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1 A Hybrid Indexing Method for Approximate String Matching Journal of Discrete Algorithms, No. 1, Vol. 1, 2000, pp. 205-239, Gonzalo Navarro and Ricardo.

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Presentation on theme: "1 A Hybrid Indexing Method for Approximate String Matching Journal of Discrete Algorithms, No. 1, Vol. 1, 2000, pp. 205-239, Gonzalo Navarro and Ricardo."— Presentation transcript:

1 1 A Hybrid Indexing Method for Approximate String Matching Journal of Discrete Algorithms, No. 1, Vol. 1, 2000, pp , Gonzalo Navarro and Ricardo Baeza-Yates Advisor: Prof. R. C. T. Lee Speaker: Y. K. Shieh

2 2 The approximate string matching problem is: Given a text T of length n, a pattern P of length m (n > m), and a threshold k to the number of "errors" in the matches, find all occurrences of a pattern in a text with k errors.

3 3 This paper uses an exhaustive searching mechanism. We open a window T in T with size m+k (Rule 2) and try to determine whether we are sure that every prefix T of this window T has ed(T,P) > k. If the answer is yes, we ignore this window; otherwise, we use dynamic programming to examine whether any prefix T of the window T has ed(T,P) k.

4 4 We use dynamic programming to compute the edit distance between two strings. A matrix C 0…|m|,0…|n| is filled, where C j,i represents the minimum number of operations need to match T 1…i to P 1…j. This is computed as follows C j,0 and C 0,i represent the edit distance between a string of length j or i and the empty string.

5 5 example: T = surgery P = survey k = 2 surgery s u r v e y There are only three prefixes of T, namely surge, surger and surgery, whose edit distances with P=survey are smaller than or equal to k=2.

6 6 Let us now see how we can be sure that for a window T with size m+k, for every prefix T of T, ed(T,P) > k. We present Lemma 1 of this paper as follows.

7 7 Lemma 1 Let T in T and P be two strings such that ed(T, P) k. Let P = P 1 x 1 P 2 x 2 … x j-1 P j, for strings P i and x i and for any j 1. Then, at least one string P i appears in T with at most errors. Thus, we always divide the pattern into j pieces. We shall point out how to divide later.

8 8 To be more precise, we may say that if ed(T,P) k, there exists a P i in P and a T in T such that ed(P i,T).

9 9 Lemma 1 tells us that if for all P i in P and every substring b in T, ed(P i,b) >, then ed(P,T) > k. Suppose that there is a window T with size m+k and for all P i in P and for every substring b in T, ed(P i,b) >. Then, we can be sure that for every prefix Tof T, for all P i in P and every substring b in T, ed(P i,b) >. b T T T PiPi P

10 10 Let us define the following condition. Condition A: For all P i in P and every substring b in T, ed(P i, b) > Thus, if Condition A is satisfied, then for every prefix T of T, ed(T,P)>k. In such a case, we ignore T and shift P one step to the right.

11 11 Question, how can we be sure that the above condition is satisfied. The approach: For each P i, we generate all possible modified strings P i whose distances with P i are smaller than or equal to k. After generating all possible modified, we may use the suffix tree of T to find all occurrences of, for all i, in T with error less than.

12 12 We still have the following questions: Question 1. How to divide P into j pieces? Question 2. How to generate all modified P i s? Question 3. How to find the occurrences of P i s in T with edit distance less than or equal to.

13 13 Question 1: How to divide P into j pieces? It can be proved that an optimal method is to partition P into j pieces with, where σ is the alphabet size. We can get j pieces of P, and the size of every piece is around log σ n.

14 14 Question 2. How to generate all modified P i s? The generation of all modified strings whose distances with P can be done trivially. One method can be found in [HHLS2006] which was reported by C. W. Lu. Another method can be found in [HM2007] reported By L. C. Chen. In this paper, the authors used the second method mentioned in [HM2007].

15 15 We can use non-deterministic finite automatons (NFA). A NFA is a five-tuple M=(Q, Σ, δ, q 0, F), where Q is a finite set of states, Σ is a finite input alphabet, δ is a mapping from Q×(Σ {ε}) into the set of subsets of Q, q 0 Q is an initial state, and F Q is a set of final states.

16 16 P = abac, k = 2. The finite automaton M accepts L k (P). L k (P)={aa, ab, ac, ba, bc, aaa, aab, aac, aba, abb, abc, acc, baa, bab, bac, bbc, bcc, aaaa, aaab, aaac, aaba, aabc, aaca, aacb, aacc, abaa, abab, abac, abba, abbb, abbc, abca, abcb, abcc, baac, babc, bbac, bbbc, bcac}.

17 17 P = abac, k = 2. The finite automaton M accepts L k (P). L k (P)={aa, ab, ac, ba, bc, aaa, aab, aac, aba, abb, abc, acc, baa, bab, bac, bbc, bcc, aaaa, aaab, aaac, aaba, aabc, aaca, aacb, aacc, abaa, abab, abac, abba, abbb, abbc, abca, abcb, abcc, baac, babc, bbac, bbbc, bcac}. Recognize aa

18 18 Full example: T = GACACAGACCAAAGCAGn = 17 P = CAAGm = 4 k = 1

19 19 P = CAAG j = (m + k) / log σ n = (4 + 1) / log 3 17 = Therefore, we partition P into two pieces. P 1 = CA P 2 = AG According to Lemma 1, at least one piece appears in substrings of T with at most = 0 error. This means that we want to find exact matching of P 1 and P 2.

20 20 NFA with k = 1 of P 1 = CA: NFA with k = 1 of P 2 = AG:

21 21 T = GACACGGACCAAAGCAG We construct the suffix tree of T. A C G C G CAAAGCAG$CAAAGCAG$ GGACCAAAGCAG$GGACCAAAGCAG$ ACGGACCAAAGCAG$ACGGACCAAAGCAG$ A AAGCAG$AAGCAG$ CGGACCAAAGCAG$CGGACCAAAGCAG$ G$G$ CAAAGCAG$CAAAGCAG$ GGACCAAAGCAG$GGACCAAAGCAG$ ACAC ACGGACCAAAGCAG$ACGGACCAAAGCAG$ CAAAGCAG$CAAAGCAG$ CAG$CAG$ GACCAAAGCAG$GACCAAAGCAG$ A GCAG$GCAG$ AGCAG$AGCAG$ $ CAG$CAG$ $

22 22 We only need to consider the tree level from root to = 3. A C G C G C G A A A C G CACA G ACAC CACA GAGA A G A $ C $ ,7 T = GACACGGACCAAAGCAG

23 23 A C G C G C G A A A C G CACA G ACAC CACA GAGA A G A $ C $ ,7 NFA of P 1 : NFA of P 2 T = GACACGGACCAAAGCAG k = 1

24 24 A C G C G C G A A A C G CACA G ACAC CACA GAGA A G A $ C $ ,7 (not exact match) T = GACACGGACCAAAGCAG k = 1

25 25 A C G C G C G A A A C G CACA G ACAC CACA GAGA A G A $ C $ ,7 Out of active states. (not exact match) T = GACACGGACCAAAGCAG k = 1

26 26 A C G C G C G A A A C G CACA G ACAC CACA GAGA A G A $ C $ ,7 (exact match) Out of active states. We record positions 13 and 16 where AG occurs. T = GACACGGACCAAAGCAG k = 1

27 27 A C G C G C G A A A C G CACA G ACAC CACA GAGA A G A $ C $ ,7 T = GACACGGACCAAAGCAG k = 1

28 28 A C G C G C G A A A C G CACA G ACAC CACA GAGA A G A $ C $ ,7 Out of active states. (exact match) We record positions 3, 10 and 15 where CA occurs. T = GACACGGACCAAAGCAG k = 1

29 29 A C G C G C G A A A C G CACA G ACAC CACA GAGA A G A $ C $ ,7 Out of active states. (not exact match) T = GACACGGACCAAAGCAG k = 1

30 30 A C G C G C G A A A C G CACA G ACAC CACA GAGA A G A $ C $ ,7 (not exact match) T = GACACGGACCAAAGCAG k = 1

31 31 A C G C G C G A A A C G CACA G ACAC CACA GAGA A G A $ C $ ,7 T = GACACGGACCAAAGCAG k = 1

32 32 A C G C G C G A A A C G CACA G ACAC CACA GAGA A G A $ C $ ,7 Out of active states. T = GACACGGACCAAAGCAG k = 1

33 33 A C G C G C G A A A C G CACA G ACAC CACA GAGA A G A $ C $ ,7 Out of active states. (not exact match) T = GACACGGACCAAAGCAG k = 1

34 34 A C G C G C G A A A C G CACA G ACAC CACA GAGA A G A $ C $ ,7 Out of active states. T = GACACGGACCAAAGCAG k = 1

35 35 A C G C G C G A A A C G CACA G ACAC CACA GAGA A G A $ C $ ,7 Out of active states. (not exact match) T = GACACGGACCAAAGCAG k = 1

36 36 After we find all probable positions in T, we verify every substring of those positions. The probable positions of T are: 3, 10, 13, 15, 16 We use the dynamic program to verify whether any approximate string matching occurs between T and P at the above locations.

37 37 The probable positions of T are 3, 10, 13, 15, 16 m+k GACAC C A A G k = 1 No approximate matching with k=1 found.

38 38 m+k ACACG C A A G The probable positions of T are: 3, 10, 13, 15, 16 k = 1 No approximate matching with k=1 found.

39 39 m+k CACGG C A A G The probable positions of T are: 3, 10, 13, 15, 16 CACG is found. k = 1

40 40 m+k The probable positions of T are: 3, 10, 13, 15, 16 This window does not include any probable position. Therefore we can ignore this window.

41 41 m+k The probable positions of T are: 3, 10, 13, 15, 16 The window does not include any probable position. Therefore we can shift the window directly.

42 42 m+k GGACC C A A G The probable positions of T are: 3, 10, 13, 15, 16 k = 1 No approximate matching with k=1 found.

43 43 m+k GACCA C A A G The probable positions of T are: 3, 10, 13, 15, 16 k = 1 No approximate matching with k=1 found.

44 44 m+k ACCAA C A A G The probable positions of T are: 3, 10, 13, 15, 16 k = 1 No approximate matching with k=1 found.

45 45 m+k CCAAA C A A G The probable positions of T are: 3, 10, 13, 15, 16 k = 1 No approximate matching with k=1 found.

46 46 m+k CAAAG C A A G The probable positions of T are: 3, 10, 13, 15, 16 CAA, CAAA and CAAAG are found. k = 1

47 47 m+k AAAGC C A A G The probable positions of T are: 3, 10, 13, 15, 16 k = 1 AAAG is found.

48 48 m+k AAGCA C A A G The probable positions of T are: 3, 10, 13, 15, 16 k = 1 AAG is found.

49 49 m+k AGCAG C A A G The probable positions of T are: 3, 10, 13, 15, 16 k = 1 No approximate matching with k=1 found.

50 50 m GCAG C11123 A22212 A33322 G43332 The probable positions of T are: 3, 10, 13, 15, 16 k = 1 No approximate matching with k=1 found.

51 51 m-k CAG 0123 C1012 A2101 A3211 G4321 The probable positions of T are: 3, 10, 13, 15, 16 k = 1 CAG is found.

52 52 Time complexity The preprocessing time complexity of constructing automatons and a suffix tree of T is O(|N|*|m|) and O(n) respectively, |N| is the number of states in a NFA and |m| is the length of m. The search time obtained using the partitioning scheme is O(n λ logn), where λ < 1 when error tolerated α < 1-e/, where e = 2.718….

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57 57 Thank you


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