CS 6293 Advanced Topics: Current Bioinformatics

CS 6293 Advanced Topics: Current Bioinformatics

Lecture 5

Exact String Matching Algorithms

Overview

• Sequence alignment: two sub-problems:– How to score an alignment with errors– How to find an alignment with the best score

• Today: exact string matching – Does not allow any errors– Efficiency becomes the sole consideration

• Time and space

Why exact string matching?

• The most fundamental string comparison problem

• 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

Definitions

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

abayababaxababb abayababaxababb

aba aba

T

P

length m

length n

The naïve algorithm

abayababaxababb abayababaxababb

aba aba

aba aba

aba aba

aba aba

aba aba

aba aba

aba aba

aba aba

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?

Average case time complexity

P(mismatch at 1st position): ¾P(mismatch at 2nd position): ¼ * ¾ P(mismatch at 3nd position): (¼)2 * ¾P(mismatch at kth position): (¼)k-1 * ¾Expected 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/3Not as bad as you thought it might be

Biological sequences are not random

T: aaaaaaaaaaaaaaaaaaaaaaaaaP: 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 casesub-linear in practice

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

Cost for exact string matching

Total cost = cost (preprocessing)

+ cost(comparison)

+ cost(output)

Constant

Minimize

Overhead

Hope: gain > overhead

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) T’s for common patterns• Q: Which one to pre-process?• A: Always pre-process the shorter seq, or the

one that is repeatedly used

Pre-processing algs

• Pattern preprocessing– Knuth-Morris-Pratt algorithm (KMP)– Aho-Corasick algorithm

• Multiple patterns

– Boyer – Moore algorithm (discuss only if have time)• The choice of most cases• Typically sub-linear time

• Text preprocessing– Suffix tree

• Very useful for many purposes

Algorithm KMP: Intuitive example 1

• Observation: by reasoning on the pattern alone, we can determine that if a mismatch happened when comparing P[8] with T[i], we can shift P by four chars, and compare P[4] with T[i], without missing any possible matches.

• Number of comparisons saved: 6

abcxabcT

abcxabcdePmismatch

abcxabcT

abcxabcde

Naïve approach:

abcxabcdeabcxabcdeabcxabcde?

?

Intuitive example 2

• Observation: by reasoning on the pattern alone, we can determine that if a mismatch happened between P[7] and T[j], we can shift P by six chars and compare T[j] with P[1] without missing any possible matches

• Number of comparisons saved: 7

abcxabcT

abcxabcdePmismatch

abcxabcT

abcxabcde

Naïve approach:

abcxabcdeabcxabcdeabcxabcde

Should not be a c

abcxabcdeabcxabcde?

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.

tt’P

t xT

y

tt’P y

z

z

Pre-processing: for any position i in P, find P[1..i]’s longest proper suffix, t = P[j..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)

ij

ij

KMP algorithm: shift rule

tt’P

t xT

y

tt’P y

z

z

Shift rule: when a mismatch occurred between P[i+1] and T[k], 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 P[i+1], resume comparison between x and P[sp(i)+1].

ij

ijsp(i)1

Failure Link Example

P: aataac

a a t a a c

sp(i) 0 1 0 0 2 0

aaat

aataac

If a char in T fails to match at pos 6, re-compare it with the

char at pos 3 (= 2 + 1)

Another example

P: abababc

a b a b a b c

Sp(i) 0 0 0 0 0 4 0

ababaababc

If a char in T fails to match at pos 7, re-compare it with the char at pos 5 (= 4 + 1)

abab

abababab

KMP Example using Failure Link

a a t a a c

aataac^^*

T: aacaataaaaataaccttacta

aataac.*aataac^^^^^*

aataac..*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)

Implicitcomparison

KMP algorithm using DFA (Deterministic Finite Automata)

P: aataac

1 2 3 4 50a a t a a c

6

a t

If the next char in T is t after matching 5 chars, go to state 3

a a t a a c

If a char in T fails to match at pos 6, re-compare it with

the char at pos 3

a

Failure link

DFA

a

All other inputs goes to state 0.

DFA Example

T: aacaataataataaccttacta

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.

1201234534534560001001

1 2 3 4 50a a t a a c

6

a t

a

DFA

a

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.

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 practice sub-linear

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

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.

Keyword tree for spell checking

• 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?

p

o

t

a

t

o

e

tr

y

t

er

y

s

c

i

e

n

c

e

h o o l

1

2

3

4

5

This version of the potato gun was inspired by the Weird Science team out of Illinois

Aho-Corasick algorithm

• Basis of the fgrep algorithm

• Generalizing KMP– Using failure links

• Example: given the following 4 patterns:– potato– tattoo– theater– other

Keyword tree

p

o

t

a

t

o

t

e

r

0t

he

r

1

2 3

4

a

t

t

o

o

h

a

t

e

Keyword tree

p

o

t

a

t

o

t

e

r

0t

he

r

1

2 3

4

a

t

t

o

o

h

a

t

e

potherotathxythopotattooattoo

Keyword tree

p

o

t

a

t

o

t

e

r

0t

he

r

1

2 3

4

a

t

t

o

o

h

a

t

e

O(mn) m: length of text. n: length of longest pattern

potherotathxythopotattooattoo

Keyword Tree with a failure link

p

o

t

a

t

o

t

e

r

0t

he

r

1

2 3

4

a

t

t

o

o

h

a

t

e

potherotathxythopotattooattoo

Keyword Tree with a failure link

p

o

t

a

t

o

t

e

r

0t

he

r

1

2 3

4

a

t

t

o

o

h

a

t

e

potherotathxythopotattooattoo

Keyword Tree with all failure links

p

o

t

a

t

o

t

e

r

0t

he

r

1

2 3

4

a

t

t

o

o

h

a

t

e

Example

p

o

t

a

t

o

t

e

r

0t

he

r

1

2 3

4

a

t

t

o

o

h

a

t

e

potherotathxythopotattooattoo

Example

p

o

t

a

t

o

t

e

r

0t

he

r

1

2 3

4

a

t

t

o

o

h

a

t

e

potherotathxythopotattooattoo

Example

p

o

t

a

t

o

t

e

r

0t

he

r

1

2 3

4

a

t

t

o

o

h

a

t

e

potherotathxythopotattooattoo

Example

p

o

t

a

t

o

t

e

r

0t

he

r

1

2 3

4

a

t

t

o

o

h

a

t

e

potherotathxythopotattooattoo

Example

p

o

t

a

t

o

t

e

r

0t

he

r

1

2 3

4

a

t

t

o

o

h

a

t

e

potherotathxythopotattooattoo

Aho-Corasick algorithm

• O(n) preprocessing, and O(m+k) 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

Suffix Tree

• All algorithms we talked about so far preprocess pattern(s)– 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

Suffix tree

• T: xabxac

• Suffixes:1. xabxac

2. abxac

3. bxac

4. xac

5. ac

6. c

a

bx

ac

bxa

c

c

c

x a b x a cc 1

2 3

4

5

6

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

Suffix tree implementation

• Explicitly labeling sequence end

• T: xabxa$

a

bx

a

bxa

x a b x a1

2 3

a

bx

a

bxa

x a b x a1

2 3

$

$$

$

$4

5

• One-to-one correspondence of leaves and suffixes

• |T| leaves, hence < |T| internal nodes

Suffix tree implementation

• Implicitly labeling edges

• T: xabxa$

a

bx

a

bxa

x a b x a1

2 3

$

$$

$

$4

5

2:2

3:$ 3:$

1

2 3

$

$4

5

1:23:$

• |Tree(T)| = O(|T| + size(edge labels))

Suffix links

• Similar to failure link in a keyword tree

• Only link internal nodes having branchesx

ab

cd

ef

g

h

ij

ab

c

de

fg

h

i

j

P: xabcff

ST Application 1: pattern matching

• Find all occurrence of P=xa in T– Find node v in the ST that

matches to P– Traverse the subtree

rooted at v to get the locations

a

bx

ac

bxa

c

c

c

x a b x a cc 1

2 3

4

5

6

T: xabxac

• 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.

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

a

bx

ac

bxa

c

c

c

x a b x a cc 1

2 3

4

5

6

T: xabxacP: xab

• O(m) to construct ST (large constant factor)

• O(n) to find v – linear to total length of P’s

• 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 P’s are fixed. Otherwise depending on relative size.

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

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

(L, N)

For each node with L >= k, and N >= p, print all leaves

O(m) to traverse tree

Maximal repeats finding

1. Right-maximal repeat– S[i+1..i+k] = S[j+1..j+k], – but S[i+k+1] != S[j+k+1]

2. Left-maximal repeat– S[i+1..i+k] = S[j+1..j+k]– But S[i] != S[j]

3. Maximal repeat– S[i+1..i+k] = S[j+1..j+k]– But S[i] != S[j], and S[i+k+1] != S[j+k+1]

acatgacatt

1. cat2. aca3. acat

Maximal repeats finding

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

5:e

2

5:e

4

1234567890acatgacatt

5:e 5cat

t

7

ca

t

t

6

a

5:e

3

5:e

1

t

8

tt

t

9

10$

Maximal repeats finding

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

5:e

2

5:e

4

1234567890acatgacatt

5:e 5cat

t

7

ca

t

t

6

a

5:e

3

5:e

1

t

8

tt

t

9

10$

Left char = [] g c c a a

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 L>=k, and L(parent)<k, and N>=p

– Traverse sub-tree rooted at v to get the locations

L<k

L>=k, N>=p

L = KL=k

This can be used in many applications. For example, to find words that appeared frequently in a genome or a document

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 (why?)

Joint suffix tree example

• S1 = abcd

• S2 = abca

• S* = abcd&abca$a

bcd

&ab

ca

bc

d&abca

c

d&

abc

d

d & ab c

d

& a b c d

a aa

$

1,1

2,1

1,2

1,3

1,4

2,2

2,32,4

(2, 0)useless

Seq ID

Suffix ID

To Simplify

• We don’t really need to do anything, since all edge labels were implicit.

• The right hand side is more convenient to look at

abc

d&

abc

a

bc

d&abca

c

d&

abc

d

d & ab c

d

& a b c d

a aa

$

1,1

2,1

1,2

1,3

1,4

2,2

2,32,4

uselessa

bcd

bc

d

c

d

d

a aa

$

1,12,1

1,21,3

1,4

2,2

2,32,4

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– B[1] = 1 if a child of v is a suffix of S1– Bottom-up: find all internal nodes with

B[1] = B[2] = 1 (green nodes)– Report a green node with the longest

label– Can be extended to k sequences. Just

use a bit vector of size k.

abc

d

bc

d

c

d

d

a aa

$

1,12,1

1,21,3

1,4

2,2

2,32,4

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

L< k

L >= k B = BitOR(1010, 0011) = 1011cardinal(B) = 3

3,x 3,x 4,x

B = 0011

1,x

B = 1010

cardinal(B) >= 3

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 doesn’t 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

abc

d

bc

d

c

d

d

a aa

$

1,12,1

1,21,3

1,4

2,2

2,32,4

Seqs: abcd, abcaP1: cdP2: bc

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

abc

d

bc

d

c

d

d

a aa

$

1,12,1

1,21,3

1,4

2,2

2,32,4

Contamination sources: abcd, abca

Sequence: dbcgaabctacgtctagt

Suffix Tree Memory Footprint

• The space requirements of suffix trees can become prohibitive– |Tree(T)| is about 20|T| in practice

• Suffix arrays provide one solution.

Suffix Arrays• Very space efficient (m integers)• Pattern lookup is nearly O(n) in practice

– O(n + log2 m) worst case with 2m additional integers

– Independent of alphabet size!

• Easiest to describe (and construct) using suffix trees– Other (slower) methods exist

a

bxa

bxa

x a b x a1

5

3

$

$

$$

$

4

2

5 2 3 4 1

abxa$a$ bxa$ xa$ xabxa$

1. xabxa2. abxa3. bxa4. xa5. a

Suffix array construction

• Build suffix tree for T$

• Perform “lexical” depth-first search of suffix tree– output the suffix label of each leaf

encountered

• Therefore suffix array can be constructed in O(m) time.

Suffix array pattern search

• If P is in T, then all the locations of P are consecutive suffixes in Pos.

• Do binary search in Pos to find P!– Compare P with suffix Pos(m/2)– If lexicographically less, P is in first half of T– If lexicographically more, P is in second half of T– Iterate!

Suffix array pattern search

• T: xabxa$

• P: abx

a

bxa

bxa

x a b x a1

5

3

$

$

$$

$

4

2

5 2 3 4 1

abxa$a$ bxa$ xa$ xabxa$

L RMR

M

Suffix array binary search

• How long to compare P with suffix of T?– O(n) worst case!

• Binary search on Pos takes O(n log m) time• Worst case will be rare

– occur if many long prefixes of P appear in T• In random or large alphabet strings

– expect to do less than log m comparisons• O(n + log m) running time when combined with

LCP table– suffix tree = suffix array + LCP table

Summary

• One T, one P– Boyer-Moore is the choice– KMP works but not the best

• One T, many P– Aho-Corasick– Suffix Tree (array)

• One fixed T, many varying P– Suffix tree (array)

• Two or more T’s– Suffix tree, joint suffix tree

Alphabet independent

Alphabet dependent

Boyer – Moore algorithm

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

Boyer – Moore algorithm

• Right to left comparison

x

y

y

Skip some chars without missing any occurrence.

Resume comparison here

Bad character rule

0 1 12345678901234567T:xpbctbxabpqqaabpqP: tpabxab *^^^^What would you do now?

Bad character rule

0 1 12345678901234567T:xpbctbxabpqqaabpqP: tpabxab *^^^^P: tpabxab

Bad character rule

0 1 123456789012345678T:xpbctbxabpqqaabpqzP: tpabxab *^^^^P: tpabxab *P: tpabxab

Basic bad character rule

char Right-most-position in P

a 6

b 7

p 2

t 1

x 5

tpabxab

Pre-processing:O(n)

Basic bad character rule

char Right-most-position in P

a 6

b 7

p 2

t 1

x 5

T: xpbctbxabpqqaabpqzP: tpabxab

*^^^^

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

k

i = 3 Shift 3 – 1 = 2

Basic bad character rule

char Right-most-position in P

a 6

b 7

p 2

t 1

x 5

T: xpbctbxabpqqaabpqzP: tpabxab *

P: tpabxab

When T(k) is not in P, shift left end of P to align with T(k+1)

k

i = 7 Shift 7 – 0 = 7

Basic bad character rule

char Right-most-position in P

a 6

b 7

p 2

t 1

x 5

T: xpbctbxabpqqaabpqz

P: tpabxab *^^

P: tpabxab

When rightmost T(k) in P is right to i, shift pattern P by 1

k

i = 5 5 – 6 < 0. so shift 1

Extended bad character rule

char Position in P

a 6, 3

b 7, 4

p 2

t 1

x 5

T: xpbctbxabpqqaabpqz

P: tpabxab *^^

P: tpabxab

Find T(k) in P that is immediately left to i, shift P to align T(k) with that position

k

i = 5 5 – 3 = 2. so shift 2

Preprocessing still O(n)

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

0 1 123456789012345678T:prstabstubabvqxrstP: qcabdabdab *^^

P: qcabdabdab

According to extended bad character rule

(weak) good suffix rule

0 1 123456789012345678T:prstabstubabvqxrstP: qcabdabdab *^^

P: qcabdabdab

(Weak) good suffix rule

tx

tyt’

tyt’

Preprocessing: For any suffix t of P, find the rightmost copy of t, denoted by t’.How to find t’ efficiently?

T

P

P

(Strong) good suffix rule

0 1 123456789012345678T:prstabstubabvqxrstP: qcabdabdab *^^

(Strong) good suffix rule

0 1 123456789012345678T:prstabstubabvqxrstP: qcabdabdab *^^

P: qcabdabdab

(Strong) good suffix rule

tx

tyt’

tyt’

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’

T

P

P

z

z

z ≠ y

Example preprocessing

qcabdabdab

char Positions in P

a 9, 6, 3

b 10, 7, 4

c 2

d 8, 5

q 1

q c a b d a b d a b1 2 3 4 5 6 7 8 9 10

0 0 0 0 2 0 0 2 0 0dabcab

Bad char rule Good suffix rule

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.

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 (Galil’s rule)

• Proofs are technical. Skip.

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

tt’P yxKMP

tyt’P xB-M

i

ij

j For each i, find a j. similar to DP. Start from i = 2

Fundamental pre-processing

• 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.

aabcaabxaazZ = 01003100210

• How to compute Z-values in linear time?

tt’Pi

x yi+zi-1zi1

Computing Z in Linear time

tt’Pl

x yrk

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.

tt’Pl

x yrk

We know that t = t’, therefore the Z-value at k-l+1 may be helpful to us.

1

k-l+1

Computing Z in Linear time

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

Pk

The previous r is smaller than k. i.e., no previous match extends beyond k. do explicit comparison.

Pl

x yrk

Zk-l+1 <= r-k+1. Zk = Zk-l+1 No comparison is needed.1

k-l+1

Case 1:

Case 2:

Pl rk

Zk-l+1 > r-k+1. Zk = Zk-l+1

Comparison start from r1

k-l+1

Case 3:

Z-preprocessing for B-M and KMP

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

tt’i

x y

j = i+zi-1zi1

tt’ yxKMP

tyt’xB-Mij

Z j

ijFor each j sp’(j+zj-1) = z(j)

Use Z backwards