![Introduction
• It makes use of elementary number-theoretic notions such as the Equivalence of Two
Numbers Modulo a Third Number.
• ∑ ={ 0, 1, . . . , 9} - Each character is a decimal digit in radix-d notation.
• d = | ∑ |.
• k consecutive characters represent length-k decimal number.
• P [1. . m] : Pattern
• p : Decimal Value corresponding to pattern P [1. . m].
• T [1 . . n] : Text
• ts : Decimal value of the length-m substring T [s + 1 . . s + m], for
s = 0, 1, . . . , n – m.](https://image.slidesharecdn.com/rabin-karpalgorithm-200422174416/85/Rabin-Karp-Algorithm-2-320.jpg)
![Introduction…
• ts = p if and only if T [s + 1 . . s + m] = P [1. . m]
→ s is a valid shift iff ts = p
Computing p:
• p can be computed in Θ (m) time using Horner’s Rule:
p = P [m] + 10 (P [m – 1] + 10 (P [m – 2] + . . . + 10 P [2] + P [1]…))
Computing (t0 , t1 , , , tn – m):
• t0 can be computed from T [1 . . m] in Θ (m) time using Horner’s rule.
• However, ts + 1 can be computed from ts in constant time:
ts + 1 = 10 (ts – 10 m – 1 T [s + 1]) + T [s + m + 1]
→ t1 , t2 , , , tn - m can be computed in Θ (n – m) time.](https://image.slidesharecdn.com/rabin-karpalgorithm-200422174416/85/Rabin-Karp-Algorithm-3-320.jpg)
![Introduction…
Eg.: T [1 . . 6] = 314152 and pattern length m = 5
Computing t0 using Horner’s Rule:
t0 = T [5] + 10 (T [4] + 10 (T [3] + 10 (T [2] + 10 (T [1])))))
= 5 + 10 (1 + 10 (4 + 10 (1 + 10 (3))))
= 5 + 10 (1 + 10 (4 + 10 (1 + 30)))
= 5 + 10 (1 + 10 (4 + 10 (31)))
= 5 + 10 (1 + 10 (4 + 310))
= 5 + 10 (1 + 10 (314))
= 5 + 10 (1 + 3140)
= 5 + 10 (3141)
= 5 + 31410
= 31415](https://image.slidesharecdn.com/rabin-karpalgorithm-200422174416/85/Rabin-Karp-Algorithm-4-320.jpg)
![Introduction…
Computing t1 using : ts + 1 = 10 (ts – 10 m – 1 T [s + 1]) + T [s + m + 1]
s = 0, m = 5, T [1 . . 6] = 314152, t0 = 31415
t1 = 10 (t0 – 10 5 – 1 (T [1])) + T [0 + 5 + 1]
= 10 (t0 – 10 4 (T [1])) + T [6]
= 10 (31415 – 10000 (3)) + 2
= 10 (31415 – 30000) + 2
= 10 (1415) + 2
= 14150 + 2
= 14152](https://image.slidesharecdn.com/rabin-karpalgorithm-200422174416/85/Rabin-Karp-Algorithm-5-320.jpg)
![Introduction…
• If 10 m – 1 is precomputed, then computing t1 , t2 , , , tn - m will take a constant number
of arithmetic operations.
• Running Time:
p can be computed in Θ (m) time.
t0 , t1 , , , tn - m can be computed in Θ (n – m + 1) time.
Hence, all occurrences of pattern P [1 . . m] in the text T [1 . . n] can be found
with Θ (m) preprocessing time and Θ (n – m + 1) matching time.](https://image.slidesharecdn.com/rabin-karpalgorithm-200422174416/85/Rabin-Karp-Algorithm-6-320.jpg)
![Introduction…
Note:
If p and ts are too large then computing p and ts may not happen in a constant time.
Solution:
• Compute p and the ts values modulo a suitable modulus q.
• If q is chosen to be a prime such that 10q fits within one computer word, then all
necessary computations can be performed with single-precision arithmetic.
• In general with a d-ary alphabet {0, 1, . . . , d – 1} q has to be chosen so that dq fits
within a computer word.
ts + 1 can be computed as follows:
ts + 1 = (d (ts – T [s + 1] h) + T [s + m + 1]) mod q
h = d m – 1 (mod q) - value of most significant digit in the m-digit text window.](https://image.slidesharecdn.com/rabin-karpalgorithm-200422174416/85/Rabin-Karp-Algorithm-7-320.jpg)
![Example
Text : 2359023141526739921
n : 19
Pattern : 31415
M : 5
q : 13
Computing p and t0 using Horner’s Rule:
P [m] + 10 (P [m - 1] + 10 (P [m - 2] + . . . + 10 P [2] + P [1]…))
p = 5 + 10 (1 + 10 (4 + 10 (1 + 10 (3)))) = 31415
t0 = 0 + 10 (9 + 10 (5 + 10 (3 + 10 (2)))) (mod 13) = 23590 mod 13 = 8](https://image.slidesharecdn.com/rabin-karpalgorithm-200422174416/85/Rabin-Karp-Algorithm-8-320.jpg)
![Example…
Computing ts + 1 :
ts + 1 = (d (ts – T [s + 1] h) + T [s + m + 1]) mod q, h = d m – 1 (mod q)
h = 10 4 (mod 13) = 3
s = 0, t0 = 8, T [1 . . 6] = 235902
t1 = (10 (8 – 2 (3)) + 2) (mod 13) = (10 (2) + 2) (mod 13) = 22 mod (13) = 9
s = 1, t1 = 9, T [2 . . 7] = 359023
t2 = (10 (9 – 3 (3)) + 3) (mod 13) = (10 (0) + 3) (mod 13) = 3 mod (13) = 3
s = 2, t2 = 3, T [3 . . 8] = 590231
t3 = (10 (3 – 5 (3)) + 1) (mod 13) = (10 (-12) + 1) (mod 13) = -119 mod (13) = 11](https://image.slidesharecdn.com/rabin-karpalgorithm-200422174416/85/Rabin-Karp-Algorithm-9-320.jpg)
![Example…
• ts ≡ p (mod q) - Hit
• ts ≡ p (mod q) but ts ≠ p - Spurious Hit
• ts ≠ p (mod q) → ts ≠ p - Invalid Shift
• ts ≡ p (mod q) and P [1. . m] = T [s + 1 . . s + m] - Valid Shift, s.](https://image.slidesharecdn.com/rabin-karpalgorithm-200422174416/85/Rabin-Karp-Algorithm-11-320.jpg)
![Algorithm
RABIN – KARP MATCHER (T, P, d, q)
n = T.length
m = P.length
h = d m - 1 (mod q)
p = 0
t0 = 0
For (i =1 to m)
p = (dp + P [i]) (mod q)
t0 = (dt0 + T [i]) (mod q)
For (s = 0 to n – m)
If (p == ts)
If (P [1 . . m] == T [s + 1 . . s + m])
Print “Pattern occurs with shift ” s
If (s < n – m)
ts + 1 = (d (ts – T [s + 1] h ) + T [s + m + 1]) (mod q)
Running Time:
Preprocessing Time: Θ (m)
Matching Time: Θ ((n – m + 1) m)](https://image.slidesharecdn.com/rabin-karpalgorithm-200422174416/85/Rabin-Karp-Algorithm-12-320.jpg)
Rabin Karp Algorithm
- 1. Rabin-Karp Algorithm Dr. Kiran K Assistant Professor Department of CSE UVCE Bengaluru, India.
- 2. Introduction • It makes use of elementary number-theoretic notions such as the Equivalence of Two Numbers Modulo a Third Number. • ∑ ={ 0, 1, . . . , 9} - Each character is a decimal digit in radix-d notation. • d = | ∑ |. • k consecutive characters represent length-k decimal number. • P [1. . m] : Pattern • p : Decimal Value corresponding to pattern P [1. . m]. • T [1 . . n] : Text • ts : Decimal value of the length-m substring T [s + 1 . . s + m], for s = 0, 1, . . . , n – m.
- 3. Introduction… • ts = p if and only if T [s + 1 . . s + m] = P [1. . m] → s is a valid shift iff ts = p Computing p: • p can be computed in Θ (m) time using Horner’s Rule: p = P [m] + 10 (P [m – 1] + 10 (P [m – 2] + . . . + 10 P [2] + P [1]…)) Computing (t0 , t1 , , , tn – m): • t0 can be computed from T [1 . . m] in Θ (m) time using Horner’s rule. • However, ts + 1 can be computed from ts in constant time: ts + 1 = 10 (ts – 10 m – 1 T [s + 1]) + T [s + m + 1] → t1 , t2 , , , tn - m can be computed in Θ (n – m) time.
- 4. Introduction… Eg.: T [1 . . 6] = 314152 and pattern length m = 5 Computing t0 using Horner’s Rule: t0 = T [5] + 10 (T [4] + 10 (T [3] + 10 (T [2] + 10 (T [1]))))) = 5 + 10 (1 + 10 (4 + 10 (1 + 10 (3)))) = 5 + 10 (1 + 10 (4 + 10 (1 + 30))) = 5 + 10 (1 + 10 (4 + 10 (31))) = 5 + 10 (1 + 10 (4 + 310)) = 5 + 10 (1 + 10 (314)) = 5 + 10 (1 + 3140) = 5 + 10 (3141) = 5 + 31410 = 31415
- 5. Introduction… Computing t1 using : ts + 1 = 10 (ts – 10 m – 1 T [s + 1]) + T [s + m + 1] s = 0, m = 5, T [1 . . 6] = 314152, t0 = 31415 t1 = 10 (t0 – 10 5 – 1 (T [1])) + T [0 + 5 + 1] = 10 (t0 – 10 4 (T [1])) + T [6] = 10 (31415 – 10000 (3)) + 2 = 10 (31415 – 30000) + 2 = 10 (1415) + 2 = 14150 + 2 = 14152
- 6. Introduction… • If 10 m – 1 is precomputed, then computing t1 , t2 , , , tn - m will take a constant number of arithmetic operations. • Running Time: p can be computed in Θ (m) time. t0 , t1 , , , tn - m can be computed in Θ (n – m + 1) time. Hence, all occurrences of pattern P [1 . . m] in the text T [1 . . n] can be found with Θ (m) preprocessing time and Θ (n – m + 1) matching time.
- 7. Introduction… Note: If p and ts are too large then computing p and ts may not happen in a constant time. Solution: • Compute p and the ts values modulo a suitable modulus q. • If q is chosen to be a prime such that 10q fits within one computer word, then all necessary computations can be performed with single-precision arithmetic. • In general with a d-ary alphabet {0, 1, . . . , d – 1} q has to be chosen so that dq fits within a computer word. ts + 1 can be computed as follows: ts + 1 = (d (ts – T [s + 1] h) + T [s + m + 1]) mod q h = d m – 1 (mod q) - value of most significant digit in the m-digit text window.
- 8. Example Text : 2359023141526739921 n : 19 Pattern : 31415 M : 5 q : 13 Computing p and t0 using Horner’s Rule: P [m] + 10 (P [m - 1] + 10 (P [m - 2] + . . . + 10 P [2] + P [1]…)) p = 5 + 10 (1 + 10 (4 + 10 (1 + 10 (3)))) = 31415 t0 = 0 + 10 (9 + 10 (5 + 10 (3 + 10 (2)))) (mod 13) = 23590 mod 13 = 8
- 9. Example… Computing ts + 1 : ts + 1 = (d (ts – T [s + 1] h) + T [s + m + 1]) mod q, h = d m – 1 (mod q) h = 10 4 (mod 13) = 3 s = 0, t0 = 8, T [1 . . 6] = 235902 t1 = (10 (8 – 2 (3)) + 2) (mod 13) = (10 (2) + 2) (mod 13) = 22 mod (13) = 9 s = 1, t1 = 9, T [2 . . 7] = 359023 t2 = (10 (9 – 3 (3)) + 3) (mod 13) = (10 (0) + 3) (mod 13) = 3 mod (13) = 3 s = 2, t2 = 3, T [3 . . 8] = 590231 t3 = (10 (3 – 5 (3)) + 1) (mod 13) = (10 (-12) + 1) (mod 13) = -119 mod (13) = 11
- 10. Example…
- 11. Example… • ts ≡ p (mod q) - Hit • ts ≡ p (mod q) but ts ≠ p - Spurious Hit • ts ≠ p (mod q) → ts ≠ p - Invalid Shift • ts ≡ p (mod q) and P [1. . m] = T [s + 1 . . s + m] - Valid Shift, s.
- 12. Algorithm RABIN – KARP MATCHER (T, P, d, q) n = T.length m = P.length h = d m - 1 (mod q) p = 0 t0 = 0 For (i =1 to m) p = (dp + P [i]) (mod q) t0 = (dt0 + T [i]) (mod q) For (s = 0 to n – m) If (p == ts) If (P [1 . . m] == T [s + 1 . . s + m]) Print “Pattern occurs with shift ” s If (s < n – m) ts + 1 = (d (ts – T [s + 1] h ) + T [s + m + 1]) (mod q) Running Time: Preprocessing Time: Θ (m) Matching Time: Θ ((n – m + 1) m)
- 13. References: • Thomas H Cormen. Charles E Leiserson, Ronald L Rivest, Clifford Stein, Introduction to Algorithms, Third Edition, The MIT Press Cambridge, Massachusetts London, England.