Conversion of Epsilon NFA to NFA - Examples (Part 2)

Hi, what if state P on getting input c? Is it not supposed to be {P,Q,R} also?

q->r with c kaha gya bhai

correct the error plox

Sir i there your answer is wrong because, q on iput c goes to c as well as p and p, E* is p,q,r...

When you solve worng solution so many people can notice it !

Q on input c it also go to state P

wonderful method used.

thank you sir

Such good explanation! Thank you sir!

You made a mistake somewhere Sir. A "c" input in state "Q" can either go to state "P" or remain in state "Q". Anyway Thank You Sir, You teach so well that I understand! 😎

auto meta is wrong where is he ??

Q on c goes to A also,, pls correct the mistake

its wrong

theres a big mistake

can anyone tell why p is the final state ??

Sir on getting input 'c' we can go itself and 'p' also
so the answer should be {p,q,r}

hey sir there is one mistake on getting C on Q IT go to Q as well as P . U forget this

Sir, Q on taking input c it can go to {P, Q} but in the Epsilon closure of Q, you have only taken Q and not P. Correct me if I am wrong

In the diagram.. Remove c from q to p guys

The answer is worng . Q on gitnig input goes to P and Q so on getting epsilon Q goes to Q but P goes to PQR so in NFA Q on getting input c goes to PQR.

Q'dan P'ye hem a hem c gitmesi gerekmiyor mu ??

i think that infamous C going to P was supposed to be an epsilon qn would've become more interesting that way

Q should also be start state.
Q on C will go to PQR in NFA

Q should be start state

sir there is a ,mistake check in c coloumn q on c can go to q and p as well please include it in the description.

May God bless you alot

Hello, I think you forgot to put input {c} from state Q to state P while working on that question! Thanks. @NESO ACADEMY otherwise, I really appreciate, your tutorials are neat!!!!

Bhai ye Q input c me P state ko ja raha hai per tune nahi kiya hai

At 4.00 the q can also goes to p then the closure of p is {p,q,r} so the final output of q with tha transaction c is {p,q,r}

44

Q on getting input "c" will go to both P and Q as per the diagram.

Here One mistake becomes huge mistake .
But it's okay we have got it correctly.
Thanks a lot sir for your efforts 💚

Q on c goes to itself and also P

اول يابه انت قلطان قلطان قلطان بروح محمدنا انت قلطان دروح

Sir q on getting c it can go to p also sir u did a mistake

This is wrong

Sir in P state in C input the right answer is {P,Q,R} (because Q-c->{Q},{P}-epsilon*->{Q},{P,Q,R} and union of Q and P,R,Q is {P,R,Q} )

Sir plz check for state Q on getting input C. Because epsilon closure of Q is Q. And on getting input C, we can reach to state P and Q. And again epsilon closure P and Q will be P,R,Q.
Plz reply sir. M i right or not?waiting for your reply

Q's epsilon closure for c is wrong....

Thankyou sir

Chill guys just thank him for teaching so good that now you are correcting him.

Yes there is minor mistake but procedure is correct.

There is a mistake .Q on getting c it can also go to P.

sir please correct

Thank you neso helped me a lot 🙏🙏🙏