GATE CS 2017 Set 2 — Question 39
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Theory of Computation → Finite Automata & Regular Languages → NFA
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Question
Let
δ denote the transition function and
δ^ denote the extended transition function of the
ϵ-NFA whose transition table is given below:
| δ | ϵ | a | b |
|---|
| →q0 | {q2} | {q1} | {q0} |
| q1 | {q2} | {q2} | {q3} |
| q2 | {q0} | ∅ | ∅ |
| q3 | ∅ | ∅ | {q2} |
Then
δ^(q2,aba) is
Correct answer
(C) \q₀, q₁, q₂\
Solution
First, compute the
ϵ-closures:
- E(q0)={q0,q2}
- E(q1)={q1,q2,q0}
- E(q2)={q2,q0}
- E(q3)={q3}
Now compute
δ^(q2,aba) step by step:
1.δ^(q2,ϵ)=E(q2)={q0,q2} 2.δ^(q2,a)=E(δ({q0,q2},a))=E(δ(q0,a)∪δ(q2,a))=E({q1}∪∅)=E(q1)={q0,q1,q2} 3.δ^(q2,ab)=E(δ({q0,q1,q2},b))=E(δ(q0,b)∪δ(q1,b)∪δ(q2,b))=E({q0}∪{q3}∪∅)=E({q0,q3})=E(q0)∪E(q3)={q0,q2,q3} 4.δ^(q2,aba)=E(δ({q0,q2,q3},a))=E(δ(q0,a)∪δ(q2,a)∪δ(q3,a))=E({q1}∪∅∪∅)=E(q1)={q0,q1,q2} Thus,
δ^(q2,aba)={q0,q1,q2}.
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