1. MCS 101: Algorithms Instructor Neelima Gupta ngupta@cs.du.ac.in

2. Table of Contents String Matching – Naïve Method – Finite Automata Approach – Rabin Karp – KMP

3. Pattern Matching Given a text string T[0..n-1] and a pattern P[0..m-1], find all occurrences of the pattern within the text. Example: T = ababcabdabcaabc and P = abc, the occurrences are: – first occurrence starts at T[3] – second occurrence starts at T[9] – third occurrence starts at T[13]

4. Let Σ denotes the set of alphabet. Given: A string of alphabets T[1..n] of size “n” and a pattern P[1..m] of size “m” where, m<<<n. To Find: Whether the pattern P occurs in text T or not. If it does, then give the first occurrence of P in T. The alphabets of both T and P are drawn from finite set Σ.

5. NAÏVE APPROACH T : P : a b c a b d a a b c d e a b d

6. Example ( Step – 1 ) T : P : c a b d a a b c d e d Mismatch after 3 Comparisons a b

7. Example ( Step – 2 ) T : P : a b c a b d a a b c d e a b d Mismatch after 1 Comparison

8. Example ( Step – 3 ) T : P : a b c a b d a a b c d e a b d Mismatch after 1 Comparison

9. Example ( Step – 4 ) T : P : a b c a b d Match found after 3 Comparisons a b da a b c d e Thus, after 8 comparisons the substring P is found in T.

10. Worst Case Running Time T : a a a a a……..a a f of size say “n” P : a a a f of size 4

11. Example ( Step – 1 ) T : P : a a a a..... a a f a a a f Mismatch found after 4 comparisons

12. Example ( Step – 2 ) T : P : a a a a a,,,, a a f a a a f Mismatch found after 4 comparisons

13. T : P : a a a f a a a a a.... Match found after 4 comparisons a a a f Example

14. This will continue to happen until (n-4)th alphabet in T is compared with the characters in P and thus the no. of comparisons required is (n-4)4 + 4. Worst Case Running Time

15. At every step, after ‘m’ comparisons a mismatch will be found. These ‘m’ comparisons will be done for (n- m) characters in T. Thus, the running time obtained is (n-m)m + m.

16. Finite Automata s0s0 s1s1 a f s2s2 s3s3 f aa # a ∑

17. Worst Case Running Time In finite automata, each character is scanned atmost once. Thus in the worst case, the searching time is O(n). Preprocessing time:- As for every character in ∑ an edge has to be formed, thus the preprocessing time is O(m*|∑|). Thus total running time is O(n) + O(m*|∑|).

18. Drawback:- If the alphabet set ∑ is very large, then the time required to construct the FA will be very large.

19. BRUTE FORCE STRATEGY In this strategy whenever a mismatch was found, the pattern was shifted right by 1 character. But this wasn’t an efficient strategy as it required a large number of comparisons. Hence a better algorithm was required. 19

20. An example T : a b c a b c a b c a b c a f mismatch found P : a b c a b c a b c a f 20

21. Shift by 3 T : a b c a b c a b c a b c a f match found P : a b c a b c a b c a f 21

22. Another example T : a b c d e g h a b c d e g h a b f mismatch found P : a b c d e g h a b f 22

23. Shift by 7 T : a b c d e g h a b c d e g h a b f P : a b c d e g h a b f 23

24. How to decide how much to shift? Ans: We use the information contained in the partial pattern that has matched so far.

25. T : …… t j.. …...t j+r-1 ….t j+k-r …...t j+k-2 t j+k-1 … ……………………………… P : p 1 …… p r …… ……… p k-1 p k …… p 1 …… p r p k … If t j+k-1 ≠ p k How much should we shift is equivalent to asking which character of the pattern, t j+k-1 be compared with? Thus we look for a prefix p 1 … p r of p 1 … p k-1 that matches the suffix of t j … t j+k-2, then, t j+k-1 should be compared with p r+1. 25 KMP : Knuth Morris Pratt Algorithm

26. Several such prefixes may exist T : a b c a b c a b c a b c a f mismatch found P : a b c a b c a b c a f One prefix is : a b c a b c a ( r = 7) And Another is : a b c a ( r = 4) t 11 be compared with p 8 or p 5 ? Choose the longest prefix..i.e largest such r. 26

27. KMP Contd.. Let r be the length of the longest prefix of P that matches with the matched part of P. Then the pattern can be shifted by r positions instead of 1 and t j+k-1 should be compared with p r+1. Claim 1: We have not missed any match i.e. the pattern does not exist at any position from j to j+k-r- 1. Proof: Had it been, we would have a longer prefix matching with its suffix.

28. Why LONGEST? T : a b c a b c a b c a b c a f mismatch found P : a b c a b c a b c a f 28

29. 29 T : a b c a b c a b c a b c a f mismatch found P : a b c a b c a b c a f the longest prefix. Correct alignment for the pattern will be by shifting it 3 characters right.

30. 30 T : a b c a b c a b c a b c a f P : a b c a b c a b c a f Pattern found.

31. 31 T : a b c a b c a b c a b c a f mismatch P : a b c a b c a b c a f Pattern not found. By finding a smaller prefix and aligning the pattern accordingly as shown, the pattern’s occurrence in the text got missed (that is we shifted by more positions than we should have)

32. How to find the longest such prefix? KMP algorithm 32

33. T : …… t j.. …...t j+r-1 ….t j+k-r …...t j+k-2 t j+k-1 … ……………………………… P : p 1 …… p r …… ……… p k-1 p k …… p 1 …… p r p k … If t j+k-1 ≠ p k Since t j … t j+k-1 has already been matched with p 1 … p k-1, we need to look for longest prefix p 1 … p r (r < k-1) of p 1 … p k-1 that matches with its own suffix. Thus the longest prefix can be found from the pattern itself and we do not need the text for the purpose……Note 1. 33 KMP : Knuth Morris Pratt Algorithm

34. P : p 1 ….………….…………… p k ………… Let the length of the longest prefix of p 1 … p k-1 that matches its suffix be ‘r.’ 34

35. T : …… t j.. …...t j+r-1 ….t j+k-r …...t j+k-2 t j+k-1 … ……………………………… P : p 1 …… p r …… ……… p k-1 p k …… p 1 …… p r p k … If t j+k-1 ≠ p k Let Fail[k] be a pointer which says that if a mismatch occurs for p k then what is the character in P that should come in place of p k by shifting P accordingly. Fail[k] is nothing but the length of the longest prefix plus 1. Thus the Q is How to compute Fail[k]? 35

36. P : p 1 … p r-1 p r p r+1 …….…. p k-1 p k … p 1 … p r’-1 p r’ p r’+1 p 1….... p s-1 p s p s+1 Look at fail[k-1]. Let it be r’. If p r’ = p k-1 (which has already been matched with t j+k-1 ) fail[k] = r’+1 1 else { look at fail[r’] = s, say if s>0 { if p s = p k-1 then fail[k] = s+1 else goto 1 with r’ = s } } else (i.e s = 0) fail[k] =1 36

37. EXAMPLE P: abcabcabcaf for k=1, fai[k]=0 (assumed) for k=2, s=fail[1]=0 therefore, fail[k]=0+1=1 for k=3, s=fail[2]=1 check whether p2=p1 since p2!=p1 so, s=fail[1]=0 therefore, fail[k]=0+1=1

38. P: abcabcabcaf for k=4, s=fail[3]=1 check whether p1=p3 since p1!=p3 so, s=fail[1]=0 therefore, fail[k]=0+1=1 For k=5 s=fail[4]=1 check whether p1=p4 yes therefore, fail[k]=1+1=2 Similarly, for others.

39. kfail[k] 10 21 31 41 52 63 74 85 96 107 118

40. Example : T : a b c a b c a b c a b c a f P : a b c a b c a b c a f k: 1 2 3 4 5 6 7 8 9 10 11 P : a b c a b c a b c a f k: 1 2 3 4 5 6 7 8 9 10 11 Mismatch found at k=11 position. Look at fail[11] = 8 which implies the pattern must be shifted such that p 8 comes in place of p 11 40 kFail[k] 10 21 31 41 52 63 74 85 96 107 118

41. Example : T : a b c a b c a b c a b c a f P : a b c a b c a b c a f k: 1 2 3 4 5 6 7 8 9 10 11 Pattern found 41 kFail[k] 10 21 31 41 52 63 74 85 96 107 118

42. Another Example : T : a b c b a b c b a b c a b c a b c a f P : a b c a b c a b c a f k: 1 2 3 4 5 6 7 8 9 10 11 P : a b c a b c a b c a f k: 1 2 3 4 5 6 7 8 9 10 11 Mismatch found at k=4 position. Look at fail[4] = 1 which implies the pattern must be shifted such that p 1 comes in place of p 4 42 kFail[k] 10 21 31 41 52 63 74 85 96 107 118

43. 43 Another Example : T : a b c b a b c b a b c a b c a b c a f P : a b c a b c a b c a f k: 1 2 3 4 5 6 7 8 9 10 11 P : a b c a b c a b c a f k: 1 2 3 4 5 6 7 8 9 10 11 Mismatch found at k=1 position. Look at fail[1] = 0 which implies read the next character in text. kFail[k] 10 21 31 41 52 63 74 85 96 107 118

44. 44 Another Example : T : a b c b a b c b a b c a b c a b c a f P : a b c a b c a b c a f k: 1 2 3 4 5 6 7 8 9 10 11 P : a b c a b c a b c a f k: 1 2 3 4 5 6 7 8 9 10 11 Mismatch found at k=4 position. Look at fail[4] = 1 which implies the pattern must be shifted such that p 1 comes in place of p 4 kFail[k] 10 21 31 41 52 63 74 85 96 107 118

45. 45 Another Example : T : a b c b a b c b a b c a b c a b c a f P : a b c a b c a b c a f k: 1 2 3 4 5 6 7 8 9 10 11 P : a b c a b c a b c a f k: 1 2 3 4 5 6 7 8 9 10 11 Mismatch found at k=1 position. Look at fail[1] = 0 which implies read the next character in text. kFail[k] 10 21 31 41 52 63 74 85 96 107 118

46. 46 kFail[k] 10 21 31 41 52 63 74 85 96 107 118 Another Example : T : a b c b a b c b a b c a b c a b c a f P : a b c a b c a b c a f k: 1 2 3 4 5 6 7 8 9 10 11 Pattern found

47. Analysis of KMP # of mismatch: For mismatch the pattern is shifted by at least 1 position. The maximum number of shifts is determined by the largest suffix. T:......a b c a b c a b c a b c d a f d........ P: d e b mismatch For every mismatch pattern is shifted by atleast1postion.  Total no. of shifts <= n-m  Total no. of mismatches <=n-m+1....

48. Analysis of KMP contd. # of matches: For every match, pointer in the text moves up by 1 position. T:......a b c a b c a b c a b c d a f d........ P: a b c b d e For every match pointer moves up by 1 position. P: a b c b d e => # of matches <= length of text <= n..... The complexity of KMP is linear in nature. O(m+n)

49. ACKNOWLEDGEMENTS 49 MSc (CS) 2009 Abhishek Behl(02) Aarti Sethiya(01) Akansha Aggarwal(03) Alok Prakash (04) Vibha Negi(31)