1. Lecture2 Regular Language
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2. DFA –deterministic finite automaton
A DFA has:
a finite set of states
1 special state – start (initial) state
0 or more special states - final (accept) states
input alphabet
transition table: (state, symbol)  next state q s q4 p a r
(p, a) -> r
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3. Informally -- How does a DFA work?q1 a
DFA begins in the start state.
DFA goes into a loop until the entire input string is read.
In each step, DFA consults a transition table and changes its state based on (q, a) where
q - current state
a – current symbol read
After reading the input string,
if the state is final, the input is accepted
if the state is not final, the input is rejected
Language recognized by a DFA or the language of a DFA – the set of all the strings accepted by the DFA.
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4. Pictorial representation of DFA
state q
Start state s
Final state q4
transition p a r
(p, a) -> r
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5. DFA Demo b b a b a a b b a b b a a a b b b Y N N 1 2  2004 SDU 51 2 b
multiple of 3 b's b b a b a a b b a b b
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6. DFA Demo b a b a a b b a b b a a a b b b Y N N 1 2  2004 SDU 61 2 b
multiple of 3 b's b a b a a b b a b b
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7. DFA Demo b a a a b b a b b a a a b b b Y N N 1 2  2004 SDU 71 2 b
multiple of 3 b's b a a a b b a b b
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8. DFA Demo b a a b b a b b a a a b b b Y N N 1 2  2004 SDU 81 2 b
multiple of 3 b's b a a b b a b b
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9. DFA Demo b a b b a b b a a a b b b Y N N 1 2  2004 SDU 91 2 b
multiple of 3 b's b a b b a b b
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10. DFA Demo b a b a b b a a a b b b Y N N 1 2  2004 SDU 101 2 b
multiple of 3 b's b a b a b b
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11. DFA Demo b a a b b a a a b b b Y N N 1 2  2004 SDU 111 2 b
multiple of 3 b's b a a b b
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12. DFA Demo b a b b a a a b b b Y N N 1 2  2004 SDU 12 multiple of 3 b's1 2 b
multiple of 3 b's b a b b
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13. DFA Demo b a b a a a b b b Y N N 1 2  2004 SDU 13 multiple of 3 b's1 2 b
multiple of 3 b's b a b
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14. DFA Demo ACCEPT b a a a a b b b Y N N 1 2  2004 SDU 141 2
ACCEPT b
multiple of 3 b's b a
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15. DFA Demo b a b b a a b b a a a b b b Y N N 1 2  2004 SDU 151 2 b
multiple of 3 b's b a b b a a b b
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16. DFA Demo b a b a a b b a a a b b b Y N N 1 2  2004 SDU 161 2 b
multiple of 3 b's b a b a a b b
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17. DFA Demo b a a a b b a a a b b b Y N N 1 2  2004 SDU 171 2 b
multiple of 3 b's b a a a b b
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18. DFA Demo b a a b b a a a b b b Y N N 1 2  2004 SDU 181 2 b
multiple of 3 b's b a a b b
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19. DFA Demo b a b b a a a b b b Y N N 1 2  2004 SDU 19 multiple of 3 b's1 2 b
multiple of 3 b's b a b b
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20. DFA Demo b a b a a a b b b Y N N 1 2  2004 SDU 20 multiple of 3 b's1 2 b
multiple of 3 b's b a b
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21. DFA Demo REJECT b a a a a b b b Y N N 1 2  2004 SDU 211 2
REJECT b
multiple of 3 b's b a
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22. Example: Diagram of DFAq0 q1 a
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23. Step-By-Step a q0 q1 a
aaa
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24. Step-By-Step a q0 q1 a
aaa
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25. Step-By-Step a q0 q1 a
aaa
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26. Step-By-Step a q0 q1 a
aaa
The string is accepted
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27. Formal definition of DFA
A DFA is a 5-tuple: M = (Q, , , s, F)
Where,
Q is a finite set of states
 is a finite set of input alphabet
s  Q is the start state
F  Q is the set of final states
: Q   -> Q is the transition function // note that this is a function, so, for each state and each symbol, there is an arrow that points to another state.
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28. Formal definition of Computation
Let M = (Q, , , s, F) be a finite automaton and let w = w1w2…wn be a string where each wi is a member of the alphabet . M accepts w if there is a sequence of states r0, r1, …, rn such that
1)r0 = s,
2)(ri, wi+1) = ri+1, for i=0, …, n-1.
3) rn F
We say that M recognizes language A if A = {w | M accepts w}.
Note that the language recognized by a DFA M (or the language of a DFA M) is unique! We write it as L(M).
What is language of the following machine? Its formal description? q0 q1 a
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29. Regular Languages
Definition: A Language is called regular if some DFA recognizes it.
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30. Given a language, how to define DFA?
Creative process requiring you to:
(i) Put yourself in DFA's shoes
(ii) Find finite amount of info, based on
which string accepted/rejected
(iii) From step (ii), determine number of states and then transitions.
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31. Examples L = {w | w begins with a 0 and ends with a 1} 1 1 q0 q1 q2 11 q0 q1 q2 1 r
0,1
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32. Examples (cont.) L = {w | w contains the substring 101 } 0,1 1 1 1 s 11 1 1 s 1 10
101
More in exercises
 2004 SDU