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Approximate Boyer-Moore String Matching. Source : SIAM Journal on Computing, Vol. 22, No. 2, 1993, pp.243-260 J. Tarhio and E. Ukkonen Advisor: Prof. R. C. T. Lee Speaker: Kuei-hao Chen. The k mismatches problem The k differences problem. Definition of the k mismatches problem. - PowerPoint PPT Presentation

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1

Approximate Boyer-Moore String Matching

Source : SIAM Journal on Computing, Vol. 22, No. 2, 1993, pp.243-260

J. Tarhio and E. Ukkonen

Advisor: Prof. R. C. T. Lee

Speaker: Kuei-hao Chen

2

• The k mismatches problem

• The k differences problem

3

Definition of the k mismatches problem• Given a pattern string P of length m and a

text string T of length n, we would like to find all approximate occurrences P in T with at most k mismatches.

If k=1, then

Text

Pattern

a

b

4

Consider the following situation where a pattern P is matching with a windows W of T and there are already (k+1) mismatches:

T

P

W

k+1 mismatches

5

Since there are already (k+1) mismatches, we must move the pattern. The following is obvious:

P must be moved to such an extent that there are at most k mismatches between a suffix S of W and a substring S’ of P.

T

Pk+1 mismatches

S

S’

6

Our trick is as follows: Consider the (k+1)-suffix of W. There are two cases:

7

Case 1: There is one character in this (k+1)-suffix which exists in P in such a way as shown below. Move the pattern to match these characters. Note that in such a situation, there are at most k mismatches between the (k+1)-suffix and its corresponding substring in P.

T

P

(k+1)-suffix

x

x

T

P

(k+1)-suffix

x

x

8

Case 2: No such a character exists. Move the pattern in such a way that the k-prefix of P aligns with the k-suffix of W as shown below. Under such a situation, again, there are at most k-mismatches between the k-suffix of W and k-prefix of P.

T

P

(k+1)-suffix

k-prefix

9

The generalization of the BM algorithm for the k mismatches problem will be very natural: for k=0 the generalized algorithm is exact string matching.

Recall that the k mismatches problem asks for finding all occurrences of P in T such that in at most k positions of P, T and P have different characters.

10

We just scan the pattern from right to the left until we have found k+1 mismatches (unsuccessful search) or the pattern ends (successful search).

11

Preprocessing phase for approximate matching

Dk tableThe value Dk for a particular alphabet is defined as the rightmost position of that character in the pattern – 1 and the end position i where i=[m..m-k].

Σ A C G *

D1[i=8, a]

1 6 2 8

Example : Let k=1, m=8, a ∑

j 1 2 3 4 5 6 7 8

P:

G C A G A G A G

i 1 2 3 4 5 6 7 8

P:

G C A G A G A G

Σ A C G *

D1[i=7, a]

2 5 1 8

j 1 2 3 4 5 6 7

P:

G C A G A G A

Σ A C G *

D1[i=8, a]

1 6 2 8

D1[i=7, a]

2 5 1 8

12

P = p1p2…pm,T = t1t2…tn

Preprocessing

For a ∑ Do

For j=m downto m-k Do Begin

dk[j,a] ← m

Find a character a that it is close to pj. If it is found, we calculate the distance between the position of the character a and j and insert it into dk[j,a].

Algorithm for preprocessing phase

13

P = p1p2…pm,T = t1t2…tn

Searchingj=m;While j ≦ n+ k Do Begin

h=j; i=m; mismatch=0;While i>0 and mismatch ≦ k Do Begin

d=min(dk[i, th], dk[i-1, th-1]);

If th≠pi Then mismatch=mismatch+1;i= i- 1; h= h-1 End of while;

If mismatch ≦ k Then report match at position j;j= j+ d End of while

Algorithm for searching phase

14

Complete example for approximate string matching

Example 1:Let k=1, m=4, n=17

T: T T A A C G T A A T G C A G C T A

P: A G C T

Σ A C G T

D1[i=4, a]

3 1 2 4

D1[i=3, a]

2 3 1 4

15

Example 1 (1/6)

T: T T A A C G T A A T G C A G C T A

P: A G C T

Σ A C G T

D1[i=4, a]

3 1 2 4

D1[i=3, a]

2 3 1 4

16

Example 1 (2/6)

T: T T A A C G T A A T G C A G C T A

P: A G C T

Σ A C G T

D1[i=4, a]

3 1 2 4

D1[i=3, a]

2 3 1 4

17

Example 1 (3/6)

T: T T A A C G T A A T G C A G C T A

P: A G C T

Σ A C G T

D1[i=4, a]

3 1 2 4

D1[i=3, a]

2 3 1 4

18

Example 1 (4/6)

T: T T A A C G T A A T G C A G C T A

P: A G C T

Σ A C G T

D1[i=4, a]

3 1 2 4

D1[i=3, a]

2 3 1 4

19

Example 1 (5/6)

T: T T A A C G T A A T G C A G C T A

P: A G C T

Σ A C G T

D1[i=4, a]

3 1 2 4

D1[i=3, a]

2 3 1 4

20

Example 1 (6/6)

T: T T A A C G T A A T G C A G C T A

P: A G C T

Σ A C G T

D1[i=4, a]

3 1 2 4

D1[i=3, a]

2 3 1 4

j ← 16 + p , j ← 16+ 3, j ← 19

jump out of while loop

21

Example 2:Let k=1, m=8, n=24

T: G C A T C G C A G A G A G T A T A C A G T A C G

P: G C A G A G A G

Σ A C G *

D1[i=8, a]

1 6 2 8

D1[i=7, a]

2 5 1 8

22

Example 2 (1/14)

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

T:

G C A T C G C A G A G A G T A T G C A G A G C G

P:

G C A G A G A G

Σ A C G *

D1[i=8, a]

1 6 2 8

D1[i=7, a]

2 5 1 8

23

Example 2 (3/14)

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

T:

G C A T C G C A G A G A G T A T G C A G A G C G

P:

G C A G A G A G

Σ A C G *

D[i=8, a] 1 6 2 8

D[i=7, a] 2 5 1 8

24

Example 2 (4/14)

Σ A C G *

D[i=8, a] 1 6 2 8

D[i=7, a] 2 5 1 8

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

T:

G C A T C G C A G A G A G T A T G C A G A G C G

P:

G C A G A G A G

25

Example 2 (5/14)

Σ A C G *

D[i=8, a] 1 6 2 8

D[i=7, a] 2 5 1 8

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

T:

G C A T C G C A G A G A G T A T G C A G A G C G

P:

G C A G A G A GThen report match at position j;

j ← 13 + p , j ← 13+ 2, j ← 15

26

Example 2 (6/14)

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

T:

G C A T C G C A G A G A G T A T G C A G A G C G

P:

G C A G A G A G

Σ A C G *

D[i=8, a] 1 6 2 8

D[i=7, a] 2 5 1 8

27

Example 2 (7/14)

Σ A C G *

D[i=8, a] 1 6 2 8

D[i=7, a] 2 5 1 8

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

T:

G C A T C G C A G A G A G T A T G C A G A G C G

P:

G C A G A G A G

28

Example 2 (8/14)

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

T:

G C A T C G C A G A G A G T A T G C A G A G C G

P:

G C A G A G A G

Σ A C G *

D[i=8, a] 1 6 2 8

D[i=7, a] 2 5 1 8

29

Example 2 (9/14)

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

T:

G C A T C G C A G A G A G T A T G C A G A G C G

P:

G C A G A G A G

Σ A C G *

D[i=8, a] 1 6 2 8

D[i=7, a] 2 5 1 8

30

Example 2 (11/14)

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

T:

G C A T C G C A G A G A G T A T G C A G A G C G

P:

G C A G A G A G

Σ A C G *

D[i=8, a] 1 6 2 8

D[i=7, a] 2 5 1 8

31

Example 2 (13/14)

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

T:

G C A T C G C A G A G A G T A T G C A G A G C G

P:

G C A G A G A G

Σ A C G *

D[i=8, a] 1 6 2 8

D[i=7, a] 2 5 1 8

32

Example 2 (14/14)

Σ A C G *

D[i=8, a] 1 6 2 8

D[i=7, a] 2 5 1 8

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

T:

G C A T C G C A G A G A G T A T G C A G A G C G

P:

G C A G A G A GIf h = 0 Then report match at position j;

j ← 24 + p , j ← 24+ 2, j ← 26

jump out of while loop

33

Time complexity

• preprocessing phase in O(m+ kc) time and O(kc) space complexity.

• searching phase in O(mn) time complexity.

34

Definition of the k differences problem• Given a pattern string P of length m and a

text string T of length n, we would like to find all approximate occurrences P in T with edit distance not larger than k.

35

The basic approach to solve the problem is to find the edit distance for T(1, i) and P for every i [Ukk85b] :

Let Edit be an m+1 by n+1 table such that Edit(i, j) is the minimum edit distance between p1p2

…pj and any substring of T ending at ti.

niiEdit

tpjiEdit

jiEdit

jiEdit

minjiEdit

ij

0 ,0)0,(

1 0 1)- 1,- (

11)- , (

1) 1,- (

) , (

elsethenif

36

Table Edit must be completely evaluated column-by-column in time O(mn).

37

If we can find out all occurrences of i where Edit(T(1, i), P) cannot be smaller than k. We may skip this i.

This paper is based upon Rule 7 proposed by Professor Lee.

38

Rule 7

• If k characters in String A do not appear in String B, Distance(A,B) is not smaller than k.

39

In the scanning phase, we define some terms first.

A diagonal h of Edit for h=-m,…, n, consists of all Edit(i, j) such that i- j=h.

For every Edit(i, j), there is a minimizing arc from Edit(i-1, j) to Edit(i, j) if Edit(i, j)=Edit(i-1, j)+1, from Edit(i, j-1) to Edit(i, j) if Edit(i, j-1)+1, and from Edit(i-1, j-1) to Edit(i, j) if Edit(i, j)=Edit(i-1, j-1) where pj=ti or if Edit(i, j)=Edit(i-1, j-1)+1 where pj≠ti. The costs of the arcs are 1, 1, 0 and 1, respectively. Edit(i, j-1)

Edit(i-1, j)

Edit(i-1, j-1)

Edit(i, j)

pj≠ti

pj≠ti

pj≠ti

pj=ti

Minimizing arc

Deletion

Insertion

Substitution

40

A minimizing path is any path that consists of minimizing arcs and leads from an entry Edit(i, 0) on the first row of Edit to an entry Edit(h, m) on the last row of Edit.

A minimizing path is successful if it leads to an entry Edit(h, m)≤k.

41

Lemma 1: The entries on a successful minimizing path M are contained in ≤ k+1 successive diagonals of Edit.

Proof : Each addition of a diagonal comes from either an insertion or deletion. If there are more than (k+1) diagonals, there must be more than (k+1) operations, either deletions or insertions. Thus there cannot be more than (k+1) diagonals.

Text

Pattern

t1t2... ...tn-1tn

pm

p1

p2

pm-1...

...

A successful minimizing pathSuccessive diagonals

42

A B C A B B A

C

B

A

B

A

C

0 0 0 0 0 0 0 0

1 1 1 0 1 1 1 1

2 2 1 1 1 1 1 2

3 2 2 2 1 2 2 1

4 3 2 3 2 1 2 2

5 4 3 3 3 2 2 2

6 5 4 3 4 3 3 3

T:ABCABBAP:CBABAC

S:C-AB--P:CBABAC

EDIT(P, S)=3

There are (k+1)=3+1=4 successive diagonals because there are three deletions.

Successive diagonals

43

B C A B D A B

C

B

A

D

B

0 0 0 0 0 0 0 0

1 1 0 1 1 1 1 1

2 1 1 1 1 2 2 1

3 2 2 1 2 2 2 2

4 3 3 2 2 2 3 3

5 4 4 3 3 3 3 3

T:BCABDABP:CBADBk =3

S:C-ABDABP:CBA-D-B

EDIT(P, S)=3

There are 1+2=3 <(k+1) =3+1=4 successive diagonals because there are one deletion and two insertions.

Successive diagonals

44

By Lemma 1, for each diagonal d, any successful minimizing path starting at the top of this diagonal will have a bandwidth of 1+k+k=2k+1

t1t2... ...tn

pm

p1

p2

pm-1

......

M

Bandwidth ≤ k of Edit

h

k

k

2k+1

45

A B C A B B A

C

B

A

B

A

C

0 0 0 0 0 0 0 0

1 1 1 0 1 1 1 1

2 2 1 1 1 1 1 2

3 2 2 2 1 2 2 1

4 3 2 3 2 1 2 2

5 4 3 3 3 2 2 2

6 5 4 3 4 3 3 3

T:ABCABBAP:CBABACk=3

S:C-AB--P:CBABAC

EDIT(P, S)=3

Result

Successive diagonals

The successful minimizing path is only in the bandwidth ≤ 7 of Edit.

k=3

k=3

46

For the width of bandwidth ≤ k of Edit, we give it a name, call k-environment.

For each j=1, …, m, let the k-environment of the pattern symbol pj be the string Cj=pj-k…pj+k, where pa=ε for a<1 and a> m.

Ppj+kpj-1pjpj+1...pj-k...

k-environment

47

The longest vertical path in any minimizing path has length not greater than 2k+1. We only have to determine whether ti appears in the k environment of pj.

t1t2... ...tn

pm

p1

p2

pm-1

......

Bandwidth ≤ k of Edit

h

2k+1

ti

pj

48

P

AT

t5 t8 t11

T C G C A G A G A T

G C A G A G A G A T G

p5 p8 p11

Given T=ATGCGAGAGAT, P=GCAGAGAGATG, and k=2. We select t5, t8 and t11 three characters.

The 2-environment of t5 is C5=p3p4p5p6p7=AGAGA.The 2-environment of t8 is C8=p6p7p8p9p10=GAGAT.The 2-environment of t11 is C11=p9p10p11=ATG.

49

We now obtain a stronger version of Rule 7.

Lemma 2: Let a successful minimizing path M go through some entry on a diagonal h of Edit. Then for at most k indexes j, 1≤j ≤m, character th+j does not occur in the k environment of Cj.

A formal proof can be found in the paper. In the following, we give some physical feeling of it.

50

In this case, although there are two mismatches, by deleting a which mismatches x, we may achieve a perfect match. Thus the edit distance between T and P may still be 1.

T

P

y

a

x

x by

k=1

51

In this case, it can be seen that deleting one character in P will not result in a perfect match. Thus, the edit distance between T and P must be larger than 1.

k=1

T

P

y

b

x

ca bca

52

The shift table is based on table Dk. We determine the first diagonal after h, say h+d, where at least one of the characters th+m, th+m-1, …, th+m-k matches with corresponding character of P. Finally, the maximum of k+1 and d is the length of the shift.

53

The algorithm explains when a possible occurrence of P in T was found, DP approach is immediately used to find alignment result.

54

Input: P = p1p2…pm,T = t1t2…tn and kOutput: All occurrence P in TInitially, the start position h of T =0, i=h+m;While i≤ n+ k do begin j=m; bad=0; While i>k and bad ≤ k do begin

If ti does not occur in Cj then bad=bad+1j=j-1;i=i-1 end;

If bad ≤ k then

W is a sequence from th-k to th+m.Using dynamic programming to align W with POutput alignment result.

We calculate shift steps d=min(Dk[i, tr], Dk[i-1, tr-1],); h=h+max(k+1,d) end;

Algorithm

55

Complete example for approximate string matching

For example :Let k=1, m=8, n=24

T: G C A T C G C A G A G A G T A T G C A G A G C G

P: G C A G A G A G

Σ A C G *

D1[i=8, a]

1 6 2 8

D1[i=7, a]

2 5 1 8

56

Example(1/15)

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

T:

G C A T C G C A G A G A G T A T G C A G A G C G

P:

G C A G A G A G

Σ A C G *

D[i=8, a] 1 6 2 8

D[i=7, a] 2 5 1 8

k=1>k

t8=A appears in P(7,8)t7=C does not appear in P(6,8)t6=G appears in P(5,7)t5=C does not appear in P(4,6)

Shifting is needed now.

57

Example(2/15)

Σ A C G *

D[i=8, a] 1 6 2 8

D[i=7, a] 2 5 1 8

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

T:

G C A T C G C A G A G A G T A T G C A G A G C G

P:

G C A G A G A G

>kk=1

t9=G appears in P(7,8)t8=A appears in P(6,8)t7=C does not appear in P(5,7)t6=G appears in P(4,6)t5=C does not appear in P(3,5)

Shifting is needed now.

58

Example(3/15)

Σ A C G *

D[i=8, a] 1 6 2 8

D[i=7, a] 2 5 1 8

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

T:

G C A T C G C A G A G A G T A T G C A G A G C G

P:

G C A G A G A G

>kk=1

t11=G appears in P(7,8)t10=A appears in P(6,8)t9=G appears in P(5,7)t8=A appears in P(4,6)t7=C does not appear in P(3,5)t6=G appears in P(2,4)t5=C appears in P(1,3)t4=T does not appear in P(1,2) Shifting is needed now.

59

Example(4/15)

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

T:

G C A T C G C A G A G A G T A T G C A G A G C G

P:

G C A G A G A G

Σ A C G *

D[i=8, a] 1 6 2 8

D[i=7, a] 2 5 1 8

GCAGAGAGGCAGAGAG

Output :

C G C A G A G A G T

G

C

A

G

A

G

A

G

0 0 0 0 0 0 0 0 0 0 0

1 1 0 1 1 0 1 0 1 0 1

2 1 1 0 1 1 1 1 1 1 1

3 2 2 1 0 1 1 2 1 2 2

4 2 1 0 1 1 2 1 2

5 2 1 0 1 1 2 2

6 2 1 0 1 1 2

7 2 1 0 1 2

8 2 1 0 1

W= CGCAGAGAGTP= GCAGAGAG

k=1

60

Example(5/15)

GCAGAGA-GCAGAGAG

Output :

C G C A G A G A G T

G

C

A

G

A

G

A

G

0 0 0 0 0 0 0 0 0 0 0

1 1 0 1 1 0 1 0 1 0 1

2 1 1 0 1 1 1 1 1 1 1

3 2 2 1 0 1 1 2 1 2 2

4 2 1 0 1 1 2 1 2

5 2 1 0 1 1 2 2

6 2 1 0 1 1 2

7 2 1 0 1 2

8 2 1 0 1

W= CGCAGAGAGTP= GCAGAGAG

61

Example(6/15)

-CAGAGAGGCAGAGAG

Output :

C G C A G A G A G T

G

C

A

G

A

G

A

G

0 0 0 0 0 0 0 0 0 0 0

1 1 0 1 1 0 1 0 1 0 1

2 1 1 0 1 1 1 1 1 1 1

3 2 2 1 0 1 1 2 1 2 2

4 2 1 0 1 1 2 1 2

5 2 1 0 1 1 2 2

6 2 1 0 1 1 2

7 2 1 0 1 2

8 2 1 0 1

W= CGCAGAGAGTP= GCAGAGAG

62

Example(7/15)

CGCAGAGAG-GCAGAGAG

Output :

C G C A G A G A G T

G

C

A

G

A

G

A

G

0 0 0 0 0 0 0 0 0 0 0

1 1 0 1 1 0 1 0 1 0 1

2 1 1 0 1 1 1 1 1 1 1

3 2 2 1 0 1 1 2 1 2 2

4 2 1 0 1 1 2 1 2

5 2 1 0 1 1 2 2

6 2 1 0 1 1 2

7 2 1 0 1 2

8 2 1 0 1

W= CGCAGAGAGTP= GCAGAGAG

63

Example(8/15)

GCAGAGAGTGCAGAGAG-

Output :

C G C A G A G A G T

G

C

A

G

A

G

A

G

0 0 0 0 0 0 0 0 0 0 0

1 1 0 1 1 0 1 0 1 0 1

2 1 1 0 1 1 1 1 1 1 1

3 2 2 1 0 1 1 2 1 2 2

4 2 1 0 1 1 2 1 2

5 2 1 0 1 1 2 2

6 2 1 0 1 1 2

7 2 1 0 1 2

8 2 1 0 1

W= CGCAGAGAGTP= GCAGAGAG

64

Example(9/15)

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

T:

G C A T C G C A G A G A G T A T G C A G A G C G

P:

G C A G A G A G

a A C G *

D[i=8, a] 1 6 2 8

D[i=7, a] 2 5 1 8

>kk=1

t15=A appears in P(7,8)t14=T does not appear in P(6,8)t13=G appears in P(5,7)t12=A appears in P(4,6)t11=G appears in P(3,5)t10=A appears in P(2,4)t9=G appears in P(1,3)t8=A does not appear in P(1,2) Shifting is needed now.

65

Example(10/15)

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

T:

G C A T C G C A G A G A G T A T G C A G A G C G

P:

G C A G A G A G

a A C G *

D[i=8, a] 1 6 2 8

D[i=7, a] 2 5 1 8

>kk=1

t16=T does not appear in P(7,8)t15=A appears in P(6,8)t14=T does not appear in P(5,7)

Shifting is needed now.

66

Example(11/15)

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

T:

G C A T C G C A G A G A G T A T G C A G A G C G

P:

G C A G A G A G

a A C G *

D[i=8, a] 1 6 2 8

D[i=7, a] 2 5 1 8

>kk=1

t18=C does not appear in P(7,8)t17=G appears in P(6,8)t16=T does not appear in P(5,7)

Shifting is needed now.

67

Example(12/15)

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

T:

G C A T C G C A G A G A G T A T G C A G A G C G

P:

G C A G A G A G

a A C G *

D[i=8, a] 1 6 2 8

D[i=7, a] 2 5 1 8

>kk=1

t19=A appears in P(7,8)t18=C does not appear in P(6,8)t17=G appears in P(5,7)t16=T does not appear in P(4,6)

Shifting is needed now.

68

Example(13/15)

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

T:

G C A T C G C A G A G A G T A T G C A G A G C G

P:

G C A G A G A G

a A C G *

D[i=8, a] 1 6 2 8

D[i=7, a] 2 5 1 8

>kk=1

t20=G appears in P(7,8)t19=A appears in P(6,8)t18=C does not appear in P(5,7)t17=G appears in P(4,6)t16=T does not appear in P(3,5)

Shifting is needed now.

69

Example(14/15)

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

T:

G C A T C G C A G A G A G T A T G C A G A G C G

P:

G C A G A G A G

a A C G *

D[i=8, a] 1 6 2 8

D[i=7, a] 2 5 1 8

>kk=1

t22=G appears in P(7,8)t21=A appears in P(6,8)t20=G appears in P(5,7)t19=A appears in P(4,6)t18=G does not appear in P(3,5)t17=G appears in P(2,4)t16=T does not appear in P(1,3)Shifting is needed now.

70

Example(15/15)

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

T:

G C A T C G C A G A G A G T A T G C A G A G C G

P:

G C A G A G A G

a A C G *

D[i=8, a] 1 6 2 8

D[i=7, a] 2 5 1 8

GCAGAGCGGCAGAGAG

jump out of while loop

T G C A G A G C G

G

C

A

G

A

G

A

G

0 0 0 0 0 0 0 0 0 0

1 1 0 1 1 0 1 0 1 0

2 2 1 0 1 1 1 1 0 1

3 2 1 0 1 1 2 1 1

4 2 1 0 1 1 2 1

5 2 1 0 1 2 2

6 2 1 0 1 2

7 2 1 1 2

8 2 2 1

W= TGCAGAGCGP= GCAGAGAGResult :

k=1

71

Time complexity

• preprocessing phase and searching phase in O(mn/k) time and O(|Σ|n) space complexity.

72

References

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