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Approximate String Matching Using Compressed Suffix Arrays

Trinh N. D. Huynh, W. K. Hon, T. W. Lam and W. K. Sung, Theoretical Computer Science, Vol. 352, 2006, pp. 240-249

Advisor: Prof. R. C. T. Lee

Speaker: C. W. Lu

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• Let x and y be two strings. Edit distance d(x, y) is the minimum number of character insertions, deletions, and replacements to covert string x to y.

• k-difference string matching problem:– Given a text T with length n, a pattern P with lengt

h m, and an error bound k.– Find all position i of T such that there exists an suf

fix S of T(1, i), d(S, P) ≦ k.

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• The approach of this paper is as the follows:

• Given a pattern P and an error bound k, we generate all possible P’s which contain (≦k) errors deduced from P.

• Then we conduct an exact match of all such P’s against T.

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• Example:

T=abbaaa,

P=aba and k=1.

From P and k, we generate the following P’s:

ba, aaba, baba, bba, aa, abba, aaa, ab, abaa, abb, aba.

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• Then we conduct an exact matching of all P’s against T. Any success indicates that there is a substring S in T such that d(S,T)≦k.

• How can we generate all P’s which we want?

• We use the following observation.

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T

P

S2

Let S be a substring of T, and S= S1S2.

P = P1P2.

If d(S1, P1) ≦k, and Dist(S2, P2) = 0,

d(S, P) ≦ k.

S1

S

P1 P2

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Example:

T A C A C A A A A A C A C C

1 2 3 4 5 6 7 8 9 10 11 12 13

A G A B C AP1 2 3 4 5 6

k = 2

Consider the substring S = T(6, 11) = AAAACA,

Let S1 = T(6, 9) = AAAA, and S2 = T(10, 11) = CA.

Dist(S1, P1) = 2 ≦k, and Dist(S2, P2) = 0.

We have Dist(S, P) = 2 ≦k.

S1

P1

S2

P2

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Example:

T A C A C A A A A A C A C C

1 2 3 4 5 6 7 8 9 10 11 12 13

A G A B C AP1 2 3 4 5 6

k = 2

Consider the substring S = T(8, 11) = AACA,

Let S1 = T(8, 9) = AA, and S2 = T(10, 11) = CA.

Dist(S1, P1) = 2 ≦k, and Dist(S2, P2) = 0.

We have Dist(S, P) = 2 ≦k.

S1

P1

S2

P2

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• Based upon the above observation, we can generate all edited pattern P’s by editing the prefix and keeping the suffix untouched, in some manner.

• Consider P=aba, k=1.

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• P=aba, k=1.

P = aba

ba (Deletion) k = 1

i = 1 aaba (Insertion) k = 1baba (Insertion) k = 1

bba (Substution) k = 1

aba k = 0

i = 2

aa (Deletion) k = 1aaba (Insertion) k = 1abba (Insertion) k = 1

aaa (Substution) k = 1

aba k = 0ab (Deletion) k = 1abaa (Insertion) k = 1abba (Insertion) k = 1

abb (Substution) k = 1

aba k = 0

i = 3

i = 4

abaa (Insertion) k = 1abab (Insertion) k = 1

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• P=aba, k=2.

P = aba

ba (Deletion) k = 1

i = 1 aaba (Insertion) k = 1baba (Insertion) k = 1

bba (Substution) k = 1

aba k = 0

i = 2

aa (Deletion) k = 1aaba (Insertion) k = 1abba (Insertion) k = 1

aaa (Substution) k = 1

aba k = 0ab (Deletion) k = 1abaa (Insertion) k = 1abba (Insertion) k = 1

abb (Substution) k = 1

aba k = 0

i = 3

i = 4

abaa (Insertion) k = 1abab (Insertion) k = 1

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• P=aba, k=2.

ba

(k = 1)

a (Deletion) k = 2i = 2 aba (Insertion) k = 2

bba (Insertion) k = 2

aa (Substution) k = 2

ba k = 1

i = 3

b (Deletion) k = 2baa (Insertion) k = 2bba (Insertion) k = 2

bb (Substution) k = 2

ba k = 1

i = 4

baa (Insertion) k = 2

bab (Insertion) k = 2

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For i=1 to m+1

PL’ PR’P’

k’=Dist(PL’, PL)≦k.

Dist(PR’, PR) = 0

iPL’ PR’

P’

iPL

PR

P

Deletion, k’++

A

PL’ PR’

P’

CP’…

Replacement , k’++

A

PL’ PR’

P’

CP’…

Insertion, k’++

PL’ PR’

P’ No operation.

i

Terminate if k’ > k.

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• Our problem now becomes the following: Given a pattern P, we produce a modified pattern P’. Our job is to determine whether P’ exactly matches some substring of T or not.

• For example, Suppose P=aba. We have ba as one of the modified patterns. So, we like to find out whether ba matches exactly with a substring in T.

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• This exact matching can be found by using the suffix array and the inverse suffix array.

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Suffix Array

• Let , where t0, t1, …tn-1 an alphabet A and tn=$ is a special symbol that is not in A and smaller than any symbol in A.

• The jth suffix of T is defined as T(j, n) = tj…tn and is denoted by Tj.

• The suffix array SA[0..n] of T is an array of integers j that represent suffix Tj and the integers are sorted in lexicographic order of corresponding suffixes.

nn- t...tttT 110

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Example:

T G A C A G T T C G $

0 1 2 3 4 5 6 7 8 9

Suffixes of T:

{GACAGTTCG$, ACAGTTCG$, CAGTTCG$, AGTTCG$, GTTCG$, TTCG$, TCG$, CG$, G$, $}

Lexicographic order:

$, ACAGTTCG$, AGTTCG$, CAGTTCG$, CG$, G$, GACAGTTCG$, GTTCG$, TCG$, TTCG$.

= T9, T1, T3, T2, T7, T8, T0, T4, T6, T5

SA[i]

9 1 3 2 7 8 0 4 6 5

0 1 2 3 4 5 6 7 8 9i

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Inverse Suffix Array

• The inverse suffix array of T is denoted as SA-1[i].• SA-1[i] equals the number of suffix which are

lexicographically smaller then Ti.

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Example:

T G A C A G T T C G $

0 1 2 3 4 5 6 7 8 9

Lexicographic order: $

(T9)ACAGTTCG$ (T1)AGTTCG$

(T3)CAGTTCG$

(T2)CG$

(T7)G$

(T8)GACAGTTCG$

(T0)GTTCG$

(T4)TCG$

(T6)TTCG$.

(T5)

SA[i]9

1

3

2

7

8

0

4

6

5

0

1

2

3

4

5

6

7

8

9

i SA-1[i]6

1

3

2

7

9

8

4

5

0

SA-1[SA[x] ] = x.

SA-1[0]=6 because there are 6 suffixes smaller than T0=

GACAGTTCG.

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• The size of SA and SA-1 are O(nlogn) bits. Both data structures can be constructed in linear time[13, 15, 17].

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• In this paper, an interval [st..ed] is called the range of the suffix array of T corresponding to a string P if [st..ed] is the largest interval such that P is a prefix of every suffix Tj for j = SA[st], SA[st+1], …, SA[ed].

We write [st..ed ] = range(T, P).

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Example:

T G A C A G T T C G $

0 1 2 3 4 5 6 7 8 9

Lexicographic order: $

(T9)ACAGTTCG$ (T1)AGTTCG$

(T3)CAGTTCG$

(T2)CG$

(T7)G$

(T8)GACAGTTCG$

(T0)GTTCG$

(T4)TCG$

(T6)TTCG$.

(T5)

SA[i]9

1

3

2

7

8

0

4

6

5

0

1

2

3

4

5

6

7

8

9

i P = G.

G is a prefix of T8, T0 and T4.

T8 = TSA[5]

T0 = TSA[6]

T4 = TSA[7]

st=5, ed=7,

range(T, P) = [5..7].

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Lemma 1 (Gusfild [12])

Given a text T together with its suffix array, assume [st..ed] = range(T, P). Then, for any character c, the interval[st’..ed’] = range(T, Pc) can be computed in O(logn) time.

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Lemma 2

Given the interval [st1..ed1] = range(T , P1) and the interval [st2..ed2] = range(T , P2), we can find the interval [st..ed] = range(T , P1P2) in O(logn) time using the suffix array and the inverse suffix array of T.

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Let [st1..ed1] = range(T , P1),

[st2..ed2] = range(T , P2),

[st..ed] = range(T , P1P2).

[st..ed] is a subinterval of [st1..ed1].

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Example:

T G A C A G T T C G $

0 1 2 3 4 5 6 7 8 9

Lexicographic order: $

(T9)ACAGTTCG$ (T1)AGTTCG$

(T3)CAGTTCG$

(T2)CG$

(T7)G$

(T8)GACAGTTCG$

(T0)GTTCG$

(T4)TCG$

(T6)TTCG$.

(T5)

SA[i]9

1

3

2

7

8

0

4

6

5

0

1

2

3

4

5

6

7

8

9

iP1 = G. P2 = A.

range(T, P1) = [5..7].

range(T, P1P2) must be

within [5..7].

How can we find the

exact interval with [5..7]?

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• By the definition of suffix array, the lexicographic order of are increasing.

• The lexicographic order of

are also increasing.

][]1[][ 111 edSAstSAstSA , ..., T, TT

||][||]1[||][ 111111 PedSAPstSAPstSA , ..., T, TT

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Lexicographic order: $

(T9)ACAGTTCG$ (T1)AGTTCG$

(T3)CAGTTCG$

(T2)CG$

(T7)G$

(T8)GACAGTTCG$

(T0)GTTCG$

(T4)TCG$

(T6)TTCG$.

(T5)

T2 = CAGTTCG$

T2+1 = T3 = AGTTCG$

T2+1 is obtained by deleting the prefix with length 1 from T2.

In general, Ti+1 can be obtained by deleting the prefix with length 1 from Ti.

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Example:T G A C A G T T C G $

0 1 2 3 4 5 6 7 8 9

Lexicographic order: $

(T9)ACAGTTCG$

(T1)AGTTCG$

(T3)CAGTTCG$

(T2)CG$

(T7)G$

(T8)GACAGTTCG$

(T0)GTTCG$

(T4)TCG$

(T6)TTCG$. (T5)

SA[i]9

1

3

2

7

8

0

4

6

5

0

1

2

3

4

5

6

7

8

9

i P1 = G. P2 = A.

range(T, P1) = [5..7].

][]1[][ 111 edSAstSAstSA , ..., T, TT

T8 < T0 < T4

||][||]1[||][ 111111 PedSAPstSAPstSA , ..., T, TT

T8+1, T0+1, T4+1

T9 < T1 < T5

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• The lexicographic order of

are also increasing.

• Thus

• To find st and ed, we find the smallest st such that and the largest ed such that

||][||]1[||][ 111111 PedSAPstSAPstSA , ..., T, TT

|]|][[ ... |]|1][[ |]|][[ 11-1

11-1

11-1 PedSASAPstSASAPstSASA

21-1

2 |]|][[ edPstSASAst . |]|][[ 21

-12 edPedSASAst

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Example:T G A C A G A T C G $

0 1 2 3 4 5 6 7 8 9

Lexicographic order: $

(T9)ACAGTTCG$

(T1)AGTTCG$

(T3)ATCG$. (T5)CAGTTCG$

(T2)CG$

(T7)G$

(T8)GACAGTTCG$

(T0)GATCG$

(T4)TCG$

(T6)

SA[i]9

1

3

5

2

7

8

0

4

6

0

1

2

3

4

5

6

7

8

9

i P1 = G. P2 = A.

range(T, P1) = [6..8].

6 ≦ st, ed ≦ 8

SA-1[i]7

1

4

2

8

3

9

5

6

0

range(T, P2) = [1..3].

range(T, P1P2) = [st..ed].

st = 7 and ed = 8.

3 1 1 1, 1][7][-1 SASA

3 3 1 3, 1][8][-1 SASA

1 0 0, 1][6][-1 SASA

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• To find the interval of the first character of P:

We construct an array C such that for any c in A, C[c] stores the total number of occurrences of all c’ in T, where c’ ≦ c.

range(T, p1) = [C[c2]+1 … C[c]] where c2 is a character immediately before c in A.

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Example:T G A C A G T T C G $

0 1 2 3 4 5 6 7 8 9

Lexicographic order: $

(T9)ACAGTTCG$

(T1)AGTTCG$

(T3)CAGTTCG$

(T2)CG$

(T7)G$

(T8)GACAGTTCG$

(T0)GTTCG$

(T4)TCG$

(T6)TTCG$. (T5)

SA[i]9

1

3

2

7

8

0

4

6

5

0

1

2

3

4

5

6

7

8

9

i

P = GACAGCA

C[A] = 2

C[C] = 4

C[G] = 7

C[T] = 9

range(T, p1)

= [C[C]+1…C[G] ]

= [5…7].

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• Lemma 3

Given the suffix array and the inverse suffix array of T, assume [st..ed] = range(T, P). For any character c, assume we have in advance the array C, we can find the interval [st’..ed’] = range(T, cP) in O(logn) time.

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I Construct Fst [1..m+1] and Fed [1..m+1] such that [Fst [i]..Fed [i]]= range(T ,P[i..m]).II Call kapproximate([0..n], 1, 0, ε, ε).

kapproximate([s’..e’], i, k’, PL’, Υ )begin 1. Given [Fst [i]..Fed [i]] = range(T , P[i..m]) and [s’..e’] = range(T , PL’), by Lemma 2 find [st..ed] = range(T , PL’P[i..m]). 2. Report occurrences of P∗ = PL’P[i..m] in [st..ed] if the interval exists. 3. If (k’ = k) return. 4. For j :=i to m+1 (a) (when j ≦m, deletion at j) Call kapproximate([s’..e’], j+1, k’+1, PL’, dΥ). (b) (when j ≦ m, replacement at j ) for each c in A i. Given [s’..e’] = range(T , PL’), by Lemma 1 find [s’’..e’’] = range(T , PL’c). ii. Call kapproximate([s’’..e’’], j+1, k’+1, PL’c, rΥ). (c) (insertion at j) for each c in A i. Given [s’..e’] = range(T , PL’), by Lemma 1 find [s’’..e’’] = range(T , PL’c). ii. Call kapproximate([s’’..e’’], j, k’+1, PL’c, iΥ). (d) (when j≦m) Given [s’..e’] = range(T , PL’), by Lemma 1 find [s’’..e’’] = range(T , PL’P[j]). s’ := s’’; e’ := e’’; PL’ := PL’P[j]; Υ := uΥ;end

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• After an O(n) time preprocessing the text T into an O(nlogn)-bit data structure, the algorithm solves the k-difference problem in O(|A|kmklogn + outputtime) time.

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Thank you!