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

[@jennyfisher6397]

I think there is one mistake: when you calculate Q at C direction, it can go to C itself and P as well. Look at the first step P.

[@aryanpatel8666]

and also final NFA is wrong.

[@mehdikatani3312]

True

[@relaxationvibs]

and p on input c, that part has q which on input goes to P, and e* on p goes to p,q,r so p o input all (a,b,c) goes to itself also

[@05-cs-adarshgautam16]

yes exactly

[@suborno9249]

true

[@HemantKumar-ws4xl]

sir there is missing a p in third table. Q on input C it goes to Q and P.

[@awaifgadekaai5432]

yes its missing.

[@KulasangarGowrisangar]

yes it's missing out.

[@kalpesh7668]

yes it is missing , minor mistake there

[@TLCHobbies]

@@kalpesh7668 I think that's quite a major mistake.

[@sayanbanerjee1047]

@@TLCHobbies Exactly, the mistake is huge, it has changed the whole automata!

[@4orth-dimensionalamansagar349]

He teaches so good So his students could find thier mistakes And correct thier answer THANK YOU NESO .

[@4orth-dimensionalamansagar349]

Mistake* it's their

[@rishabgangwar9901]

@@4orth-dimensionalamansagar349 He teaches so good So his students could find 'their' own committed mistakes And correct them.

[@tarle_engineer9372]

@@rishabgangwar9901 hahaha lack of concentration 2:30

[@2271masoud]

Thanks for the tutorial . Q on input c goes to P and Q and in it's equivalent NFA, Q on c goes to {P,Q,R}

[@raiyandiwan-sim]

4:05 'c' also goes from Q to P, so Q to {P,Q,R} for input c

[@inderjeetchawla527]

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

[@funmilayobisola6906]

i think it should be like that though. We learnt from him and understand well enough to spot errors

[@Nirala_414]

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

[@amrikdutta]

Thank you so much ...you guys teach so good .. i am learning from this channel for past 2 semesters.. thank you so much neso

[@digamberkhandebharad3587]

Sir on getting input 'c' we can go itself and 'p' also

[@abdullahsaid4765]

so the answer should be {p,q,r}

[@aditya21006]

Yes correct👍

[@programming_hut]

@@abdullahsaid4765 yes

[@Piyush-yq9iu]

Yes correct

[@danielwintermeyer4642]

When you check the c input for State Q then on checking the 2nd epsilon why did you not include state P. State Q in c input will have {P,Q,R} in it's final states. Please Verify.

[@prabhnoorsingh4907]

Someone confirm please ASAP

[@kunwarashutoshsingh4405]

yes it is true

[@jatinanand3726]

correct bro

[@bishalmukherjee9794]

Daniel Wintermeyer exactly i was going to say the same

[@rishavbadola7357]

Yeah you're right

[@abhishekbhardwaj2518]

Bro taught everyone so well that he himself got behind😂

[@yatridavda3972]

Q on input C it goes to P as well that not considered in this video so the null closure of Q for C is {P,Q,R}

[@vikram5569]

3:59 Q goes to P also

[@BilluAck]

Thank you neso helped me a lot 🙏🙏🙏

[@imritam97]

Sir...state Q on getting the input c it goes to Q itself and P , and P on getting input c it goes to R...So the 2nd €* will be P,Q,R...

[@JP-qp9ht]

exactly

[@saswatasarkar7792]

Agreed. Either the transition table has to have P,Q R at 2nd epsilon for state Q on input c ; or there should be no transition from Q to P on c, in the epsilon NFA diagram itself.

[@pratiknichite1024]

agreed.

[@elyasaf755]

Sir, Q on getting C to Q & P. so in the NFA, P on getting C goes to {P,R,Q} Thank you so much for the videos! you are a saviour!!

[@dushyantjakhar1716]

who could thought, 2 year after an another new student going to find same mistake , and when he did comment he sow mine and get soure about it

[@vyomaudichya]

Q on getting c goes to P also...

[@pankushmahajan4598]

Transition(Q,c) results in P&Q instead of Q only Isnt it?

[@anushreevirtualgaming226]

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

[@anushreevirtualgaming226]

U taught the concepts so well that even after solving just 1 Q we are able to solve nearly all Q , thanks for such a great teaching

[@divyaanshuagrawal3383]

Such good explanation! Thank you sir!

[@alabiabdullahi9054]

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

[@ashukashyap7142]

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}

[@kingprince2604]

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! 😎

[@rajuneupane1488]

Sir I follow your videos very much ....sir thank u for making these videos for us these videos are d only hope sir ...sir during third transition u didn't went to p state wen se got c as a input ....y sir. .....I think it should be Q->C = p,q,r

[@bipinkumartiwary3122]

Sir in epsilon- NFA diagram Q on input c going to Q as well as P ....so there must be some change in epsilon clouser ... what's should be that change??

[@riyapandey4798]

3:56 "Q on input 'c' goes to P and Q"

[@farazjawaid2982]

wonderful method used.

[@philip_vo]

Correct! All of you said correct. There is a missing here

[@devanshydv6904]

2:50 Sir there is a mistake P on input C....

[@OneStopShorts16]

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} )

[@ev7joe]

May God bless you alot

[@RahulSikarwar99]

Sir , Please Start To Make and Upload Videos for Microprocessors and Microcontrollers

[@sabesan]

3:58 Q on input C it goes to P and Q and then we need to update in table {P,Q,R} .....

[@Jerdz]

3:06 I think there is a mistake, State Q with the input c gives Q but also P ! Because of the arrow a,c from Q to P. Right ?

[@Deathtonewworldorder]

Exactly at 2:45 and 4:00

[@vellaviper79]

thank you sir

[@vijaykumarsinha30]

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

[@the_ocean_eyes]

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

[@ev7joe]

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!!!!

[@priyanshugupta3816]

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

[@a.y5742]

C input from Q should go to P and R (I've not seen anyone mention R) as well since from Q if you input C and Epsilon, you have C which you can reach R with.

[@prateekkaushik5847]

Q after €* ---> Q ... Q after C ----> P,Q ... P,Q after €* --- > P,Q for C in table ... correction ;)

[@Ramakrishna-nb8oq]

ya its wrong there

[@arsha1625]

Sir at 2:46 Q on getting 'c' goes to P & Q. Same mistake at 3:57 So in the transition table, We get P on getting c goes to P,Q&R. Q on getting c goes to P,Q&R.

[@dhanushsivajaya1356]

Thankyou sir

[@Mahi15820]

sir there is a P is missing in 3rd table of P. Q on input C it goes to P&Q both.The epsilon closure of P is also included in the 3rd table of P

[@Dacapoelcodaa]

i mean its ok to make mistakes while teaching, but not ok to discover them and not publish! i spent time doubting and searching in the comments

[@SutraNewsNetwork]

true

[@rohitbharati4374]

Sir please start microprocessor and microcontroller

[@hoangan7613]

If english was my mother's language I would play it in 1.5x

[@manikanta-oh3qy]

sir plz network theory lectures

[@Stoicbanda]

nice ..can u plss make a playlists of network and system for gate & ese

[@deependrasingh7685]

Can you do a video on, Epsilon NFA to DFA.

[@rakeshreddynallamilli3233]

now as we know how to convert E NFA to NFA , we can obtain DFA from the resultant NFA .

[@premjaiswal3935]

State Q on getting input c goes to Q itself and also on state P.

[@001-q1s]

Can anyone answer this question please , my university asks enfa to dfa. So if I do first enfa to nfa and again nfa to dfa will this be fine? It wont be minimized though

[@SutraNewsNetwork]

Yes, it will be fine. You can also minimize the dfa in the later step.

[@fact-jr9tt]

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

[@rithanyabalamurali6936]

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.

[@LaibaRao-x5g]

P on c goes to p,q,r

[@Mohammad-hg8ty]

Dear sir, state Q when getting c as an input it goes to states Q and P

[@nitigyajoshi4658]

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

[@tapanmahata8330]

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

[@ahmadazeem7785]

Sir you have made a mistake there...P on input C will go on {P,Q,R} while you wrote {Q,R}.

[@kshitijkumre9895]

Ahmad Azeem bro.. P on input C will go to only R. This is not the mistake.. He is correct. But.. There is a mistake.. In Q on input C will go to Q and P also.. He wrote only Q. This is the mistake. So €* there will be {P, Q, R}

[@mdmuquimakhter5145]

@@kshitijkumre9895 I didn't get u .. why only R ??

[@_zoro]

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

[@atkareakash]

Q can reach the final state through only eplsilon.....by going to P first then to R??

[@amanchamoli3796]

Q on input c it also go to state P

[@rohithchinna1105]

from q using c we can also Travers to p

[@omraut_gaming]

Q on c goes to q and P both. But u have mentioned only q

[@5UNIL]

Is all these video notes available in pdf file. ?

[@aditya_s_10]

can anyone tell why p is the final state ??

[@ekinkale6243]

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

[@alperkaya8919]

Q can go to P with c too??

[@shrishkumari1255]

In previous one ,for one state you draw the phi state but in this why you can't draw phi state ...you mention phi state in transition table for R state ...

[@chocolateboyanubhab7340]

sir kindly check this video again...there are a lot of errors

[@SaiTeja-wp3qp]

Q on c goes to itself and also P

[@yogeshvishnole1264]

Yes there is minor mistake but procedure is correct.

[@devanshusingh8665]

Sir plzz re-check the process, u made a mistake while mapping using input c

[@universalknowledge7304]

sir Q on input c goes to Q and P

[@arnabdana2273]

I think u don't consider Q input c

[@RohitSingh-ps5cy]

There is a mistake while calculating Q on input c. And final NFA is wrong.

[@SimpleMind34]

I thought the closer is only when you see epsilon and nothing else with it?

[@ankitsisodiya8408]

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.

[@parasgupta6540]

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

[@bachelorsdesk]

4:16 wrong on Q at c

[@devanshydv6904]

Sai to ha bhai

[@bachelorsdesk]

@@devanshydv6904 look at all the comments, i am not going to watch the video again. dear

[@anuragverma906]

q->c goes to pqr

[@sambitpati6129]

sir here Q on input C it goes to Q and P both.

[@muthanaisawi3482]

اول يابه انت قلطان قلطان قلطان بروح محمدنا انت قلطان دروح

[@sulfikharshamsuddin3916]

Epsilon Closure of P on Q is not correct. Please correct it

[@Mohammed.1471]

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...

[@nirajdev6940]

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

[@prakharjain2707]

Q on getting c it go P also

[@anuragsahu7852]

The Answer is wrong there must be C also included in the Q-e* transistion.

[@lokeshmishra3179]

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

[@veerenkonchada2540]

sir .... q on getting c goes to p also,:p

[@abcmedia9044]

Please indicate the error in the video. Thanks.

[@mkgroup6079]

there is some error in it

[@eminence_Shadow]

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

[@poojaharidas759]

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

[@mrfab740]

Hlw sir i have query in this video

[@prithhviraajchattopadhyay7698]

Q on input c goes to both Q and P. I believe you missed an alphabet.

[@srikarbokka8046]

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

[@aaryanraina4995]

Actually, Its a kinda drawing error. He wanted to draw (a,epsilon) for the transition from Q to P.

[@AUSTINJOHNSdeigra]

If it was Epsilon then Q on epsilon should have gone to P, which isn't done either.

[@aayushsingh5382]

Sir in 3 rd state 0f conversion you misses the P part in Q state...

[@charanguggulla6899]

sir,there you have c which goes to p

[@charanguggulla6899]

about Q