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asked in Theory of Computation by Active (1.1k points) | 44 views
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leads to {q2,q3,q4}, q3 is final state therefore string accepted
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how {q2,q3,q4}, what δ* [email protected]Shaik Masthan

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δ(Qi,x) = Qj where x is a string and it have only one input symbol,

δ*(Qi,x) = Qj where x is a string and it have more than input symbol, i.e., it means you will be reach Qj state after processing the string x.

δ*(Qi,"ABC") = δ ( δ( δ(Qi,"A") ,"B") ,"C")

1 Answer

+1 vote

String ababab is accepted by NFA.

answered by Boss (22.6k points)
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@abhishek  @shaik

Please suggest how from (q2, abab) ---> it reached (q1,bab) as you missed epsilon movement to Q3.

 

Please suggest on epsilon movement and rules related to it.

Thanks
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Why $\delta^*$ is used here why not $\delta$
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answer is given {q2,q3,q4} @abhishekmehta4u....what is the significance of delta*?
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@Mayankprakash,

Please suggest on epsilon movement and rules related to it.

if a transition from Qi to Qj  due to ∈

then you can interchange the Qi with Qj at any Place.



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