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Advisor: Prof. R. C. T. Lee Speaker: C. W. Lu

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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Advisor: Prof. R. C. T. Lee Speaker: C. W. Lu

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  1. Approximate String Matching Using Compressed Suffix ArraysTrinh 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

  2. 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 length m, and an error bound k. • Find all position i of T such that there exists an suffix S of T(1, i), d(S, P) ≦ k.

  3. 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.

  4. 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.

  5. 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.

  6. S S1 S2 T P P1 P2 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.

  7. k = 2 1 2 3 4 5 6 7 8 9 10 11 12 13 T Example: A C A C A A A A A C A C C S1 S2 1 2 3 4 5 6 P A G A B C A P1 P2 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.

  8. k = 2 1 2 3 4 5 6 7 8 9 10 11 12 13 T Example: A C A C A A A A A C A C C S1 S2 1 2 3 4 5 6 P A G A B C A P1 P2 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.

  9. 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.

  10. ba (Deletion) k = 1 aaba (Insertion) k = 1 i = 1 baba (Insertion) k = 1 P = aba • P=aba, k=1. bba (Substution) k = 1 aa (Deletion) k = 1 aba k = 0 aaba (Insertion) k = 1 abba (Insertion) k = 1 i = 2 aaa (Substution) k = 1 ab (Deletion) k = 1 aba k = 0 abaa (Insertion) k = 1 abba (Insertion) k = 1 i = 3 abb (Substution) k = 1 aba k = 0 abaa (Insertion) k = 1 abab (Insertion) k = 1 i = 4

  11. ba (Deletion) k = 1 aaba (Insertion) k = 1 i = 1 baba (Insertion) k = 1 P = aba • P=aba, k=2. bba (Substution) k = 1 aa (Deletion) k = 1 aba k = 0 aaba (Insertion) k = 1 abba (Insertion) k = 1 i = 2 aaa (Substution) k = 1 ab (Deletion) k = 1 aba k = 0 abaa (Insertion) k = 1 abba (Insertion) k = 1 i = 3 abb (Substution) k = 1 aba k = 0 abaa (Insertion) k = 1 abab (Insertion) k = 1 i = 4

  12. a (Deletion) k = 2 i = 2 • P=aba, k=2. aba (Insertion) k = 2 bba (Insertion) k = 2 ba (k = 1) aa (Substution) k = 2 b (Deletion) k = 2 ba k = 1 baa (Insertion) k = 2 bba (Insertion) k = 2 i = 3 bb (Substution) k = 2 ba k = 1 baa (Insertion) k = 2 bab (Insertion) k = 2 i = 4

  13. PR’ PL’ i For i=1 to m+1 Deletion, k’++ P’ PR PL i P PL’ PR’ P’ PL’ PR’ A Replacement , k’++ P’ P’ C … PL’ PR’ k’=Dist(PL’, PL)≦k. Dist(PR’, PR) = 0 P’ Insertion, k’++ A P’ C … PL’ PR’ No operation. P’ i Terminate if k’ > k.

  14. 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.

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

  16. 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.

  17. Example: 0 1 2 3 4 5 6 7 8 9 T G A C A G T T C G $ 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 i 0 1 2 3 4 5 6 7 8 9  SA[i] 9 1 3 2 7 8 0 4 6 5

  18. 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.

  19. Example: 0 1 2 3 4 5 6 7 8 9 T G A C A G T T C G $ Lexicographic order: $ (T9) ACAGTTCG$ (T1) AGTTCG$ (T3) CAGTTCG$ (T2) CG$ (T7) G$ (T8) GACAGTTCG$ (T0) GTTCG$ (T4) TCG$ (T6) TTCG$. (T5) i SA[i] SA-1[i] SA-1[0]=6 because there are 6 suffixes smaller than T0= GACAGTTCG. 0 9 6 1 1 1 2 3 3 3 2 2 4 7 7 5 8 9 6 0 8 7 4 4 SA-1[SA[x] ] = x. 8 6 5 9 5 0

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

  21. 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).

  22. Example: 0 1 2 3 4 5 6 7 8 9 T G A C A G T T C G $ i SA[i] P = G. Lexicographic order: $ (T9) ACAGTTCG$ (T1) AGTTCG$ (T3) CAGTTCG$ (T2) CG$ (T7) G$ (T8) GACAGTTCG$ (T0) GTTCG$ (T4) TCG$ (T6) TTCG$. (T5) 0 9 G is a prefix of T8, T0 and T4. 1 1 2 3 T8 = TSA[5] T0 = TSA[6] T4 = TSA[7]  st=5, ed=7, range(T, P) = [5..7]. 3 2 4 7 5 8 6 0 7 4 8 6 9 5

  23. 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.

  24. 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.

  25. Let [st1..ed1] = range(T , P1), [st2..ed2] = range(T , P2), [st..ed] = range(T , P1P2). • [st..ed] is a subinterval of [st1..ed1].

  26. Example: 0 1 2 3 4 5 6 7 8 9 T G A C A G T T C G $ i SA[i] Lexicographic order: $ (T9) ACAGTTCG$ (T1) AGTTCG$ (T3) CAGTTCG$ (T2) CG$ (T7) G$ (T8) GACAGTTCG$ (T0) GTTCG$ (T4) TCG$ (T6) TTCG$. (T5) P1 = G. P2 = A. 0 9 1 1 range(T, P1) = [5..7]. 2 3 3 2 range(T, P1P2) must be within [5..7]. How can we find the exact interval with [5..7]? 4 7 5 8 6 0 7 4 8 6 9 5

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

  28. 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. Lexicographic order: $ (T9) ACAGTTCG$ (T1) AGTTCG$ (T3) CAGTTCG$ (T2) CG$ (T7) G$ (T8) GACAGTTCG$ (T0) GTTCG$ (T4) TCG$ (T6) TTCG$. (T5)

  29. Example: 0 1 2 3 4 5 6 7 8 9 T G A C A G T T C G $ P1 = G. P2 = A. i SA[i] Lexicographic order: $ (T9) ACAGTTCG$ (T1) AGTTCG$ (T3) CAGTTCG$ (T2) CG$ (T7) G$ (T8) GACAGTTCG$ (T0) GTTCG$ (T4) TCG$ (T6) TTCG$. (T5) 0 9 range(T, P1) = [5..7]. 1 1 2 3 3 2  T8 < T0 < T4 4 7 5 8 6 0 • T8+1, T0+1, T4+1 • T9 < T1 < T5 7 4 8 6 9 5

  30. 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

  31. Example: 0 1 2 3 4 5 6 7 8 9 T G A C A G A T C G $ P1 = G. P2 = A. i SA[i] SA-1[i] Lexicographic order: $ (T9) ACAGTTCG$ (T1) AGTTCG$ (T3) ATCG$. (T5) CAGTTCG$ (T2) CG$ (T7) G$ (T8) GACAGTTCG$ (T0) GATCG$ (T4) TCG$ (T6) 0 9 7 range(T, P1) = [6..8]. 1 1 1 range(T, P2) = [1..3]. 2 3 4 range(T, P1P2) = [st..ed]. 3 5 2 4 2 8 6≦ st, ed ≦8 5 7 3 6 8 9 7 0 5 8 4 6 9 6 0  st = 7and ed =8.

  32. 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.

  33. Example: 0 1 2 3 4 5 6 7 8 9 T G A C A G T T C G $ i SA[i] C[A] = 2 C[C] = 4 C[G] = 7 C[T] = 9 Lexicographic order: $ (T9) ACAGTTCG$ (T1) AGTTCG$ (T3) CAGTTCG$ (T2) CG$ (T7) G$ (T8) GACAGTTCG$ (T0) GTTCG$ (T4) TCG$ (T6) TTCG$. (T5) 0 9 1 1 2 3 3 2 4 7 5 8 P = GACAGCA 6 0 7 4 range(T, p1) = [C[C]+1…C[G] ] = [5…7]. 8 6 9 5

  34. 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.

  35. 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

  36. 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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  40. Thank you!

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