1.  Author: Ricardo A. Baeza-Yates, Gaston H. Gonnet  Publisher: 1992 Communications of the ACM  Presenter: Yuen-Shuo Li  Date: 2013/08/14 1

2.  String searching is a very important component of many problems, including text editing, bibliographic retrieval, and symbol manipulation.

3. T[a] = 11010 T[b] = 10101 T[c] = 01111 T[d] = 11111 cbaba T[a] = 11010

4. 50301 State cbbabababcaba… text T[a] = 11010 T[b] = 10101 T[c] = 01111 T[d] = 11111

5. 0301 State cbbabababcaba… text T[a] = 11010 T[b] = 10101 T[c] = 01111 T[d] = 11111

6. 14020 State cbbabababcaba… text T[a] = 11010 T[b] = 10101 T[c] = 01111 T[d] = 11111

7. 14020 State cbbabababcaba… text T[a] = 11010 T[b] = 10101 T[c] = 01111 T[d] = 11111

8. 14020 State cbbabababcaba… text T[a] = 11010 T[b] = 10101 T[c] = 01111 T[d] = 11111

9. 4020 State cbbabababcaba… text T[a] = 11010 T[b] = 10101 T[c] = 01111 T[d] = 11111

10. 50301 State cbbabababcaba… text T[a] = 11010 T[b] = 10101 T[c] = 01111 T[d] = 11111

11. 04121 State cbbabababcaba… text T[a] = 11010 T[b] = 10101 T[c] = 01111 T[d] = 11111

12.  To update the state after reading a new character on the text, we must  Shift the vector state b bits to the left to reflect that we have advanced one position in the text.  Update the individual states according to the new character.

13. The number of mismatches

14. 0 or 1 b = 1

15. Let {a, b, c, d} be the alphabet, and ababc the pattern. T[a] = 11010 T[b] = 10101 T[c] = 01111 T[d] = 11111

16.  The initial state is 11111 11111 State abdabababc text T[a] = 11010 T[b] = 10101 T[c] = 01111 T[d] = 11111

17.  The initial state is 11111 11111 State abdabababc text T[a] = 11010 T[b] = 10101 T[c] = 01111 T[d] = 11111

18.  The initial state is 11111 1111 State abdabababc text T[a] = 11010 T[b] = 10101 T[c] = 01111 T[d] = 11111

19.  The initial state is 11111 11110 State abdabababc text T[a] = 11010 T[b] = 10101 T[c] = 01111 T[d] = 11111

20.  The initial state is 11111 11101 State abdabababc text T[a] = 11010 T[b] = 10101 T[c] = 01111 T[d] = 11111

21.  The initial state is 11111 11111 State abdabababc text T[a] = 11010 T[b] = 10101 T[c] = 01111 T[d] = 11111

22.  The initial state is 11111 11110 State abdabababc text T[a] = 11010 T[b] = 10101 T[c] = 01111 T[d] = 11111

23.  The initial state is 11111 11101 State abdabababc text T[a] = 11010 T[b] = 10101 T[c] = 01111 T[d] = 11111

24.  The initial state is 11111 11010 State abdabababc text T[a] = 11010 T[b] = 10101 T[c] = 01111 T[d] = 11111

25.  The initial state is 11111 10101 State abdabababc text T[a] = 11010 T[b] = 10101 T[c] = 01111 T[d] = 11111

26.  The initial state is 11111 11010 State abdabababc text T[a] = 11010 T[b] = 10101 T[c] = 01111 T[d] = 11111

27.  The initial state is 11111 10101 State abdabababc text T[a] = 11010 T[b] = 10101 T[c] = 01111 T[d] = 11111

28.  The initial state is 11111 01111 State abdabababc text T[a] = 11010 T[b] = 10101 T[c] = 01111 T[d] = 11111 The match at the end of the text is indicated by the value 0 in the leftmost bit of the state

30. m: pattern size w: word size

32. T[a] = 11000 T[b] = 10011 T[c] = 11101 T[d] = 01101

34.  We allow up to k characters of the pattern to mismatch with the corresponding text. For example, if k = 2, the pattern mismatch: mismatch (match) dispatch (match) respatch (mismatch)

36. At each step we record the overflow bits in an overflow state, and we reset the overflow bits of all individual states.

37.  We want to search for all occurrences of ababc with at most 2 mismatch. Because the value of b is 3 for 2 mismatches, every position in the state is represented by a number in the range 0- 4.  Initial state: 00000  Initial overflow: 44444 We report a match when the sum of the leftmost digits of the state and the overflow is less than 3

41.  Experimental results for searching 100 times for all possible matches of a pattern in a 50,000 character English text(a legal document)

42. BMH: Boyer-Moore, as suggested by Horspool

43.  The execution time while search 1,000 words chosen at random from the same English text