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

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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.

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3 The approach of this paper is as the follows: Given a pattern P and an error bound k, we generate all possible Ps which contain ( k) errors deduced from P. Then we conduct an exact match of all such Ps against T.

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4 Example: T=abbaaa, P=aba and k=1. From P and k, we generate the following Ps: ba, aaba, baba, bba, aa, abba, aaa, ab, abaa, abb, aba.

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5 Then we conduct an exact matching of all Ps against T. Any success indicates that there is a substring S in T such that d(S,T) k. How can we generate all Ps which we want? We use the following observation.

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6 T P S2S2 Let S be a substring of T, and S= S 1 S 2. P = P 1 P 2. If d(S 1, P 1 ) k, and Dist(S 2, P 2 ) = 0, d(S, P) k. S1S1 S P1P1 P2P2

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7 Example: T ACACAAAAACACC AGABCA P k = 2 Consider the substring S = T(6, 11) = AAAACA, Let S 1 = T(6, 9) = AAAA, and S 2 = T(10, 11) = CA. Dist(S 1, P 1 ) = 2 k, and Dist(S 2, P 2 ) = 0. We have Dist(S, P) = 2 k. S1S1 P1P1 S2S2 P2P2

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8 Example: T ACACAAAAACACC AGABCA P k = 2 Consider the substring S = T(8, 11) = AACA, Let S 1 = T(8, 9) = AA, and S 2 = T(10, 11) = CA. Dist(S 1, P 1 ) = 2 k, and Dist(S 2, P 2 ) = 0. We have Dist(S, P) = 2 k. S1S1 P1P1 S2S2 P2P2

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9 Based upon the above observation, we can generate all edited pattern Ps by editing the prefix and keeping the suffix untouched, in some manner. Consider P=aba, k=1.

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

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

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

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13 For i=1 to m+1 P L P R P k=Dist(P L, P L ) k. Dist(P R, P R ) = 0 i P L P R P i PLPL PRPR P Deletion, k++ A P L P R P C P … Replacement, k++ A P L P R P C P … Insertion, k++ P L P R P No operation. i Terminate if k > k.

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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.

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

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16 Suffix Array Let, where t 0, t 1, …t n-1 an alphabet A and t n =$ 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) = t j …t n and is denoted by T j. The suffix array SA[0..n] of T is an array of integers j that represent suffix T j and the integers are sorted in lexicographic order of corresponding suffixes.

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17 Example: T GACAGTTCG$ Suffixes of T: {GACAGTTCG$, ACAGTTCG$, CAGTTCG$, AGTTCG$, GTTCG$, TTCG$, TCG$, CG$, G$, $} Lexicographic order: $, ACAGTTCG$, AGTTCG$, CAGTTCG$, CG$, G$, GACAGTTCG$, GTTCG$, TCG$, TTCG$. = T 9, T 1, T 3, T 2, T 7, T 8, T 0, T 4, T 6, T 5 SA[i] i

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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 T i.

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19 Example: T GACAGTTCG$ Lexicographic order: $(T 9 ) ACAGTTCG$(T 1 ) AGTTCG$(T 3 ) CAGTTCG$(T 2 ) CG$(T 7 ) G$(T 8 ) GACAGTTCG$(T 0 ) GTTCG$(T 4 ) TCG$(T 6 ) TTCG$.(T 5 ) SA[i] i SA -1 [i] SA -1 [SA[x] ] = x. SA -1 [0]=6 because there are 6 suffixes smaller than T 0 = GACAGTTCG.

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20 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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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 T j for j = SA[st], SA[st+1], …, SA[ed]. We write [st..ed ] = range(T, P).

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22 Example: T GACAGTTCG$ Lexicographic order: $(T 9 ) ACAGTTCG$(T 1 ) AGTTCG$(T 3 ) CAGTTCG$(T 2 ) CG$(T 7 ) G$(T 8 ) GACAGTTCG$(T 0 ) GTTCG$(T 4 ) TCG$(T 6 ) TTCG$.(T 5 ) SA[i] i P = G. G is a prefix of T 8, T 0 and T 4. T 8 = T SA[5] T 0 = T SA[6] T 4 = T SA[7] st=5, ed=7, range(T, P) = [5..7].

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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.

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24 Lemma 2 Given the interval [st 1..ed 1 ] = range(T, P 1 ) and the interval [st 2..ed 2 ] = range(T, P 2 ), we can find the interval [st..ed] = range(T, P 1 P 2 ) in O(logn) time using the suffix array and the inverse suffix array of T.

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25 Let [st 1..ed 1 ] = range(T, P 1 ), [st 2..ed 2 ] = range(T, P 2 ), [st..ed] = range(T, P 1 P 2 ). [st..ed] is a subinterval of [st 1..ed 1 ].

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26 Example: T GACAGTTCG$ Lexicographic order: $(T 9 ) ACAGTTCG$(T 1 ) AGTTCG$(T 3 ) CAGTTCG$(T 2 ) CG$(T 7 ) G$(T 8 ) GACAGTTCG$(T 0 ) GTTCG$(T 4 ) TCG$(T 6 ) TTCG$.(T 5 ) SA[i] i P 1 = G. P 2 = A. range(T, P 1 ) = [5..7]. range(T, P 1 P 2 ) must be within [5..7]. How can we find the exact interval with [5..7]?

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

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28 Lexicographic order: $(T 9 ) ACAGTTCG$(T 1 ) AGTTCG$(T 3 ) CAGTTCG$(T 2 ) CG$(T 7 ) G$(T 8 ) GACAGTTCG$(T 0 ) GTTCG$(T 4 ) TCG$(T 6 ) TTCG$.(T 5 ) T 2 = CAGTTCG$ T 2+1 = T 3 = AGTTCG$ T 2+1 is obtained by deleting the prefix with length 1 from T 2. In general, T i+1 can be obtained by deleting the prefix with length 1 from T i.

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29 Example: T GACAGTTCG$ Lexicographic order: $(T 9 ) ACAGTTCG$(T 1 ) AGTTCG$(T 3 ) CAGTTCG$(T 2 ) CG$(T 7 ) G$(T 8 ) GACAGTTCG$(T 0 ) GTTCG$(T 4 ) TCG$(T 6 ) TTCG$.(T 5 ) SA[i] i P 1 = G. P 2 = A. range(T, P 1 ) = [5..7]. T 8 < T 0 < T 4 T 8+1, T 0+1, T 4+1 T 9 < T 1 < T 5

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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

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31 Example: T GACAGATCG$ Lexicographic order: $(T 9 ) ACAGTTCG$(T 1 ) AGTTCG$(T 3 ) ATCG$.(T 5 ) CAGTTCG$(T 2 ) CG$(T 7 ) G$(T 8 ) GACAGTTCG$(T 0 ) GATCG$(T 4 ) TCG$(T 6 ) SA[i] i P 1 = G. P 2 = A. range(T, P 1 ) = [6..8]. 6 st, ed 8 SA -1 [i] range(T, P 2 ) = [1..3]. range(T, P 1 P 2 ) = [st..ed]. st = 7 and ed = 8.

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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, p 1 ) = [C[c 2 ]+1 … C[c]] where c 2 is a character immediately before c in A.

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33 Example: T GACAGTTCG$ Lexicographic order: $(T 9 ) ACAGTTCG$(T 1 ) AGTTCG$(T 3 ) CAGTTCG$(T 2 ) CG$(T 7 ) G$(T 8 ) GACAGTTCG$(T 0 ) GTTCG$(T 4 ) TCG$(T 6 ) TTCG$.(T 5 ) SA[i] i P = GACAGCA C[A] = 2 C[C] = 4 C[G] = 7 C[T] = 9 range(T, p 1 ) = [C[C]+1…C[G] ] = [5…7].

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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.

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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, P L, Υ ) begin 1. Given [Fst [i]..Fed [i]] = range(T, P[i..m]) and [s..e] = range(T, P L ), by Lemma 2 find [st..ed] = range(T, P LP[i..m]). 2. Report occurrences of P = P LP[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, P L, dΥ). (b) (when j m, replacement at j ) for each c in A i. Given [s..e] = range(T, P L ), by Lemma 1 find [s..e] = range(T, P Lc). ii. Call kapproximate([s..e], j+1, k+1, P Lc, rΥ). (c) (insertion at j) for each c in A i. Given [s..e] = range(T, P L ), by Lemma 1 find [s..e] = range(T, P Lc). ii. Call kapproximate([s..e], j, k+1, P Lc, iΥ). (d) (when j m) Given [s..e] = range(T, P L ), by Lemma 1 find [s..e] = range(T, P LP[j]). s := s; e := e; P L := P LP[j]; Υ := uΥ; end

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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| k m k logn + outputtime) time.

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40 Thank you!

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