String matching algorithm, Lecture notes of Design and Analysis of Algorithms

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String Matching

Finding all occurrences of a pattern in

the text.

Applications

• Spell Checkers.

• Search Engines.

• Spam Filters.

• Intrusion Detection System.

• Plagiarism Detection.

• Bioinformatics – DNA Sequencing.

• Digital Forensics.

• Information Retrieval, etc.

Contd…

shift = 0

shift = 1

shift = 2

shift = 3

shift = 4

shift = 5

shift = 6

shift = 7

shift = 8

shift = 9

T a b c a b a a b c a b a c

P a b a a

P a b a a

P a b a a

P a b a a

P a b a a

P a b a a

P a b a a

P a b a a

P a b a a

P a b a a

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

Contd…

• Shift s a valid shift, if P occurs with shift s in T.

– 0 ≤ s ≤ n – m and T [ s + 1.. s + m ] = P [1.. m ].

• Otherwise, shift s is an invalid shift.

• String-matching problem means finding all valid

shifts with which a given pattern P occurs in a given

text T.

Example

• Text T = acaabc , and pattern P = aab.

Rabin-Karp Algorithm

• Uses elementary number-theoretic notions.

– Equivalence of two numbers modulo a third

number.

• Let, Σ = { 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 }.

• String of k consecutive characters represents a

length- k decimal number.

– Thus, character string 31415 corresponds to the

decimal number 31,415.

• Note:

– In the general case, each character is a digit in

radix- d notation, where d = |Σ|.)

A few calculations…

• For a pattern P [1.. m ], let p denote its corresponding

value in radix- d notation.

• Using Horner’s rule, p can be computed in time ϴ ( m ).

p = P [ m ] + d ( P [ m – 1] + d ( P [ m – 2] + … + d ( P [2] + dP [1]) …)).

• Similarly, for a text T [1.. n ], let t

s

denotes the radix- d

notation value of the length- m substring T [ s + 1.. s +

m ], for s = 0, 1, …, n – m.

• Again, t

0

can be computed from T [1.. m ] in ϴ ( m ).

Contd…

• Each of the remaining values t

1

, t

2

, …, t

n – m

can be

computed in constant time.

– Subtracting d

m – 1

T [ s + 1] removes the high-order digit

from t

s

, multiplying the result by d shifts the number left

by one digit position, and adding T [ s + m + 1] brings in

the appropriate low-order digit.

t

s +

= d ( t

s

– d

m – 1

T [ s + 1]) + T [ s + m + 1].

• Let h = d

m – 1

, then

t

s +

= d ( t

s

– hT [ s + 1]) + T [ s + m + 1].

• Modulus:

– p modulo q takes ϴ ( m ) time.

– For all t

s

, t

s

modulo q takes ϴ ( n – m + 1) time.

Example – 1

Index 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

Text (T) 2 3 5 9 0 2 3 1 4 1 5 2 6 7 3 9 9 2

Pattern: 3 1 4 1 5

Σ = {0, 1, …, 9} d = |Σ| = 10 q = 13

• n = 19, m = 5, n – m = 14.
• h = 10 5- mod 13 = 10 4 mod 13 = 3.
• p = (3 × 10 4 + 1 × 10 3 + 4 × 10 2 + 1 × 10 1 + 5 × 10 0 ) mod 13 = 7.
• t 0 = (2 × 10 4 + 3 × 10 3 + 5 × 10 2 + 9 × 10 1 + 0 × 10 0 ) mod 13 = 8. Step 1: s = 0, t 0 = 8, p = 7. p == t 0 → No. s < 14 → Yes. ts +1 = d ( ts – hT [ s + 1]) + T [ s + m + 1] (mod 13) t 1 = 10 (8 – 3 (2)) + 2 (mod 13) t 1 = 10 (2) + 2 (mod 13) = 22 (mod 13) = 9.

Contd…

Index 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

Text (T) 2 3 5 9 0 2 3 1 4 1 5 2 6 7 3 9 9 2

Pattern: 3 1 4 1 5

Σ = {0, 1, …, 9} d = |Σ| = 10 q = 13

• n = 19, m = 5, n – m = 14.
• h = 10 5- mod 13 = 10 4 mod 13 = 3.
• p = (3 × 10 4 + 1 × 10 3 + 4 × 10 2 + 1 × 10 1 + 5 × 10 0 ) mod 13 = 7.
• t 0 = (2 × 10 4 + 3 × 10 3 + 5 × 10 2 + 9 × 10 1 + 0 × 10 0 ) mod 13 = 8. Step 2: s = 1, t 1 = 9, p = 7. p == t 1 → No. s < 14 → Yes. ts +1 = d ( ts – hT [ s + 1]) + T [ s + m + 1] (mod 13) t 2 = 10 (9 – 3 (3)) + 3 (mod 13) t 2 = 10 (0) + 3 (mod 13) = 3 (mod 13) = 3.

Contd…

Index 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

Text (T) 2 3 5 9 0 2 3 1 4 1 5 2 6 7 3 9 9 2

Pattern: 3 1 4 1 5

Σ = {0, 1, …, 9} d = |Σ| = 10 q = 13

• n = 19, m = 5, n – m = 14.
• h = 10 5- mod 13 = 10 4 mod 13 = 3.
• p = (3 × 10 4 + 1 × 10 3 + 4 × 10 2 + 1 × 10 1 + 5 × 10 0 ) mod 13 = 7.
• t 0 = (2 × 10 4 + 3 × 10 3 + 5 × 10 2 + 9 × 10 1 + 0 × 10 0 ) mod 13 = 8. Step 4: s = 3, t 3 = 11, p = 7. p == t 3 → No. s < 14 → Yes. ts + 1 = d ( ts – hT [ s + 1 ]) + T [ s + m + 1 ] (mod 13 ) t 4 = 10 (11 – 3 (9)) + 4 (mod 13) = 10 (-16) + 4 (mod 13) Because -16 mod 13 = 10^ t 4 = 10 (10) + 4 (mod 13)^ = 104 (mod 13) = 0.

Contd…

Index 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

Text (T) 2 3 5 9 0 2 3 1 4 1 5 2 6 7 3 9 9 2

Pattern: 3 1 4 1 5

Σ = {0, 1, …, 9} d = |Σ| = 10 q = 13

• n = 19, m = 5, n – m = 14.
• h = 10 5- mod 13 = 10 4 mod 13 = 3.
• p = (3 × 10 4 + 1 × 10 3 + 4 × 10 2 + 1 × 10 1 + 5 × 10 0 ) mod 13 = 7.
• t 0 = (2 × 10 4 + 3 × 10 3 + 5 × 10 2 + 9 × 10 1 + 0 × 10 0 ) mod 13 = 8. Step 5: s = 4, t 4 = 0, p = 7. p == t 4 → No. s < 14 → Yes. ts + 1 = d ( ts – hT [ s + 1 ]) + T [ s + m + 1 ] (mod 13 ) t 5 = 10 (0 – 3 (0)) + 1 (mod 13) t 5 = 1 (mod 13) = 1.

Contd…

Index 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

Text (T) 2 3 5 9 0 2 3 1 4 1 5 2 6 7 3 9 9 2

Pattern: 3 1 4 1 5

Σ = {0, 1, …, 9} d = |Σ| = 10 q = 13

• n = 19, m = 5, n – m = 14.
• h = 10 5- mod 13 = 10 4 mod 13 = 3.
• p = (3 × 10 4 + 1 × 10 3 + 4 × 10 2 + 1 × 10 1 + 5 × 10 0 ) mod 13 = 7.
• t 0 = (2 × 10 4 + 3 × 10 3 + 5 × 10 2 + 9 × 10 1 + 0 × 10 0 ) mod 13 = 8. Step 7: s = 6, t 6 = 7, p = 7. p == t 6 → Yes. Character by character matching p[1..5] == T[7..11]. {3 1 4 1 5} == {3 1 4 1 5}. Match, hence s = 6 is a valid shift. s < 14 → Yes. ts +1 = d ( ts – hT [ s + 1]) + T [ s + m + 1] (mod 13) t 7 = 10 (7 – 3 (3)) + 2 (mod 13) = 10 (-2) + 2 (mod 13) t 7 = 10 (11) + 2 (mod 13) = 112 (mod 13) = 8. Because -2 mod 13 = 11

Contd…

Index 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

Text (T) 2 3 5 9 0 2 3 1 4 1 5 2 6 7 3 9 9 2

Pattern: 3 1 4 1 5

Σ = {0, 1, …, 9} d = |Σ| = 10 q = 13

• n = 19, m = 5, n – m = 14.
• h = 10 5- mod 13 = 10 4 mod 13 = 3.
• p = (3 × 10 4 + 1 × 10 3 + 4 × 10 2 + 1 × 10 1 + 5 × 10 0 ) mod 13 = 7.
• t 0 = (2 × 10 4 + 3 × 10 3 + 5 × 10 2 + 9 × 10 1 + 0 × 10 0 ) mod 13 = 8. Step 8: s = 7, t 7 = 8, p = 7. p == t 7 → No. s < 14 → Yes. ts +1 = d ( ts – hT [ s + 1]) + T [ s + m + 1] (mod 13) t 8 = 10 (8 – 3 (1)) + 6 (mod 13) t 8 = 10 (5) + 6 (mod 13) = 56 (mod 13) = 4.