CONVERSION OF NFA WITH EPSILON TO NFA WITHOUT EPSILON IN AUTOMATA THEORY || TOC
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- เผยแพร่เมื่อ 25 ส.ค. 2024
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I'm sure you guys are here before exams🙂
😁
Yes obviously without exams who will study🥲
No, I am here after exams, having tails
My exam is tomorrow and i just started studying. 😅🥲
No
So best for understanding. I was having many doubts in this numerical but now I get to understand each and every part. Thank you for this
Then q1 should also be final state by above explanation.
Same doubt
he did mistake
Sir you are such an amazing teacher u explain very well with proper understanding thank you so much !!!!!
Thank you for this whole series. You make our study easier than ever.
easy explanation also cover Finite Automata with Outputs
good teaching sir easily understanding
and we want to more examples on this method
thank you sir, your explaintion rule is good, i have understand in better way.
very informative lecture with easy process.
i am in now love with this concept
Good for understanding. Teacher forgot to mark initial state here. For the above example epsilon closure of q1 also contains q2 which is final state. Can we make q1 also as final state?
good question bro
yes any state which has F (final state) in its epsilon closure is also a final state. And q0 is also initial state because initial state of bot nfa with and without epsilon is the same
@@beckhamroshan3655 The part with F confused me, because F is a set. So for anyone watching this in the future: The epsilon closure does not need to contain all of F. It just needs to have a non-empty intersection with F (i.e. contain at least one final state).
very clear explanation
thank you sir, you have my respect
Explanation awesome sir
Directions and Editing was so nice 🎉❤keep rocking guys👥💥
very well done!
Thanks a lot for clearly mentioning abt final states!!
Teaching vere level anthe
Sir thankyou for your explanation sir but please explain all the topics in flat and also please make a video how to get the good score in semister
Thank you so much 🙏
Thanks for Everything sir🤗🤗
Super explanation sir
finals states should be q0,q1, and q2 then it is fully correct.
sir what will be the answer if it would have been epsilon-closure(q1Uq2)
supper sir ,thanks
q1 is also final state sir
Nice
Listening 1 hour before exam
Sir
is there any need to draw any transition graph after conversion
Sir there is a mistake q1 is also a final state
How can you say q1 is final state
Op sir
Q1 also should be final state
right ? since q2 is present on the epsilon closure of q1 also
sir, why q0 is final state with q2? I don't understund.
if u consider the e-closure of q0 we have (q0,q1,q2) where we have q2 in it right...so any closure of particular state having a final state ..then that state also becomes final state.
@@sushmithapampari342 same for q1 also
@@taniasaha3518 i have same doubt
Why Q1 is not marked a final state ?
Sir how q0 is final state????
if u consider the e-closure of q0 we have (q0,q1,q2) where we have q2 in it right...so any closure of particular state having a final state ..then that state also becomes final state.
@@sushmithapampari342 mam epilson closure (Q1) = { Q1,Q2}
Q2 present in epsilon closure Q0 , Q1
But sir marked final state only Q0
We should mark Q1 also final state ?