Theory of Computation: Conversion of Epsilon-NFA to DFA
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- Опубликовано: 7 фев 2025
- If the DFA doesn’t have a transition for any symbol, it can go to a dead/trap state, qD. So in the example explained in video, the following transitions can be included, instead of ϕ.
δ({ q1, q2 },a) = qD
δ({ q2 },a) = qD
δ({ q2 },b) = qD
From the state qD, all input symbols go to same state.
δ (qD, a) = qD
δ (qD, b) = qD
δ (qD, c) = qD
It's really so helpful for us to solve the questions at the peak time. The way you are giving the explanation is clear and crisp. Thank you so much for providing us this wonderful content.
less than 6 hrs left for sem exam and I'm here watching this....lmao
Would be better if you watch it in the examination hall itself..............
Same
Less than half an hour left for mid sem exam and I'm watching this.
For me just 2hours remaining for exam
Last 1 he
I got 98/100 in toc( bca), thanks you and other yt channels for uploading videos.
From where means which book you practice the question and please also tell which channel you follow
Also tell for maths 4
Scholar mare
Solllleyyy😂
bca bahii bca
Thanks a lot miss i have my FLAT exam in about 3 days and our teacher made this shit so much harder than it should be.
many explanations in the youtube where really complicated and hard to remember but the table method you introduced in this tutorial was damn easy and is easilty rememberedd thankyou for mam for making this so easy
Yes it's easy, but konni videos lo verela undhi ee mathod, ye method follow avvalo teliyatledu
I want to clear sem exam exam is starting by Tuesday please prayer for me i pass all subject😢😢😢
Me too bro...
@@mallikarjunavb-gi2hi best of luck my good wishes with you
U definitely fail good luck
@@wilxer5878 why r uh telling like this ??
Ma'am video was very helpful...
But one point is "In DFA there should be a transition for EVERY input character"
So You must have shown that using a dead/trap state
Yes... I think so
Yaah we have to add Dead state when there is Phi in above DFA table
If there is no transition, it is assumed the input is rejected there. There is no transition for every character in any FA afaik
You are saying right. In the "DFA" transition diagram dead state is missing.
Thankyou mam, this video made me Confident for such numericals
this lec solved every doubts i had in epsilon nfa to dfa Thank you so much
Very honest and precise explanation,
Thanks
Thanks a lot for this clear explainations.
I was really looking very badly for a suitable example for epsilon NFA to DFA conversion.
And finally found it. 💖💖
Tqsm for clear explanation ❤
Ma'am you're truly a blessing
5:22 state a value is only q0 kindly check
Epsilon closure( union of (q0,a), (q1,a), (q2,a) ) gives {q0, q1, q2}
🎉 Ma'am this video is life saving before exam 😊
Thanks a lot mam, i have my exams in an hour will get 10marks because of you :)
Thank you so much, it was really great
This is fundamentally wrong. You are just converting e-NFA to NFA. in DFA each state must have transitions equal to the number of inputs.
I got so confused just because of this video
same question here !! if i"ll add dead states, will it be a DFA ¿¿
yes, there is no definition for the transition of state b on input of a, and transition of state c on input of a and b!?
but in DFA we have make transition for every symbol on each state....
Thanks mam you are teaching awesom every student cal easily learn keet up mam
Thank you for giving such good explanation ❤ it really save my many hours ❤🎉
Less than 12 hours left and i am watching😅
Ended with easy one example not even entered intermediate problem also ,But any how you have explained well ,appreciated for that
Thank you ma'am.
Easy to understand!🙏
ghantaa
Best video for understanding this concept simply
Very Good up to the point of NOT including the Dead State for the null transitions
Clearly understood the concept thank you ❤
Excellent teaching....thank you mam
Many people comment here to add "dead state". But dead/trap state is not mandatory to add for 'phi'.
What will happen if i got two e-closure for One input like. For "a" i got q1,q2 e-closure. Then what will happen?
She is not a robot to answer machine like questions ask politely
How can a dfa have no transition for some symbols ??
Thanks you ma'am this made my life peaceful 🥰
Easy to understand with no doubts.thank you mam.
Thank you mam for making us understand more with this tutorial.Thank you and God bless you always.❤😇
mam DFA always consists one way for each input symbol from every state , then why you don't make path for another inputs like from {q1,q2} where is the transition for "a". Please explain is it you missed or it is not necessary . Although you explain very well thank you so much .
This helped me a lot ❤
Thank you
Very clear explanation ma'am,thanks a lot ♥️🙌
Thank you maam 💓 but what is that sound in 7:22 😅😅
Mai gali nhi de saktha! RUclips me sab log dekhte hai
what happens when we have more than one states inside the e-closure? for example, I gotta find the e-closure of {q0,q1}, should I take the union of e-closure of q0 and q1?
Yes. eclosure(q0) U eclosure(q1)
@@anitar8483 thanks a lot
Less than 2 hours for internal, and I'm watching this here
Thanksss 🎉🙏
Very Good Explanation Madam
This is so helpful. Thank you ma'am
there"s a doubt a null move to itself is always present even if its not marked right then in E-nfa table you havent marked null move for q2
Is it possible that a DFA contains only the final state
For Dfa all i/p should have one path know maam..here we dont have all i/p for all states.how can we call this as a DFA?
Thank you for your detailed explanation!
Very nice n in short explanatiin
Mam I did the same problem by converting first epsilon nfa to nfa and then nfa to dfa....but I got different answer? Can you help me to solve this issue
Very well explained ma'am
I believe for a DFA must have an exiting transition arrow for every symbol in its alphabet but good video
Yes u r right bro
We should take a dead state
This is epsilon nfa to without epsilon nfa, but not a dfa, since all of the alphabets are not used in each states.
but in dfa there should be transition for every input in a state
Guys please read the description ma'am has mentioned the dead state...🙂
girl your are the best
Mam do we need to draw the transitions for ɸ also in DFA?
could you please teach in detail so that people can understand the calculations in dfa, i cant understand that. the q0 union pi, etc are unable understand.please teach properly
thank you mam, very easily explained
Less than 6 hrs left to declare my Sem result....and I'm watching this video...lmao
this was great, saved me a lot of time, thank you
very cool stuffs
amazing video maam
thank you very much maam very helpful
great explanation
thank you!! Finally understood
i have a doubt if {q2 a}= null but in dfa we can give null right? then a should go on deadth state?
What, in DFA it is impossible, that from one state, you don't know where to go after some symbol. For example in your last DFA, from state B you have to go somewhere after symbol a, if it is not included, it is not a DFA...
Awesomeeee explaination madammm
Less than 9 hours remaining for sessional exam and i am watching this
Less than 1 hour left from exam, agar ye paper me aya to maja aayega pure marks ☺
Mam is there possible 'pi' in DFA
Why didn't use dead state
Do I need to write formula explaination for all ...or tables are ok??
SUPER EXPLANATION MAM TO UNDERSTAND😇
But dfa must use all the output(a,b,c) symbol,here B,C don't use all output symbol
This is completely incorrect. When you get an E-transition, you follow it, until you can't follow it anymore and you include the state itself.
thankyou mam, nice explanation
Construct NFA with E which accepts a language Consisting the strings of any numb of o's followed by any number of 1'sfollowed by any number 2's, Also convert into witn E hansition. Explain this pbm mam
Mam if there is no input symbols then what were we do
but isnt that a NFA at the end ???
2 hours before sem exam watching this....🙃🫣
Same
you forgot the transitions to the dead state for the DFA
I lost in the DFA first state {q0,q1,q2}a -> {q0,q1,q2}, Can anyone explain that why ?
self looping
Supr mam ....understood clearly.
Thnx a lot mam..it's really helpful
Where are the trap sttaes ma'am
Mam plz Focus the paper properly...it keeps getting blurry
Mam how to find the union i we have q1,q2
Thank you mam ☺️ clearly explained
Amazing 🎉
Thank you 👏
Thankyou so much ra : )
Nice explanation mam 👏🏻
Thanks a lot mam it saved me
isnt one of the rules for a dfa that each state must have a transition for every symbol? B has no transition for a. C has no transition for either a or b. This wouldn't be a DFA.
The best explanation, period.
Usumularsey in the background 🔥
Guys in place of phi put (qt) and keep it as dead state
There should be square brackets not curly braces in DFA table
So there is no need to mark phi as qtrap in dfa?
you have not make a final sign in A B C