String matching with finite state automata
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The document discusses the string matching algorithm that builds a finite automaton to scan a text string for occurrences of a pattern string. It explains the suffix function and transition function that are used to specify the string matching automaton corresponding to a given pattern. It then provides an example of computing the transition function for a given text and pattern string through multiple iterations.
![STRING
MATCHING
AUTOMATA
(SUFFIX
FUNCTION):• In order to specify the string-matching automaton
corresponding to a given pattern P[1…m]
• An auxiliary function σ, called the suffix function
corresponding to P
• The function σ maps * to {0,1,…..,m} such thatƩ
σ(x) is the length of the longest prefix of P that is
also a suffix of x:
• σ(x) = max{k: Pk ⊐x}](https://image.slidesharecdn.com/stringmatchingwithfsa-140622093143-phpapp01/85/String-matching-with-finite-state-automata-7-320.jpg)
![COMPUTE TRANSITION
FUNCTION:
• TEXT=abababacaba
• PATTERN=ababaca
• ={a,b,c}Ʃ
Compute transition
function(P, )Ʃ
1 m← length[P]
m← 7
2 for q←0 to m
for q←0 to 7
3 do for each character aєƩ
// ={a,b,c}Ʃ](https://image.slidesharecdn.com/stringmatchingwithfsa-140622093143-phpapp01/85/String-matching-with-finite-state-automata-11-320.jpg)
![FINITE AUTOMATON
MATCHER
FINITE-AUTOMATON-MATCHER(T,δ,m)
1.n=T . length
2.q=0
3.for I =1 to n
4. q= δ( q, T[ i ] )
5. If q==m
6. Print “Pattern occurs with shift” i -m](https://image.slidesharecdn.com/stringmatchingwithfsa-140622093143-phpapp01/85/String-matching-with-finite-state-automata-38-320.jpg)
![FINITE-AUTOMATON
MATCHER
i= 1 2 3 4 5 6 7 8 9 10 11
T= a b a b a b a c a b a
1.n← length[T]
n←11
2. q←0
3. for i←1 to n
for i←1 to 11](https://image.slidesharecdn.com/stringmatchingwithfsa-140622093143-phpapp01/85/String-matching-with-finite-state-automata-40-320.jpg)
![1st
ITERATION:
3. for i←1 to 11
4. do q←δ(q,T[i])
q←δ(0,T[1])=δ(0,a)
q←(0,a)=1
5. if q=m
if 1=7 FALSE
0 1 2 3 4 5 6 7
a b a b a c a
a
b
a
b
a
a](https://image.slidesharecdn.com/stringmatchingwithfsa-140622093143-phpapp01/85/String-matching-with-finite-state-automata-41-320.jpg)
![2ND
ITERATION:
i=2,q=2
4. do q←δ(1,T[2])
=δ(1,b)
q←2
5 if 2=7 FALSE
0 1 2 3 4 5 6 7
a b a b a c a
a
b
a
b
a
a](https://image.slidesharecdn.com/stringmatchingwithfsa-140622093143-phpapp01/85/String-matching-with-finite-state-automata-42-320.jpg)
![3RD
ITERATION:
i=3,q=2
4. do q←δ(2,T[3])
=δ(2,a)
q←3
5 if 3=7 FALSE
0 1 2 3 4 5 6 7
a b a b a c a
a
b
a
b
a
a](https://image.slidesharecdn.com/stringmatchingwithfsa-140622093143-phpapp01/85/String-matching-with-finite-state-automata-43-320.jpg)
![4th
ITERATION:
i=4,q=3
4. do q←δ(3,T[4]) =δ(3,b)
q←4
5 if 4=7 FALSE
0 1 2 3 4 5 6 7
a b a b a c a
a
b
a
b
a
a](https://image.slidesharecdn.com/stringmatchingwithfsa-140622093143-phpapp01/85/String-matching-with-finite-state-automata-44-320.jpg)
![5th
ITERATION:
i=5,q=4
4. do q←δ(4,T[5])
=δ(4,a)=5
q←5
5 if 5=7 FALSE
0 1 2 3 4 5 6 7
a b a b a c a
a
b
a
b
a
a](https://image.slidesharecdn.com/stringmatchingwithfsa-140622093143-phpapp01/85/String-matching-with-finite-state-automata-45-320.jpg)
![6TH
ITERATION:
i=6,q=5
4. do q←δ(5,T[6])
=δ(5,b)=4
q←5
5 if 4=7 FALSE
0 1 2 3 4 5 6 7
a b a b a c a
a
b
a
b
a
a](https://image.slidesharecdn.com/stringmatchingwithfsa-140622093143-phpapp01/85/String-matching-with-finite-state-automata-46-320.jpg)
![7TH
ITERATION:
i=7,q=4
4. do q←δ(4,T[7])
=δ(4,a)=5
q←5
5 if 5=7 FALSE
0 1 2 3 4 5 6 7
a b a b a c a
a
b
a
b
a
a](https://image.slidesharecdn.com/stringmatchingwithfsa-140622093143-phpapp01/85/String-matching-with-finite-state-automata-47-320.jpg)
![8TH
ITERATION:
i=8,q=5
4. do q←δ(5,T[8])
=δ(5,c)=6
q←6
5 if 6=7 FALSE
0 1 2 3 4 5 6 7
a b a b a c a
a
b
a
b
a
a](https://image.slidesharecdn.com/stringmatchingwithfsa-140622093143-phpapp01/85/String-matching-with-finite-state-automata-48-320.jpg)
![9TH
ITERATION:
i=9,q=6
4. do q←δ(6,T[9])
=δ(6,a)=7
q←7
5 if 7=7 TRUE
Then print “pattern occurs with
shift”i-m
Print” pattern occurs with
shift”9-7=2
0 1 2 3 4 5 6 7
a b a b a c a
a
b
a
b
a
a](https://image.slidesharecdn.com/stringmatchingwithfsa-140622093143-phpapp01/85/String-matching-with-finite-state-automata-49-320.jpg)
![10th
ITEARTION:
i=10,q=7
4. do q←δ(7,T[10])
=δ(7,b)=2
q←2
5 if 2=7 FALSE
0 1 2 3 4 5 6 7
a b a b a c a
a
b
a
b
a
a](https://image.slidesharecdn.com/stringmatchingwithfsa-140622093143-phpapp01/85/String-matching-with-finite-state-automata-50-320.jpg)
![11TH
ITEARTION:
i=11,q=2
4. do
q←δ(2,T[11])=δ(2,a)=3
q←3
5 if 3=7 FALSE
0 1 2 3 4 5 6 7
a b a b a c a
a
b
a
b
a
a](https://image.slidesharecdn.com/stringmatchingwithfsa-140622093143-phpapp01/85/String-matching-with-finite-state-automata-51-320.jpg)
![CONT..
State a b c P
0 1 0 0 a
1 1 2 0 b
2 3 0 0 a
3 1 4 0 b
4 5 0 0 a
5 1 4 6 c
6 7 0 0 a
7 1 2 0
i= - 1 2 3 4 5 6 7 8 9 10
11
T[i] - a b a b a b a c a b a
state ΦT[i] 0 1 2 3 4 5 4 5 6 77 2 3](https://image.slidesharecdn.com/stringmatchingwithfsa-140622093143-phpapp01/85/String-matching-with-finite-state-automata-53-320.jpg)
String matching with finite state automata
- 1. STRING MATCHING WITH FINITE AUTOMATA SUBMITTED TO : MAM MAIMOONA SUBMITTED BY : IQRA MUNIR ANMOL HAMID ANALYSIS OF ALGORITHM
- 3. STRING MATCHING WITH FINITE AUTOMATA: • String matching algorithm builds a finite automaton • A simple machine for processing information that scans the text string T for all occurrences of the pattern P
- 7. STRING MATCHING AUTOMATA (SUFFIX FUNCTION):• In order to specify the string-matching automaton corresponding to a given pattern P[1…m] • An auxiliary function σ, called the suffix function corresponding to P • The function σ maps * to {0,1,…..,m} such thatƩ σ(x) is the length of the longest prefix of P that is also a suffix of x: • σ(x) = max{k: Pk ⊐x}
- 8. ALGORITHMS…
- 9. COMPUTE TRANSITION FUNCTION: COMPUTE-TRANSITION-FUNCTION(P, )Ʃ 1.m-P.length 2.for q=0 to m 3. for each character a є Ʃ 4. k= min(m+1,q+2) 5. repeat 6. K=k-1 7. until P k P q a⊐ 8. δ(q ,a )=k 9.return δ
- 10. DRY RUNNING…
- 11. COMPUTE TRANSITION FUNCTION: • TEXT=abababacaba • PATTERN=ababaca • ={a,b,c}Ʃ Compute transition function(P, )Ʃ 1 m← length[P] m← 7 2 for q←0 to m for q←0 to 7 3 do for each character aєƩ // ={a,b,c}Ʃ
- 12. 1st ITERATION: q=0, a=a 3. for each character aєa 4. k←min(m+1,q+2) = min(7+1,0+2)=2 K←2 7. Pk Pqa= P2 P0a⊐ ⊐ (ab a) FALSE⊐ K←1 7.Pk Pqa= P1 P0a⊐ ⊐ (a a) TRUE⊐ 8. δ(q,a)←k δ(0,a)←1 state a b c P 0 1 a 1 b 2 a 3 b 4 a 5 c 6 a 7
- 13. CONT… q=0 3. for each character aєb 4. K=2 7. Pk Pqa=P2 P0a⊐ ⊐ (ab b) FALSE⊐ K=1 7.Pk Pqa=P1 P0a⊐ ⊐ (a b) FALSE⊐ K=0 Pk Pqa=P0 P0b⊐ ⊐ (є b) TRUE⊐ 8.δ(q,a)←k δ(0,b)←0 state a b c P 0 1 0 a 1 b 2 a 3 b 4 a 5 c 6 a 7
- 14. CONT… q=0 3. for each character aєc 4. K=2 7. Pk Pqa=P2 P0a⊐ ⊐ (ab c) FALSE⊐ K=1 7.Pk Pqa=P1 P0a⊐ ⊐ (a c) FALSE⊐ K=0 Pk Pqa=P0 P0s⊐ ⊐ (є c) TRUE⊐ 8.δ(q,a)←k δ(0,c)←0 state a b c P 0 1 0 0 a 1 b 2 a 3 b 4 a 5 c 6 a 7
- 15. 2nd ITERATION: 2.for q←1 to 7 3.for each character aє(a,b,c) q=1,a=a 4.k←min(m+1,q+2)=min(7+1,1 +2)=3 K←3 Pk Pqa=P3 P1a=(aba aa)⊐ ⊐ ⊐ FALSE k←2 7. Pk Pqa=P2 P1a⊐ ⊐ (ab aa) FALSE⊐ K=1 7.Pk Pqa=P1 P1a⊐ ⊐ (a aa) TRUE⊐ 8.δ(q,a)←k δ(1,a)←1 State a b c P 0 1 0 0 a 1 1 b 2 a 3 b 4 a 5 c 6 a 7
- 16. CONT… q=1 for each character aєb 4.K←3 Pk Pqa=P3 P1a=(aba ab)⊐ ⊐ ⊐ FALSE k←2 7. Pk Pqa=P2 P1a⊐ ⊐ (ab ab) TRUE⊐ 8.δ(1,b) 2← State a b c P 0 1 0 0 a 1 1 2 b 2 a 3 b 4 a 5 c 6 a 7
- 17. CONT… q=1 for each character aєc 4.K←3 Pk Pqa=P3 P1a=(aba ac)⊐ ⊐ ⊐ FALSE k←2 7. Pk Pqa=P2 P1a⊐ ⊐ (ab ac) FALSE⊐ k 1← Pk Pqa=P1 P1a⊐ ⊐ (a ac) FALSE⊐ k 0← Pk Pqa=P0 P1a⊐ ⊐ (є ac) TRUE⊐ 8.δ(1,c)←0 State a b c P 0 1 0 0 a 1 1 2 0 b 2 a 3 b 4 a 5 c 6 a 7
- 18. 3rd ITERATION: 2.for q←2 to 7 3.for each character aє(a,b,c) q=2,a=a 4.k←min(m+1,q+2)=min(8,4)=4 K←4 Pk Pqa=P4 P2a=(abab aba)⊐ ⊐ ⊐ FALSE Pk Pqa=P3 P2a=(aba aba)⊐ ⊐ ⊐ TRUE 8.δ(q,a)←k δ(2,a)←3 State a b c P 0 1 0 0 a 1 1 2 0 b 2 3 a 3 b 4 a 5 c 6 a 7
- 19. CONT… q=2 for each character aєb K 4← Pk Pqa=P4 P2a=(abab abb)⊐ ⊐ ⊐ FALSE K=3 Pk Pqa=P3 P2a=(aba abb)⊐ ⊐ ⊐ FALSE k←2 7. Pk Pqa=P2 P2a =(ab abb)⊐ ⊐ ⊐ FALSE K=1 7.Pk Pqa=P1 P2a⊐ ⊐ (a abb) FALSE⊐ State a b c P 0 1 0 0 a 1 1 2 0 b 2 3 0 a 3 b 4 a 5 c 6 a 7 K=0 7.Pk Pqa=P0 P2a⊐ ⊐ (є abb)TRUE⊐ 8.δ(q,a)←k δ(2,b)←0
- 20. CONT… q=2 for each character aєc K 4← Pk Pqa=P4 P2a=(abab abc)⊐ ⊐ ⊐ FALSE K=3 Pk Pqa=P3 P2a=(aba abc)⊐ ⊐ ⊐ FALSE k←2 7. Pk Pqa=P2 P2a =(ab abc)⊐ ⊐ ⊐ FALSE K=1 7.Pk Pqa=P1 P2a⊐ ⊐ (a abc) FALSE⊐ K=0 7.Pk Pqa=P0 P2a⊐ ⊐ (є abc)TRUE⊐ 8.δ(q,a)←k δ(2,c)←0 State a b c P 0 1 0 0 a 1 1 2 0 b 2 3 0 0 a 3 b 4 a 5 c 6 a 7
- 21. 4TH ITERATION: K 1← Pk Pqa=P1 P3a=(a abaa)⊐ ⊐ ⊐ TRUE 8.δ(q,a)←k δ(3,a)←1 2.for q←3 to 7 3.for each character aє(a,b,c) q=3,a=a 4.k←min(m+1,q+2)=min(8,4)=4 K←5 Pk Pqa=P5 P3a=(ababa abaa)⊐ ⊐ ⊐ FALSE k←4 Pk Pqa=P4 P3a=(abab abaa)⊐ ⊐ ⊐ FALSE K←3 Pk Pqa=P3 P3a=(aba abaa)⊐ ⊐ ⊐ FALSE k←2 Pk Pqa=P2 P3a=(ab abaa)⊐ ⊐ ⊐ FALSE State a b c P 0 1 0 0 a 1 1 2 0 b 2 3 0 0 a 3 1 b 4 a 5 c 6 a 7
- 22. CONT… 2.for q←3 3.for each character aєb K←5 Pk Pqa=P5 P3a=(ababa abab)⊐ ⊐ ⊐ FALSE k←4 Pk Pqa=P4 P3a=(abab abab)⊐ ⊐ ⊐ TRUE 8.δ(q,a)←k δ(3,b)←4 State a b c P 0 1 0 0 a 1 1 2 0 b 2 3 0 0 a 3 1 4 b 4 a 5 c 6 a 7
- 23. CONT… 2.q=3 3.for each character aєc K←5 Pk Pqa=P5 P3a=(ababa abac)⊐ ⊐ ⊐ FALSE k←4 Pk Pqa=P4 P3a=(abab abac)⊐ ⊐ ⊐ FALSE K←3 Pk Pqa=P3 P3a=(aba abac)⊐ ⊐ ⊐ FALSE k←2 Pk Pqa=P2 P3a=(ab abac)⊐ ⊐ ⊐ FALSE K←1 Pk Pqa=P1 P3a=(a abac)⊐ ⊐ ⊐ FALSE K←0 Pk Pqa=P0 P3a=(⊐ ⊐ є abac)⊐ TRUE 8.δ(q,a)←k δ(3,c)←0 State a b c P 0 1 0 0 a 1 1 2 0 b 2 3 0 0 a 3 1 4 0 b 4 a 5 c 6 a 7
- 24. 5th ITERATION: 2.for q←4to 7 3.for each character aє(a,b,c) q=4,a=a 4.k←min(m+1,q+2)=min(8,6)=6 K←6 Pk Pqa=P6 P4a=(ababac ababa)⊐ ⊐ ⊐ FALSE K←5 Pk Pqa=P5 P4a=(ababa abaBA)⊐ ⊐ ⊐ TRUE 8.δ(q,a)←k δ(4,a)←5 State a b c P 0 1 0 0 a 1 1 2 0 b 2 3 0 0 a 3 1 4 0 b 4 5 a 5 c 6 a 7
- 25. CONT… k←1 Pk Pqa=P2 P4a=(a ababb)⊐ ⊐ ⊐ FALSE k←0 Pk Pqa=P2 P4a=(є ababb)⊐ ⊐ ⊐ TRUE 8.δ(q,a)←k δ(4,b)←0 2.q=4 3.for each character aєb K←6 Pk Pqa=P6 P4a=(ababac ababb)⊐ ⊐ ⊐ FALSE K←5 Pk Pqa=P5 P4a=(ababa ababb)⊐ ⊐ ⊐ FALSE k←4 Pk Pqa=P4 P4a=(abab ababb)⊐ ⊐ ⊐ FALSE K←3 Pk Pqa=P3 P4a=(aba ababb)⊐ ⊐ ⊐ FALSE k←2 Pk Pqa=P2 P4a=(ab ababb)⊐ ⊐ ⊐ FALSE State a b c P 0 1 0 0 a 1 1 2 0 b 2 3 0 0 a 3 1 4 0 b 4 5 0 a 5 c 6 a 7
- 26. CONT… 2.q=4 3.for each character aєc K←6 Pk Pqa=P6 P4a=(ababac ababc)⊐ ⊐ ⊐ FALSE K←5 Pk Pqa=P5 P4a=(ababa ababc)⊐ ⊐ ⊐ FALSE k←4 Pk Pqa=P4 P4a=(abab ababc)⊐ ⊐ ⊐ FALSE K←3 Pk Pqa=P3 P4a=(aba ababc)⊐ ⊐ ⊐ FALSE k←2 Pk Pqa=P2 P4a=(ab ababc)⊐ ⊐ ⊐ FALSE k←1 Pk Pqa=P2 P4a=(a ababbc⊐ ⊐ ⊐ FALSE k←0 Pk Pqa=P2 P4a=(є ababc)⊐ ⊐ ⊐ TRUE 8.δ(q,a)←k δ(4,c)←0 State a b c P 0 1 0 0 a 1 1 2 0 b 2 3 0 0 a 3 1 4 0 b 4 5 0 0 a 5 c 6 a 7
- 27. 6th ITERATION: k←2 Pk Pqa=P2 P5a=(ab ababaa)⊐ ⊐ ⊐ FALSE k←1 Pk Pqa=P2 P4a=(a ababbaa⊐ ⊐ ⊐ TRUE 8.δ(q,a)←k δ(5,a)←1 2.for q←5to 7 3.for each character aє(a,b,c) q=5,a=a 4.k←min(m+1,q+2)=min(8,7)=7 K←7 Pk Pqa=P7 P5a=(ababaca ababaa)⊐ ⊐ ⊐ FALSE K←6 Pk Pqa=P6 P5a=(ababac ababaa)⊐ ⊐ ⊐ FALSE K←5 Pk Pqa=P5 P5a=(ababa ababaa) FALSE⊐ ⊐ ⊐ k←4 Pk Pqa=P4 P5a=(abab ababaa)⊐ ⊐ ⊐ FALSE K←3 Pk Pqa=P3 P5a=(aba ababaa) FALSE⊐ ⊐ ⊐ State a b c P 0 1 0 0 a 1 1 2 0 b 2 3 0 0 a 3 1 4 0 b 4 5 0 0 a 5 1 c 6 a 7
- 28. CONT… 4.q←5 3.for each character aєb K←7 Pk Pqa=P7 P5a=(ababaca ababab)⊐ ⊐ ⊐ FALSE K←6 Pk Pqa=P6 P5a=(ababac ababab)⊐ ⊐ ⊐ FALSE K←5 Pk Pqa=P5 P5a=(ababa ababab)⊐ ⊐ ⊐ FALSE k←4 Pk Pqa=P4 P5a=(abab ababab)⊐ ⊐ ⊐ TRUE 8.δ(q,a)←k δ(5,b)←4 State a b c P 0 1 0 0 a 1 1 2 0 b 2 3 0 0 a 3 1 4 0 b 4 5 0 0 a 5 1 4 c 6 a 7
- 29. CONT… 4.q←5 3.for each character aєc K←7 Pk Pqa=P7 P5a=(ababaca ababac)⊐ ⊐ ⊐ FALSE K←6 Pk Pqa=P6 P5a=(ababac ababac)⊐ ⊐ ⊐ TRUE 8.δ(q,a)←k δ(5,c)←6 State a b c P 0 1 0 0 a 1 1 2 0 b 2 3 0 0 a 3 1 4 0 b 4 5 0 0 a 5 1 4 6 c 6 a 7
- 30. 7TH ITERATION: 2.for q←6to 7 3.for each character aє(a,b,c) q=6,a=a 4.k←min(m+1,q+2)=min(8,8)=8 K←8 Pk Pqa=P8 P6a=(ababacaa ababaca)⊐ ⊐ ⊐ FALSE K←7 Pk Pqa=P7 P6a=(ababaca ababaca)⊐ ⊐ ⊐ TRUE 8.δ(q,a)←k δ(6,a)←7 State a b c P 0 1 0 0 a 1 1 2 0 b 2 3 0 0 a 3 1 4 0 b 4 5 0 0 a 5 1 4 6 c 6 7 a 7
- 31. CONT… K←1 Pk Pqa=P1 P6a=(a ababacb) FALSE⊐ ⊐ ⊐ K←0 Pk Pqa=P0 P6a=(є ababacb) TRUE⊐ ⊐ ⊐ 8.δ(q,a)←k δ(6,b)←0 2.q←6 3.for each character aєb K←8 Pk Pqa=P8 P6a=(ababacaa ababacb)⊐ ⊐ ⊐ FALSE K←7 Pk Pqa=P7 P6a=(ababaca ababacb)⊐ ⊐ ⊐ FALSE K←6 Pk Pqa=P6 P6a=(ababac ababacb) FALSE⊐ ⊐ ⊐ K←5 Pk Pqa=P5 P6a=(ababa ababacb) FALSE⊐ ⊐ ⊐ K←4 Pk Pqa=P4 P6a=(abab ababacb) FALSE⊐ ⊐ ⊐ K←3 Pk Pqa=P3 P6a=(aba ababacb) FALSE⊐ ⊐ ⊐ K←2 Pk Pqa=P2 P6a=(ab ababacb) FALSE⊐ ⊐ ⊐ State a b c P 0 1 0 0 a 1 1 2 0 b 2 3 0 0 a 3 1 4 0 b 4 5 0 0 a 5 1 4 6 c 6 7 0 a 7
- 32. CONT… 2.q←6 3.for each character aєc K←8 Pk Pqa=P8 P6a=(ababacaa ababac)⊐ ⊐ ⊐ FALSE K←7 Pk Pqa=P7 P6a=(ababaca ababac) FALSE⊐ ⊐ ⊐ K←6 Pk Pqa=P6 P6a=(ababac ababac) FALSE⊐ ⊐ ⊐ K←5 Pk Pqa=P5 P6a=(ababa ababac) FALSE⊐ ⊐ ⊐ K←4 Pk Pqa=P4 P6a=(abab ababac) FALSE⊐ ⊐ ⊐ K←3 Pk Pqa=P3 P6a=(aba ababac) FALSE⊐ ⊐ ⊐ K←2 Pk Pqa=P2 P6a=(ab ababac) FALSE⊐ ⊐ ⊐ K←1 Pk Pqa=P1 P6a=(a ababac) FALSE⊐ ⊐ ⊐ K←0 Pk Pqa=P0 P6a=(є ababac) TRUE⊐ ⊐ ⊐ 8.δ(q,a)←k δ(6,c)←0 State a b c P 0 1 0 0 a 1 1 2 0 b 2 3 0 0 a 3 1 4 0 b 4 5 0 0 a 5 1 4 6 c 6 7 0 0 a 7
- 33. 8TH ITERATION:2.q←7 to 7 3.for each character aє(a,b,c) q=7,a=a 4.k←min(m+1,q+2)=min(8,9)=8 K←8 Pk Pqa=P8 P7a=(ababacaa ababacaa)⊐ ⊐ ⊐ FALSE K←7 Pk Pqa=P7 P7a=(ababaca ababacaa) FALSE⊐ ⊐ ⊐ K←6 Pk Pqa=P6 P7a=(ababac ababacaa) FALSE⊐ ⊐ ⊐ K←5 Pk Pqa=P5 P7a=(ababa ababacaa) FALSE⊐ ⊐ ⊐ K←4 Pk Pqa=P4 P7a=(abab ababacaa) FALSE⊐ ⊐ ⊐ K←3 Pk Pqa=P3 P7a=(aba ababacaa) FALSE⊐ ⊐ ⊐ K←2 Pk Pqa=P2 P7a=(ab ababacaa) FALSE⊐ ⊐ ⊐ K←1 Pk Pqa=P1 P7a=(a ababacaa) TRUE⊐ ⊐ ⊐ 8.δ(q,a)←k δ(7,a)←1 State a b c P 0 1 0 0 a 1 1 2 0 b 2 3 0 0 a 3 1 4 0 b 4 5 0 0 a 5 1 4 6 c 6 7 0 0 a 7 1
- 34. CONT… 8.δ(q,a)←k δ(7,b)←2 2.q←7 3.for each character aєb K←8 Pk Pqa=P8 P7a=(ababacaa ababacab)⊐ ⊐ ⊐ FALSE K←7 Pk Pqa=P7 P7a=(ababaca ababacab) FALSE⊐ ⊐ ⊐ K←6 Pk Pqa=P6 P7a=(ababac ababacab) FALSE⊐ ⊐ ⊐ K←5 Pk Pqa=P5 P7a=(ababa ababacab) FALSE⊐ ⊐ ⊐ K←4 Pk Pqa=P4 P7a=(abab ababacab) FALSE⊐ ⊐ ⊐ K←3 Pk Pqa=P3 P7a=(aba ababacab) FALSE⊐ ⊐ ⊐ K←2 Pk Pqa=P2 P7a=(ab ababacab) TRUE⊐ ⊐ ⊐ State a b c P 0 1 0 0 a 1 1 2 0 b 2 3 0 0 a 3 1 4 0 b 4 5 0 0 a 5 1 4 6 c 6 7 0 0 a 7 1 2
- 35. CONT… K←1 Pk Pqa=P1 P7a=(a ababacac)⊐ ⊐ ⊐ FALSE K←0 Pk Pqa=P0 P7a=(є ababacac)⊐ ⊐ ⊐ TRUE 8.δ(q,a)←k δ(7,c)←0 2.q←7 3.for each character aєc K←8 Pk Pqa=P8 P7a=(ababacaa ababacac)⊐ ⊐ ⊐ FALSE K←7 Pk Pqa=P7 P7a=(ababaca ababacac)⊐ ⊐ ⊐ FALSE K←6 Pk Pqa=P6 P7a=(ababac ababacac) FALSE⊐ ⊐ ⊐ K←5 Pk Pqa=P5 P7a=(ababa ababacac) FALSE⊐ ⊐ ⊐ K←4 Pk Pqa=P4 P7a=(abab ababacac) FALSE⊐ ⊐ ⊐ K←3 Pk Pqa=P3 P7a=(aba ababacac) FALSE⊐ ⊐ ⊐ K←2 Pk Pqa=P2 P7a=(ab ababacac) FALSE⊐ ⊐ ⊐ State a b c P 0 1 0 0 a 1 1 2 0 b 2 3 0 0 a 3 1 4 0 b 4 5 0 0 a 5 1 4 6 c 6 7 0 0 a 7 1 2 0
- 36. REQUIRED RESULT: State a b c P 0 1 0 0 a 1 1 2 0 b 2 3 0 0 a 3 1 4 0 b 4 5 0 0 a 5 1 4 6 c 6 7 0 0 a 7 1 2 0 THE FINITE STATE AUTOMATA IS: 0 1 2 3 4 5 6 7 a b a b a c a a a b b a a
- 37. ALGORITHM…
- 38. FINITE AUTOMATON MATCHER FINITE-AUTOMATON-MATCHER(T,δ,m) 1.n=T . length 2.q=0 3.for I =1 to n 4. q= δ( q, T[ i ] ) 5. If q==m 6. Print “Pattern occurs with shift” i -m
- 39. DRY RUNNING…
- 40. FINITE-AUTOMATON MATCHER i= 1 2 3 4 5 6 7 8 9 10 11 T= a b a b a b a c a b a 1.n← length[T] n←11 2. q←0 3. for i←1 to n for i←1 to 11
- 41. 1st ITERATION: 3. for i←1 to 11 4. do q←δ(q,T[i]) q←δ(0,T[1])=δ(0,a) q←(0,a)=1 5. if q=m if 1=7 FALSE 0 1 2 3 4 5 6 7 a b a b a c a a b a b a a
- 42. 2ND ITERATION: i=2,q=2 4. do q←δ(1,T[2]) =δ(1,b) q←2 5 if 2=7 FALSE 0 1 2 3 4 5 6 7 a b a b a c a a b a b a a
- 43. 3RD ITERATION: i=3,q=2 4. do q←δ(2,T[3]) =δ(2,a) q←3 5 if 3=7 FALSE 0 1 2 3 4 5 6 7 a b a b a c a a b a b a a
- 44. 4th ITERATION: i=4,q=3 4. do q←δ(3,T[4]) =δ(3,b) q←4 5 if 4=7 FALSE 0 1 2 3 4 5 6 7 a b a b a c a a b a b a a
- 45. 5th ITERATION: i=5,q=4 4. do q←δ(4,T[5]) =δ(4,a)=5 q←5 5 if 5=7 FALSE 0 1 2 3 4 5 6 7 a b a b a c a a b a b a a
- 46. 6TH ITERATION: i=6,q=5 4. do q←δ(5,T[6]) =δ(5,b)=4 q←5 5 if 4=7 FALSE 0 1 2 3 4 5 6 7 a b a b a c a a b a b a a
- 47. 7TH ITERATION: i=7,q=4 4. do q←δ(4,T[7]) =δ(4,a)=5 q←5 5 if 5=7 FALSE 0 1 2 3 4 5 6 7 a b a b a c a a b a b a a
- 48. 8TH ITERATION: i=8,q=5 4. do q←δ(5,T[8]) =δ(5,c)=6 q←6 5 if 6=7 FALSE 0 1 2 3 4 5 6 7 a b a b a c a a b a b a a
- 49. 9TH ITERATION: i=9,q=6 4. do q←δ(6,T[9]) =δ(6,a)=7 q←7 5 if 7=7 TRUE Then print “pattern occurs with shift”i-m Print” pattern occurs with shift”9-7=2 0 1 2 3 4 5 6 7 a b a b a c a a b a b a a
- 50. 10th ITEARTION: i=10,q=7 4. do q←δ(7,T[10]) =δ(7,b)=2 q←2 5 if 2=7 FALSE 0 1 2 3 4 5 6 7 a b a b a c a a b a b a a
- 51. 11TH ITEARTION: i=11,q=2 4. do q←δ(2,T[11])=δ(2,a)=3 q←3 5 if 3=7 FALSE 0 1 2 3 4 5 6 7 a b a b a c a a b a b a a
- 52. REQUIRED OUTPUT: After removing useless edges: 0 1 2 3 4 5 6 7 a a a a ab b c b b
- 53. CONT.. State a b c P 0 1 0 0 a 1 1 2 0 b 2 3 0 0 a 3 1 4 0 b 4 5 0 0 a 5 1 4 6 c 6 7 0 0 a 7 1 2 0 i= - 1 2 3 4 5 6 7 8 9 10 11 T[i] - a b a b a b a c a b a state ΦT[i] 0 1 2 3 4 5 4 5 6 77 2 3
- 54. COMPLEXITY…
- 55. COMPLEXITY ANALYSIS: • The simple loop time of FINITE-AUTOMATON-MATCHER implies that its matching time on the text string of length is Θ(n).This matching time does not include the preprocessing time required to compute the transition function δ) • The running time of COMPUTE-TRANSITION-FUNCTION is O(m^3 ∑ ),because the outer loops contribute a factor of│ │ m ∑ ,the inner repeat loop can run atmost m+1 times, and the test│ │ Pk Pqa on line 7 can require computing upto m characters.⊐ • Much faster procedures exist; the time required to compute δ from P can be improved to O(m ∑ ) by utilizing some cleverly computed│ │ information about the pattern P. With this improved procedure for computing δ from P to O(m| |), we can find all occurrences of aƩ length-m pattern in a length-n text over an alphabet ∑ with O(m ∑ ) preprocessing time and Θ(n) matching time.│ │
- 57. ADVANTAGES & DISADVANTAGE: • String matching automaton are very efficient • Examine each text character exactly once taking constant time per text character. • The time to build the automaton can be large if Ʃ is large
- 58. THANK YOU