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What is Non-Deterministic Finite Automata (NFA)?
Non-Deterministic means that there can be several possible transitions on the same input symbol from some state. The output is non-deterministic for a given input. NFA can be represented as a nondeterministic finite state machine.
NFA can be represented by 5 tuples (Q,$\sum$,$\delta$,q0,F)
Q is a finite non-empty set of states.
$\sum$ is a finite set of input symbols.
$\delta$ is the transition function to move from the current state to the next state.
∴ $\delta$:Q x $\sum$ → 2Q
q0 ∈ Q is the initial state.
F \subseteq Q is the set of final states.
Example1 − Draw NFA for the Regular Expression a(a+b)*ab
Solution
Example2 − Draw NFA for a + b + ab
Solution
Example3 − Let M=(Q,∑,$\delta$,q0,F) be an NFA. Show that for any qϵQ and a ϵ Σ,$\delta$′(q,a)= $\delta$(q,a)
Solution
Let M=(Q,∑,$\delta$,q0,F) be NFA that recognizes the language L. Then the DFA, M′=(Q1,∑,$\delta$′,q′0,F′) that satisfies the following conditions recognizes L:Q1=2Q, that is the set of all subset of Q.
q′0={q0}
$\delta$(q,a)=∪pϵq$\delta$(p,a) for each state q in Q1 and each symbol a in Σand
F={qϵQ1|q∩F2≠ф}
To obtain DFA M′=(Q1,Σ,$\delta$′,q′0,F′) which accepts the same language, as given NFA, M=(Q,∑,$\delta$,q0,F) does. It can proceed as follows −
Step1 − Initially Q1=ф
Step2 − Put {q0} into Q1.{q0} is the initial state of the DFA M′.
Step3 − Then for each state q in Q1 do the following: add this new state, add δ(q,a)=∪pϵqδ(p,a) to δ, where δ on the right-hand side is that of NFA M’.
Step4 − Repeat step3 till new states are there to add in Q1, if there is a new state create to add in Q1, the process eliminates.
Example4 − Prove that if L is accepted by an NFA with ∈−transitions, then L is accepted by an NFA without ∈−transitions.
Solution
Suppose L = L (D) for some DFA or NFA without ∈−transitions. It can turn D into a ∈−DFA (E) by adding transitions δ(q,∈)=ф for all states q of D. It can also move the transitions of D on input symbols, for example, δD(q,a)=D into an NFA transition to the set including the only P that is δE(q,a)={P}. Therefore, the transitions of E and D are equal, but E explicitly states that there are no transitions out of any state of E.
Let E=(QE,Σ,δE,q0,FE) be a ∈−NFA. It can use the modified subset construction represented above to make the DFA.
D=(QE,∑,δE,qD,FD)
L(D)=L(E), because the extended transition functions of E and D are equal. The extended transition function is the function that takes a state q and a string w and returns a state P, the state that automation reaches when starting in state q and processing the sequence of inputs w.
δE(q0,w)=δD(qD,w)by induction on the length of w.
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