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String Matching Problem Given a text string T of length n and a  pattern string  P of length m, the exact string matching problem is to find all occurrences of  P in T . • Example: T=“AGCTTGAP=“GCT” Applications:   Searching keywords in a file   Searching engines (like Google and Openfind)   Database searching (GenBank)

String Matching Problem.ppt

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    String Matching Problem

    Given a text stringTof length n and apattern stringPof length m, the exact stringmatching problem is to find all occurrences

    ofPin T. Example: T=AGCTTGA P=GCT

    Applications:

    Searchingkeywords in a fileSearching engines (like Google and Openfind)

    Database searching (GenBank)

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    Problem/issue

    Finding occurrence of a pattern (string)

    P in String S and also finding theposition in S where the pattern match

    occurs

    What is pattern matching?

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    Brute Force algorithm

    The brute-force pattern matching algorithm comparesthe pattern Pwith the text Tfor each possible shift ofP

    relative to T,

    *until either a match is found, or

    *all placements of the pattern have been tried

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    Brute-force

    Worst O(m*n)

    Best O(n)

    algorithmbrute-force:

    input: an array of characters, T (the string to be analyzed) , length n

    an array of characters, P (the pattern to be searched for), length m

    for i := 0 to n-m do

    for j := 0 to m-1 do

    compare T[j] with P[i+j]ifnot equal, exit the inner loop

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    Compare each character of P with S if

    match continue else shift one position

    String S

    a

    b

    a

    a

    a

    b

    c

    a

    b

    a

    a

    b

    c

    a

    b

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    c

    Pattern p

    Example

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    Step 1:compare p[1] with S[1]

    a

    b

    c

    a

    b

    a

    a

    b

    c

    a

    b

    a

    c

    a

    b

    a

    a

    Step 2: compare p[2] with S[2]

    a

    b

    c

    a

    b

    a

    a

    b

    c

    a

    b

    a

    c

    a

    b

    a

    a

    S

    S

    p

    p

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    Step 3: compare p[3] with S[3]

    S a b c a b a a b c a b a c

    a b a ap

    Mismatch occurs here..

    Since mismatch is detected, shift P one position to the Right andperform steps analogous to those from step 1 to step 3. At positionwhere mismatch is detected, shift P one position to the right andrepeat matching procedure.

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    The Knuth-Morris-Pratt Algorithm

    Knuth, Morris and Pratt proposed a lineartime algorithm for the string matchingproblem.

    A matching time of O(n) is achieved byavoiding comparisons with elements of Sthat have previously been involved incomparison with some element of thepattern p to be matched. i.e.,backtracking on the string S never occurs

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    Components of KMP algorithm

    The prefix function,

    The prefix function, for a pattern encapsulatesknowledge about how the pattern matches against shiftsof itself. This information can be used to avoid uselessshifts of the pattern p. In other words, this enablesavoiding backtracking on the string S.

    The KMP Matcher

    With string S, pattern p and prefix function asinputs, finds the occurrence of p in S and returns the

    number of shifts of p after which occurrence is found.

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    Knuth-Morris-Pratt algorithm-Algorithm

    Compute-Prefix-Function(P)

    1. m length[T]

    2. [1] 0

    3. k 0

    4. forq2 tom5. do whilek> 0 and P[k+ 1] P[q]

    6. dok[k] /*ifk= 0 orP[k+ 1]

    = P[q],

    7. ifP[k+ 1] = P[q] going out of thewhile-loop.*/

    8. then kk+ 1

    9. [q]k

    10.return

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    Knuth-Morris-Pratt algorithm-Algorithm

    KMP-Matcher(T, P)

    1. n length[T]

    2. m length[P]

    3. Compute-Prefix-Function(P)

    4. q 05. fori1 ton

    6. do whileq > 0 and P[q + 1] T[i]

    7. doq [q]

    8. ifP[q + 1] = T[i]9. then qq + 1

    10. ifq = m

    11. thenprint pattern occurs with shift im

    12. q [q]

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    Compute prefix function

    P= ababababca, T= ababaababababca[1]= 0

    k= 0

    q = 2, P[k+ 1] = P[1] = a, P[q] = P[2] = b, P[k+ 1] P[q]

    [q]k([2] 0)q = 3, P[k+ 1] = P[1] = a, P[q] = P[3] = a, P[k+ 1] = P[q]

    k k+ 1, [q]k([3] 1)

    k= 1

    q = 4, P[k+ 1] = P[2] = b, P[q] = P[4] = b, P[k+ 1] = P[q]

    k k+ 1, [q]k([4] 2)

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    k= 2

    q = 5, P[k+ 1] = P[3] = a, P[q] = P[5] = a, P[k+ 1] = P[q]

    k k+ 1, [q]k([5] 3)k= 3

    q = 6, P[k+ 1] = P[4] = b, P[q] = P[6] = b, P[k+ 1] = P[q]

    k k+ 1, [q]k([6] 4)

    k= 4q = 7, P[k+ 1] = P[5] = a, P[q] = P[7] = a, P[k+ 1] = P[q]

    k k+ 1, [q]k([7] 5)

    k= 5

    q = 8, P[k+ 1] = P[6] = b, P[q] = P[8] = b, P[k+ 1] = P[q]

    k k+ 1, [q]k([8] 6)

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    k= 6

    q = 9, P[k+ 1] = P[6] = b, P[q] = P[9] = c, P[k+ 1] P[q]k[k](k[6] = 4)

    P[k+ 1] = P[5] = a, P[q] = P[9] = c, P[k+ 1] P[q]

    k[k](k[4] = 2)

    P[k+ 1] = P[3] = a, P[q] = P[9] = c, P[k+ 1] P[q]

    k[k](k[2] = 0)

    k= 0

    q = 9, P[k+ 1] = P[1] = a, P[q] = P[9] = c, P[k+ 1] P[q]

    [q]k([9] 0)

    q = 10, P[k+ 1] = P[1] = a, P[q] = P[10] = a, P[k+ 1] = P[q]k k+ 1, [q]k([10] 1)

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    After prefix computation, the table is shown below

    P= ababababca

    1 2 3 4 5 6 7 8 9 10

    a b a b a b a b c a0 0 1 2 3 4 5 6 0 1

    i

    P[i][i]

    a b a b a b a b c a

    a b a b a b

    a b a b

    a b

    a b c a

    a b a b c a

    a b a b a b c a

    a b a b a b a b c a

    P8

    P6

    P4

    P2

    P0

    [8] = 6[6] = 4

    [4] = 2

    [2] = 0

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    Another Example for KMP Algorithm

    Phase 1

    Phase 2

    f(41)+1=f(3)+1=0+1=1

    f(13-1)+1= 4+1=5matched

    First finish the prefix

    computation

    Next, Search phase computation