A Deterministic Finite Automata(DFA) is a five tuple notation can be represented as
D=(Q,𝚺,𝛿,q0,F).
Where D is the name of the DFA.
Q is the finite set of states.
𝚺 is the set of input alphabet.
𝛿 is the transition function which can be mapped from Qx𝚺->Q.
q0 is the starting state of the machine.
F is the finite set of final states Which is an acceptance state of the machine .
Example:
Design DFA which accepts the string 1100 only.
Solution
Let the DFA D=(Q,𝚺,𝛿,q0,F).
q0 be the starting sate
F be the final state which is an acceptance state i.e., according to our machine it is q4. It is represented as a double circle in transition diagram.
Acceptance of a string
Let w=1100
𝛿(q0,1100)---> q0 upon 1 transition the machine enter into q1 state which gives the result as
𝛿(q1,100).
----> Now again, q1 upon 1 transition the machine enters into q2 state which gives the result as
𝛿(q2,00)
Similarly,
----->𝛿(q3,0)
------>q4 which belongs to final state
So, the given string 1100 is accepted by the given DFA.
Not Acceptance of a string
Let choose the string other than 1100
i.e., Let w=1001
𝛿(q0,1001)---->𝛿(q1,001)
----->𝛿(q5,01)
----->𝛿(q5,1)
------>q5 which does not belongs to the set of final state.
So, the given string 1001 is not accepted by the given DFA.
D=(Q,𝚺,𝛿,q0,F).
Where D is the name of the DFA.
Q is the finite set of states.
𝚺 is the set of input alphabet.
𝛿 is the transition function which can be mapped from Qx𝚺->Q.
q0 is the starting state of the machine.
F is the finite set of final states Which is an acceptance state of the machine .
Example:
Design DFA which accepts the string 1100 only.
Solution
Let the DFA D=(Q,𝚺,𝛿,q0,F).
Where Q={q0,q1,q2,q3,q4,q5}
𝚺 be the input alphabet which contains the symbols 0 and 1 i.e.,{0,1}.q0 be the starting sate
F be the final state which is an acceptance state i.e., according to our machine it is q4. It is represented as a double circle in transition diagram.
The Transition table for the above DFA is as follows
𝛿
|
0
|
1
|
q0 (start state)
|
q5
|
q1
|
q1
|
q5
|
q2
|
q2
|
q3
|
q5
|
q3
|
q4
|
q5
|
q4(final state)
|
q5
|
q5
|
q5
|
q5
|
q5
|
Acceptance of a string
Let w=1100
𝛿(q0,1100)---> q0 upon 1 transition the machine enter into q1 state which gives the result as
𝛿(q1,100).
----> Now again, q1 upon 1 transition the machine enters into q2 state which gives the result as
𝛿(q2,00)
Similarly,
----->𝛿(q3,0)
------>q4 which belongs to final state
So, the given string 1100 is accepted by the given DFA.
Not Acceptance of a string
Let choose the string other than 1100
i.e., Let w=1001
𝛿(q0,1001)---->𝛿(q1,001)
----->𝛿(q5,01)
----->𝛿(q5,1)
------>q5 which does not belongs to the set of final state.
So, the given string 1001 is not accepted by the given DFA.