

• Often the choice of algorithm for many cases
• One T and one P
• We will talk about it later if have time
• In its original version does not guarantee linear
  time
• Some modification did it
• In practice sub-linear

28
The set matching problem

• Find all occurrences of a set of patterns in T
• First idea run KMP or BM for each P
• O(km n)
• k number of patterns
• m length of text
• n total length of patterns
• Better idea combine all patterns together and
  search in one run

29
A simpler problem spell-checking

• A dictionary contains five words
• potato
• poetry
• pottery
• science
• school
• Given a document, check if any word is (not) in
  the dictionary
• Words in document are separated by special chars.
• Relatively easy.

30
Keyword tree for spell checking
This version of the potato gun was inspired by
the Weird Science team out of Illinois
p
s
o
c
l
h
o
o
5
e
i
t
e
t
a
t
r
n
t
y
e
c
o
r
e
y
3
1
4
2

• O(n) time to construct. n total length of
  patterns.
• Search time O(m). m length of text
• Common prefix only need to be compared once.
• What if there is no space between words?

31
Aho-Corasick algorithm

• Basis of the fgrep algorithm
• Generalizing KMP
• Using failure links
• Example given the following 4 patterns
• potato
• tattoo
• theater
• other

32
Keyword tree
0
p
t
t
h
o
e
h
a
t
r
e
t
a
a
4
t
t
t
e
o
o
r
1
o
3
2
33
Keyword tree
0
p
t
t
h
o
e
h
a
t
r
e
t
a
a
4
t
t
t
e
o
o
r
1
o
3
2
potherotathxythopotattooattoo
34
Keyword tree
0
p
t
t
h
o
e
h
a
t
r
e
t
a
a
4
t
t
t
e
o
o
r
1
o
3
2
potherotathxythopotattooattoo
O(mn)
m length of text. n length of longest pattern
35
Keyword Tree with a failure link
0
p
t
t
h
o
e
h
a
t
r
e
t
a
a
4
t
t
t
e
o
o
r
1
o
3
2
potherotathxythopotattooattoo
36
Keyword Tree with a failure link
0
p
t
t
h
o
e
h
a
t
r
e
t
a
a
4
t
t
t
e
o
o
r
1
o
3
2
potherotathxythopotattooattoo
37
Keyword Tree with all failure links
0
p
t
t
h
o
e
h
a
t
r
e
t
4
a
a
t
t
t
e
o
o
r
1
o
3
2
38
Example
0
p
t
t
h
o
e
h
a
t
r
e
t
4
a
a
t
t
t
e
o
o
r
1
o
3
2
potherotathxythopotattooattoo
39
Example
0
p
t
t
h
o
e
h
a
t
r
e
t
4
a
a
t
t
t
e
o
o
r
1
o
3
2
potherotathxythopotattooattoo
40
Example
0
p
t
t
h
o
e
h
a
t
r
e
t
4
a
a
t
t
t
e
o
o
r
1
o
3
2
potherotathxythopotattooattoo
41
Example
0
p
t
t
h
o
e
h
a
t
r
e
t
4
a
a
t
t
t
e
o
o
r
1
o
3
2
potherotathxythopotattooattoo
42
Example
0
p
t
t
h
o
e
h
a
t
r
e
t
4
a
a
t
t
t
e
o
o
r
1
o
3
2
potherotathxythopotattooattoo
43
Aho-Corasick algorithm

• O(n) preprocessing, and O(mk) searching.
• n total length of patterns.
• m length of text
• k is of occurrence.
• Can create a DFA similar as in KMP.
• Requires more space,
• Preprocessing time depends on alphabet size
• Search time is constant
• A Where can this algorithm be used in previous
  topics?
• Q BLAST
• Given a query sequence, we generate many seed
  sequences (k-mers)
• Search for exact matches to these seed sequences
• Extend exact matches into longer inexact matches

44
Suffix Tree

• All algorithms we talked about so far preprocess
  pattern(s)
• Karp-Rabin small pattern, small alphabet
• Boyer-Moore fastest in practice. O(m) worst
  case.
• KMP O(m)
• Aho-Corasick O(m)
• In some cases we may prefer to pre-process T
• Fixed T, varying P
• Suffix tree basically a keyword tree of all
  suffixes

45
Suffix tree

• T xabxac
• Suffixes
• xabxac
• abxac
• bxac
• xac
• ac
• c

x
a
b
x
a
a
c
c
1
c
b
b
x
x
c
4
6
a
a
c
c
5
2
3
Naïve construction O(m2) using
Aho-Corasick. Smarter O(m). Very technical. big
constant factor Difference from a keyword tree
create an internal node only when there is a
branch
46
Suffix tree implementation

• Explicitly labeling sequence end
• T xabxa

x
a
x
a
b
x
b
a
a
x
a
a

1
1

b
b
b
b
x

x
x
x
4
a
a
a
a

5

2
2
3
3
47
Suffix tree implementation

• Implicitly labeling edges
• T xabxa

12
x
a
3
b
x
22
a
a

1
1


b
b


x
x
3
3
4
4
a
a
5

5

2
2
3
3
48
Suffix links

• Similar to failure link in a keyword tree
• Only link internal nodes having branches

x
a
b
P xabcf
a
b
c
f
c
d
d
e
e
f
f
g
g
h
h
i
i
j
j
49
Suffix tree construction
1234567890acatgacatt
1
1
50
Suffix tree construction
1234567890acatgacatt
2
1
1
2
51
Suffix tree construction
1234567890acatgacatt
a
2
2
4
3
1
2
52
Suffix tree construction
1234567890acatgacatt
a
4
2
2
4
4
3
1
2
53
Suffix tree construction
5
1234567890acatgacatt
5
a
4
2
2
4
4
3
1
2
54
Suffix tree construction
5
1234567890acatgacatt
5
a
4
c
a
2
4
t
4
t
5

3
6
1
2
55
Suffix tree construction
5
1234567890acatgacatt
5
a
c
4
a
c
t
a
4
t
4
t
5
t
5

7
3
6
1
2
56
Suffix tree construction
5
1234567890acatgacatt
5
a
c
4
a
c
t
t
a
t
4
t
5
5
t
5
t

7
3
6
8
1
2
57
Suffix tree construction
5
1234567890acatgacatt
5
t
a
c
a
5
t
c
t
t
a
9
t
4
t
5
5
t
5
t

7
3
6
8
1
2
58
Suffix tree construction
5
1234567890acatgacatt
5
t
a

c
10
a
5
t
c
t
t
a
9
t
4
t
5
5
t
5
t

7
3
6
8
1
2
59
ST Application 1 pattern matching

• Find all occurrence of Pxa in T
• Find node v in the ST that matches to P
• Traverse the subtree rooted at v to get the
  locations

x
a
b
x
a
a
c
c
1
c
b
b
x
x
c
4
6
a
a
c
c
5
T xabxac
2
3

• O(m) to construct ST (large constant factor)
• O(n) to find v linear to length of P instead of
  T!
• O(k) to get all leaves, k is the number of
  occurrence.
• Asymptotic time is the same as KMP. ST wins if T
  is fixed. KMP wins otherwise.

60
ST Application 2 set matching

• Find all occurrences of a set of patterns in T
• Build a ST from T
• Match each P to ST

x
a
b
x
a
a
c
c
1
c
b
b
x
x
c
4
6
a
a
c
c
5
T xabxac P xab
2
3

• O(m) to construct ST (large constant factor)
• O(n) to find v linear to total length of Ps
• O(k) to get all leaves, k is the number of
  occurrence.
• Asymptotic time is the same as Aho-Corasick. ST
  wins if T fixed. AC wins if Ps are fixed.
  Otherwise depending on relative size.

61
ST application 3 repeats finding

• Genome contains many repeated DNA sequences
• Repeat sequence length Varies from 1 nucleotide
  to millions
• Genes may have multiple copies (50 to 10,000)
• Highly repetitive DNA in some non-coding regions
• 6 to 10bp x 100,000 to 1,000,000 times
• Problem find all repeats that are at least
  k-residues long and appear at least p times in
  the genome

62
Repeats finding

• at least k-residues long and appear at least p
  times in the seq
• Phase 1 top-down, count label lengths (L) from
  root to each node
• Phase 2 bottom-up count of leaves descended
  from each internal node

For each node with L gt k, and N gt p, print all
leaves
O(m) to traverse tree
(L, N)
63
Maximal repeats finding

• Right-maximal repeat
• Si1..ik Sj1..jk,
• but Sik1 ! Sjk1
• Left-maximal repeat
• Si1..ik Sj1..jk
• But Si ! Sj
• Maximal repeat
• Si1..ik Sj1..jk
• But Si ! Sj, and Sik1 ! Sjk1

acatgacatt

• cat
• aca
• acat

64
Maximal repeats finding
5e
1234567890acatgacatt
5
t
a

c
10
a
5e
t
c
t
t
a
9
t
4
t
5e
5e
t
5e
t
7
3
6
8
1
2

• Find repeats with at least 3 bases and 2
  occurrence
• right-maximal cat
• Maximal acat
• left-maximal aca

65
Maximal repeats finding
5e
1234567890acatgacatt
5
t
a

c
10
a
5e
t
c
t
t
a
9
t
4
t
5e
5e
t
5e
t
7
3
6
8
1
2
Left char
g
c
c
a
a

• How to find maximal repeat?
• A right-maximal repeats with different left chars

66
ST application 4 word enumeration

• Find all k-mers that occur at least p times
• Compute (L, N) for each node
• L total label length from root to node
• N leaves
• Find nodes v with Lgtk, and L(parent)ltk, and Ngty
• Traverse sub-tree rooted at v to get the locations

Lltk
Lk
L K
Lgtk, Ngtp
This can be used in many applications. For
example, to find words that appeared frequently
in a genome or a document
67
Joint Suffix Tree (JST)

• Build a ST for more than two strings
• Two strings S1 and S2
• S S1 S2
• Build a suffix tree for S in time O(S1 S2)
• The separator will only appear in the edge ending
  in a leaf

68
Joint suffix tree example

• S1 abcd
• S2 abca
• S abcdabca

a b c d
useless
a
d

c
b c d a b c a
a
b
c
b
c
d

d
d

a

a
a
a
2,4
b
1,4
a
c
2,3
a
b
2,1
c
2,2
d
1,1
Seq ID
1,3
Suffix ID
1,2
69
To Simplify
a b c d
useless
a
d

c
b c d a b c a
a
a
b
d
c
b
c
c
b c d

b
d
d
d
c


a

d
a
d
a
1,4
a
2,4
b
a
1,4
a
a
c
a
2,4
2,3
a
b
1,1
2,3
2,1
c
2,1
2,2
1,3
d
1,1
2,2
1,2
1,3
1,2

• We dont really need to do anything, since all
  edge labels were implicit.
• The right hand side is more convenient to look at

70
Application 1 of JST

• Longest common substring between two sequences
• Using smith-waterman
• Gap mismatch -infinity.
• Quadratic time
• Using JST
• Linear time
• For each internal node v, keep a bit vector B
• B1 1 if a child of v is a suffix of S1
• Bottom-up find all internal nodes with B1
  B2 1 (green nodes)
• Report a green node with the longest label
• Can be extended to k sequences. Just use a longer
  bit vector.

a
d
c
b c d
b
c

d
d
1,4
a
a
a
2,4
1,1
2,3
2,1
1,3
2,2
1,2
71
Application 2 of JST

• Given K strings, find all k-mers that appear in
  at least (or at most) d strings
• Exact motif finding problem

Llt k
cardinal(B) gt 3
B BitOR(1010, 0011) 1011
L gt k
cardinal(B) 3
B 0011
B 1010
4,x
3,x
1,x
3,x
72
Application 3 of JST

• Substring problem for sequence databases
• Given A fixed database of sequences (e.g.,
  individual genomes)
• Given A short pattern (e.g., DNA signature)
• Q Does this DNA signature belong to any
  individual in the database?
• i.e. the pattern is a substring of some sequences
  in the database
• Aho-Corasick doesnt work
• This can also be used to design signatures for
  individuals
• Build a JST for the database seqs
• Match P to the JST
• Find seq IDs from descendents

a
d
c
b c d
b
c

d
1,4
a
d
a
a
2,4
1,1
2,3
2,1
Seqs abcd, abca P1 cd P2 ac
1,3
2,2
1,2
73
Application 4 of JST

• Detect DNA contamination
• For some reason when we try to clone and sequence
  a genome, some DNAs from other sources may
  contaminate our sample, which should be detected
  and removed
• Given A fixed database of sequences (e.g.,
  possible cantamination sources)
• Given A DNA just sequenced (e.g., DNA signature)
• Q Does this DNA contain longer enough substring
  from the seqs in the database?
• Build a JST for the database seqs
• Scan T using the JST

a
d
c
b c d
b
c

d
d
1,4
a
a
a
2,4
1,1
2,3
Contamination sources abcd, abca Sequence
dbcgaabctacgtctagt
2,1
1,3
2,2
1,2
74
Summary

• One T, one P
• Boyer-Moore is the choice
• KMP works but not the best
• One T, many P
• Aho-Corasick
• Suffix Tree
• One fixed T, many varying P
• Suffix tree
• Two or more Ts
• Suffix tree, joint suffix tree

Alphabet independent
Alphabet dependent
75
Boyer Moore algorithm

• Three ideas
• Right-to-left comparison
• Bad character rule
• Good suffix rule

76
Boyer Moore algorithm

• Right to left comparison

Resume comparison here
x
y
Skip some chars without missing any occurrence.
y
77
Bad character rule

• 0 1
• 12345678901234567
• Txpbctbxabpqqaabpq
• P tpabxab
• What would you do now?

78
Bad character rule

• 0 1
• 12345678901234567
• Txpbctbxabpqqaabpq
• P tpabxab
• P tpabxab

79
Bad character rule

• 0 1
• 123456789012345678
• Txpbctbxabpqqaabpqz
• P tpabxab
• P tpabxab
• P tpabxab

80
Basic bad character rule
tpabxab
char Right-most-position in P
a 6
b 7
p 2
t 1
x 5
Pre-processing O(n)
81
Basic bad character rule
k
T xpbctbxabpqqaabpqz
P tpabxab
When rightmost T(k) in P is left to i, shift
pattern P to align T(k) with the rightmost T(k)
in P
Shift 3 1 2
i 3
P tpabxab
char Right-most-position in P
a 6
b 7
p 2
t 1
x 5
82
Basic bad character rule
k
T xpbctbxabpqqaabpqz
P tpabxab
When T(k) is not in P, shift left end of P to
align with T(k1)
i 7
Shift 7 0 7
P tpabxab
char Right-most-position in P
a 6
b 7
p 2
t 1
x 5
83
Basic bad character rule
k
T xpbctbxabpqqaabpqz
P tpabxab
When rightmost T(k) in P is right to i, shift
pattern P by 1
i 5
5 6 lt 0. so shift 1
P tpabxab
char Right-most-position in P
a 6
b 7
p 2
t 1
x 5
84
Extended bad character rule
k
T xpbctbxabpqqaabpqz
P tpabxab
Find T(k) in P that is immediately left to i,
shift P to align T(k) with that position
i 5
5 3 2. so shift 2
P tpabxab
char Position in P
a 6, 3
b 7, 4
p 2
t 1
x 5
Preprocessing still O(n)
85
Extended bad character rule

• Best possible m / n comparisons
• Works better for large alphabet size
• In some cases the extended bad character rule is
  sufficiently good
• Worst-case O(mn)
• Expected time is sublinear

86

• 0 1
• 123456789012345678
• Tprstabstubabvqxrst
• P qcabdabdab
• P qcabdabdab

According to extended bad character rule
87
(weak) good suffix rule

• 0 1
• 123456789012345678
• Tprstabstubabvqxrst
• P qcabdabdab
• P qcabdabdab

88
(Weak) good suffix rule
x
t
T
Preprocessing For any suffix t of P, find the
rightmost copy of t, denoted by t. How to find
t efficiently?
y
t
P
t
y
t
t
P
89
(Strong) good suffix rule

• 0 1
• 123456789012345678
• Tprstabstubabvqxrst
• P qcabdabdab

90
(Strong) good suffix rule

• 0 1
• 123456789012345678
• Tprstabstubabvqxrst
• P qcabdabdab
• P qcabdabdab

91
(Strong) good suffix rule
x
t
T
In preprocessing For any suffix t of P, find
the rightmost copy of t, t, such that the char
left to t ? the char left to t
y
z
P
t
t
z ? y
y
z
t
t
P
92
Example preprocessing
qcabdabdab
Bad char rule
Good suffix rule
char Positions in P
a 9, 6, 3
b 10, 7, 4
c 2
d 8, 5
q 1
1 2 3 4 5 6 7 8 9 10
q c a b d a b d a b
0 0 0 0 2 0 0 2 0 0
dabcab
dabdabcabdab
Where to shift depends on T
Does not depend on T
Largest shift given by either the (extended) bad
char rule or the (strong) good suffix rule is
used.
93
Time complexity of BM algorithm

• Pre-processing can be done in linear time
• With strong good suffix rule, worst-case is O(m)
  if P is not in T
• If P is in T, worst-case could be O(mn)
• E.g. T m100, P m10
• unless a modification was used (Galils rule)
• Proofs are technical. Skip.

94
How to actually do pre-processing?

• Similar pre-processing for KMP and B-M
• Find matches between a suffix and a prefix
• Both can be done in linear time
• P is usually short, even a more expensive
  pre-processing may result in a gain overall

y
x
KMP
t
t
P
i
j
For each i, find a j. similar to DP. Start from i
2
y
x
B-M
t
t
P
i
j
95
Fundamental pre-processing
y
x
t
t
P
izi-1
zi
i
1

• Zi length of longest substring starting at i
  that matches a prefix of P
• i.e. t t, x ? y, Zi t
• With the Z-values computed, we can get the
  preprocessing for both KMP and B-M in linear
  time.
• aabcaabxaaz
• Z 01003100210
• How to compute Z-values in linear time?

96
Computing Z in Linear time
We already computed all Z-values up to k-1. need
to compute Zk. We also know the starting and
ending points of the previous match, l and r.
t
t
y
x
P
r
k
l
t
t
y
x
P
We know that t t, therefore the Z-value at
k-l1 may be helpful to us.
r
k
l
1
k-l1
97
Computing Z in Linear time
The previous r is smaller than k. i.e., no
previous match extends beyond k. do explicit
comparison.
P
Case 1
k
Case 2
y
x
P
Zk-l1 lt r-k1. Zk Zk-l1 No comparison is
needed.
l
r
k
1
k-l1
Zk-l1 gt r-k1. Zk Zk-l1 Comparison start from
r
Case 3
P
r
k
l
1
k-l1

• No char inside the box is compared twice. At most
  one mismatch per iteration.
• Therefore, O(n).

98
Z-preprocessing for B-M and KMP
y
x
t
t
Z
zi
i
j
1
j izi-1
y
x
KMP
t
t
For each j sp(jzj-1) z(j)
j
i
y
x
B-M
t
t
Use Z backwards
i
j

• Both KMP and B-M preprocessing can be done in O(n)