1. Backward Nondeterministic DAWG Matching Algorithm
A Bit-parallel Approach to Suffix Automata:
Fast Extended String Matching,
Navarro, G. and Raffinot, M., Lecture Notes in Computer Science, Vol.1448, 1998, pp
Advisor: Prof. R. C. T. Lee
Speaker: L. C. Chen

2. Problem Definition: Input : A text T and a pattern P.
Output : All the locations where P matches T.

3. This algorithm uses rule 1: Suffix to Prefix Rule:
For a window to have any chance to match a pattern, in some way, there must be a suffix of the window which is equal to a prefix of the pattern. T P

4. Find the longest suffix U of the window which is equal to some prefix of P. Skip the pattern as follows: U

5. Example
T = GCA TCGACAGAC TATACAGTACG
P = GACGGATCA
∵The longest suffix of the window which is equal to a prefix of P is “GAC”, slide the window by 6.
T = GCATCGACAGACTATACAGTACG
P = GACGGATCA

6. We give an example to introduce how this algorithm find the longest
suffix of the window which is equal to a prefix of P.

7. Text : ABDDCCDBADEGGGGJJ
Example:
Text : ABDDCCDBADEGGGGJJ
Pattern : BADADCEAD
We want to find the longest suffix of “BDDCCDBAD” which is also
a prefix of the pattern.

8. Text : ABDDCCDBADEGGGGJJ
Example:
Text : ABDDCCDBADEGGGGJJ
Pattern : BADADCEAD
First, we read “D”.

9. Text : ABDDCCDBADEGGGGJJ
Example:
Text : ABDDCCDBADEGGGGJJ
Pattern : BADADCEAD
We find all the substrings ”D” in the pattern.

10. Text : ABDDCCDBADEGGGGJJ
Example:
Text : ABDDCCDBADEGGGGJJ
Pattern : BADADCEAD
We read the next character “A”.
We check if the right of the substrings ”D” are “A” or not.

11. Text : ABDDCCDBADEGGGGJJ
Example:
Text : ABDDCCDBADEGGGGJJ
Pattern : BADADCEAD
Thus, we find out all the substrings ”AD” in the pattern.

12. Text : ABDDCCDBADEGGGGJJ
Example:
Text : ABDDCCDBADEGGGGJJ
Pattern : BADADCEAD
We read the next character “B”.
We check if the right of the substrings “AD” are “B” or not.

13. Text : ABDDCCDBADEGGGGJJ
Example:
Text : ABDDCCDBADEGGGGJJ
Pattern : BADADCEAD
We find that the substring ”BAD” is in the pattern. Note that “BAD”
is also a prefix of P.

14. Text : ABDDCCDBADEGGGGJJ
Example:
Text : ABDDCCDBADEGGGGJJ
Pattern : BADADCEAD
We read the next character “D”.
We can not find a character “D” in the right of the substring “BAD”.
We report that “BAD” is the longest suffix of “BDDCCDBAD”
which is equal a prefix of P.

15. Text : ABDDCCDDADEGGGGJJ
Another example:
Text : ABDDCCDDADEGGGGJJ
Pattern : ACDADCEAD
We want to find the longest suffix of “BDDCCDDAD” which is also
a substring of the pattern.

16. Text : ABDDCCDDADEGGGGJJ
Another example:
Text : ABDDCCDDADEGGGGJJ
Pattern : ACDADCEAD
First, we find all the substrings ”D” in the pattern.

17. Text : ABDDCCDDADEGGGGJJ
Another example:
Text : ABDDCCDDADEGGGGJJ
Pattern : ACDADCEAD
mismatch
Then we find out all the substrings ”AD” in the pattern.

18. Text : ABDDCCDDADEGGGGJJ
Another example:
Text : ABDDCCDDADEGGGGJJ
Pattern : ACDADCEAD
Then we find out all the substrings ”AD” in the pattern.

19. Text : ABDDCCDDADEGGGGJJ
Another example:
Text : ABDDCCDDADEGGGGJJ
Pattern : ACDADCEAD
mismatch
We find out all the substrings ”DAD” in the pattern.

20. Text : ABDDCCDDADEGGGGJJ
Another example:
Text : ABDDCCDDADEGGGGJJ
Pattern : ACDADCEAD
We find out all the substrings ”DAD” in the pattern.

21. Text : ABDDCCDDADEGGGGJJ
Another example:
Text : ABDDCCDDADEGGGGJJ
Pattern : ACDADCEAD
mismatch
We find all the substrings ”DDAD” in the pattern.

22. Text : ABDDCCDDADEGGGGJJ
Another example:
Text : ABDDCCDDADEGGGGJJ
Pattern : ACDADCEAD
mismatch
We find all the substrings ”DDAD” in the pattern. There is no substring
“DDAD” in the pattern.
There is no any suffix of “BDDCCDDAD” which is equal to a prefix
of P.

23. The idea that we explained above is the main idea of this
algorithm. And next we will use bit-parallel method to
implement this algorithm.

24. We use bits to store the positions of a character in P.
Example:
P: CABBCAD
P: CABBCAD A:
For character “A”, we store B:
For character “B”, we store
For character “C”, we store C:
For character “D”, we store D:
For the characters do not exit in P we store *:

25. Here, we explain how to use bit-parallel to find the substring
of a pattern which is equaled to a suffix of the window.
Text: ABCABCABA
,∑={A,B,C,D}
Pattern: CABBCAD
Pattern: CABCCAD A: B: C: D:
other: D:
We use a mask D to record some information.

26. <<1: left shift one bit.
Text: ABCABCABA
,∑={A,B,C,D}
Pattern: CABBCAD
Pattern: CABCCAD A: B: C: D:
other: D:
And A: D:
<<1: left shift one bit.
D= <<1 =

27. Text: ABCABCABA ,∑={A,B,C,D} Pattern: CABBCAD Pattern: CABCCAD
other: D:
And C: D:
We know “CA” is a suffix of the window which is equal to a prefix of the pattern.
D= <<1 =

28. Text: ABCABCABA ,∑={A,B,C,D} Pattern: CABBCAD Pattern: CABCCAD
other: D:
And B: D:
We know “BCA” is a substring of the pattern.
D= <<1 =

29. Text: ABCABCABA ,∑={A,B,C,D} Pattern: CABBCAD Pattern: CABCCAD
other: D:
And A:
There is no substring “ABCA” in the pattern.

30. Text: ABCABCABA ,∑={A,B,C,D} Pattern: CABBCAD
“CA” is a suffix of “BCA” which is a prefix of the pattern.

31. We take another example:
Text: ABCABCCBA
,∑={A,B,C,D}
Pattern: ACBCCBD

32. Example: Text: ABCABCCBA ,∑={A,B,C,D} Pattern: ACBCCBD
First, we build:
Pattern: ACBCCBD A: B: C: D:
others:

33. Example: Text: ABCABCCBA ,∑={A,B,C,D} Pattern: ACBCCBD
others:
Pattern: ACBCCBD D:

34. Example: Text: ABCABCCBA ,∑={A,B,C,D} Pattern: ACBCCBD
others:
Pattern: ACBCCBD D:

35. Example: Text: ABCABCCBA ,∑={A,B,C,D} Pattern: ACBCCBD
others:
Pattern: ACBCCBD D:
And C: D:
Where there is a “1”, there is a substring “C” in Pattern.
We set D =
<<1=

36. Example: Text: ABCABCCBA ,∑={A,B,C,D} Pattern: ACBCCBD
others:
Pattern: ACBCCBD D:
And C: D:
Where there is a “1”, there is a substring “CC” in Pattern.
We set D =
<<1=

37. Example: Text: ABCABCCBA ,∑={A,B,C,D} Pattern: ACBCCBD
others:
Pattern: ACBCCBD D:
And B: D:
Where there is a “1”, there is a substring “BCC” in Pattern.
We set D =
<<1=

38. Example: Text: ABCABCCBA ,∑={A,B,C,D} Pattern: ACBCCBD
others:
Pattern: ACBCCBD D:
And A: D:
There is no substring “ABCC” in Pattern.
There is no any suffix of the window which is equal to a prefix of the pattern.

39. Time Complexity:
If the length of the text is n and the length of pattern is m,
the time complexity of this algorithm is O(mn) in the worst
case.

40. Reference
[BG92]A new approach to text searching, R. Baeza-Yates and Navarro, G., CACM. Vol. 35, 1992, pp
[BEH89]Average sizes of suffix trees and dawgs., Blumer, A., Ehrenfeucht, A. and Haussler, D., Discrete Applied Mathematics, Vol. 24, 1989, pp
[BM77] A fast string searching algorithm. Boyer, R. S. and Moore, J. S., Communications of the ACM, Vol. 20, 1977, pp
[GM98] A Bit-Parallel Approach to Suffix Automata: Fast Extended String Matching, G. NAVARRO and M. RAFFINOT, In Proceedings of the 9th Annual Symposium on Combinatorial Pattern Matching, Lecture Notes in Computer Science 1448, Springer-Verlag, Berlin, 1998, pp