one ofmy university teacher was watching the TOC lectures by NESO academy. she could solve the questions butfailed to explain us the same topics she studied a while ago. I mean,the conditions of a few college teachers is really bad, and pathetic. thank you Neso academy for saving the lives of thousands of students. Mad respect for the neso academy teachers!!
I appreciate the simplicity and explanation of small things, I only wish the institute which I pay dearly for would try such an approach instead of just being smug and saying how smart they are while not teaching us.
sir im not even kidding i will bow down to your feet and do a namskar with my head. you are a legendar teacher sir. i had failed in this TOC twice sir, everytime i asked i profesor for notes or how to study this subject they always told that study from the reference book but you know that the reference book as scary as a real ghost. i think i found neso acamedy a little late but anyways finaly i found it and im understanding concepts crystal clear. shat shat naman hai guruji aapko.
Thanku for making this amazing videos on automata. It's help me a lot. Initially I have little bit confused from where it can study automata I have even buy so much books but nothing get help but as I have seen this tutorial it can get help me a lot thanks 🙏🙏
5:200 In the last video you mentioned that for completion of DFA for each state we have to give some input... Suppose here in this example is it necessary to give= A takes input ->0,1 & goes to A, same for state B, then & only then it's complete DFA?
You cannot self loop at C because it would be contradictory. C is the final state but looping again means you add another symbol, which would exceed the required length of 2. When you exceed that required length it means it would be invalid, which would not be possible as C is the final state. (Final state cannot be invalid).
If it C has a self loop an input of 3 alphabets will be accepted because the third input would end at C and C is an accept state. He extended the diagram to have a 4th state D so as that the third input can end in a dead state(D), understood?
If we self loop state C the string length becomes 3 and it also accepts the string of length 3 as C is the final state.which is contrary to the given question.
the purpose is to eliminate any strings above two to dead state so a self loop on c would not send it to dead state but to final state meaning the you want any 10 combination of infinite length
To NESO Academy Why can't the self loop be implemented in state C itself rather extending it to state D and then implementing a self loop and making it a length of 3. If self loop is implemented in state C itself it stays in state C wether or not its '0' or '1'. As done in the previous lecture DFA (Example 1)
if we apply self loop on the state C only it will accept strings of longer lengths like(00000011111) which we dont want to do, we only want to make accept strings having length 2
because we are to construct a dfa that accepts sets of all strings over the length of "2" making c loop means sets larger than 2 would be accepted because they will then be in the final state
Sir,forgive me but in DFA Rule 1:For each state in the DFA, there must be exactly one transition defined for each symbol in the alphabet So:How 0,1 going to same state ? There is an undeterministic situation here i think ? However thanks for your clear explanation.
> For each state in the DFA, there must be exactly one transition defined for each symbol in the alphabet There was exactly one transition defined for each symbol. There was a transition defined for symbol 0, it goes from A to B. Then there was a separate transition defined for symbol 1, it goes from A to B. Transitions for both symbols go to the same state. He could've used two arrows, one on top going from A to B and labeled 0 and another one at the bottom also going from A to B and labeled 1. Instead he reused the same arrow and simply labeled it 0,1 to mark that they're in fact two separate transitions.
Please help me.. I have doubt is DFA contain only one input where NFA contain two or more that's a difference between them.. Then how is this they use two inputs?
Yes, if dead state isn't present your machine would simply truncate the remaining inputs . For eg: if the input is 000 or 001( these are strings of length 3 ,and hence should be rejected by the machine, but if the dfa doesn't have a dead end every input would simply stay in the final state at the end thereby accepting all string values and our designed dfa will be wrong ) . I just hope you understand whatever I tried to explain.
wonderful explanation. Thank you so much. But i am having one doubt that why you have not put self loop to state C. In first example of DFA construction as the state ended at B you given self loop then why not in C in example 2
Because state C is a final state and a string larger than 2 should not be accepted. A self loop in C would mean that strings larger than 2 would become accepted by the automaton.
Hi I know this is late but for anyone else wondering. If you put a self loop on state C the string 001 would end on state C, which is the final state. This means it would be accepted. We need to ensure an invalid string ends in a dead state so that it is not accepted.
yeah definitely there is a problem cuz if u keep a self loop for C then it means that it accepts strings of length greater than 2 also, so to avoid this we'll send the input to a dead state
Bro this is right because if we take loop over 'C' then problem at time 9:00 L=001 will accept the string and became final state on 'C' but lenght of is two so this is not possible to loop over 'C' in this question...
I thought a finite automaton is only deterministic when the transition from one state and another contains only one symbol... you went from A to B with both 0 and 1 on the same transition. This makes this NOT A DFA...
it is not '0 and 1' it is '0or 1'. see carefully , you are taking any one input(either 0 or 1) at a time. since the input at a time is deterministic it is a DFA
I dint understand one thing !! Is it compulsory to mention Dead state for all DFA problems ... my teacher he dint gave dead state for some problems ... is it optional ..or its mistake if we dont mention it ..
What if i take only two states A and B performing self loop on A and on B also giving both of them 0,1 on the self loop..... please make me understand...quickly
These video lectures is great when you really don't know where to start with.
Explanation of the guy is amazing.
one ofmy university teacher was watching the TOC lectures by NESO academy. she could solve the questions butfailed to explain us the same topics she studied a while ago. I mean,the conditions of a few college teachers is really bad, and pathetic.
thank you Neso academy for saving the lives of thousands of students. Mad respect for the neso academy teachers!!
wouldn't fly in america
I appreciate the simplicity and explanation of small things, I only wish the institute which I pay dearly for would try such an approach instead of just being smug and saying how smart they are while not teaching us.
lol
Literally commented the words in the minds of everyone who came to Neso Academy YT channel as a last ray of hope.
I can't compare this class with my teacher...Because they teaching the whole thing in a pure disaster manner..
Your class is precious to me..❤
Finally I have started to understand automata.. Thank u sir
Thank you so much. I have exam tomorrow and i skipped lectures. You are amazing ❤
So how was the exam?
Why these videos keep explaining so amazing while our teacher is teaching like sh*t?? 😃😃😍
Asking myself the same question.
Here we are going to listen before exam in class we listen with a sleepy mood
I am surprised
Exactly
Because you are paying to your teacher.
How can people dislike this video?!
Those are college teachers, lol
@@gamar1226 lmao 🤣
Even though I have not entered the class for this I now hv a good knowledge of this.
Very good explanation, I really appreciate that :)
Thanks for making this videos :)
well, Thanks so much sir for your efforts .. simple, direct and easy to deliver the idea of the question and how to solve it..
Really helpful man, been to so many lectures and can never understand but your videos are much more helpful in understanding. Thanks
Prime minister Modi is going to call you out
Keep up the good work.
10000000 times better than my course teacher.
Your way of explanation is so good sir
Fantastic, very useful. Thank you so much.
I have a mid-exam in the afternoon. Your teaching style is a piece of cake.
sir im not even kidding i will bow down to your feet and do a namskar with my head. you are a legendar teacher sir. i had failed in this TOC twice sir, everytime i asked i profesor for notes or how to study this subject they always told that study from the reference book but you know that the reference book as scary as a real ghost. i think i found neso acamedy a little late but anyways finaly i found it and im understanding concepts crystal clear.
shat shat naman hai guruji aapko.
loved how you explained it in such simple way
easy to take in
Thanku for making this amazing videos on automata. It's help me a lot. Initially I have little bit confused from where it can study automata I have even buy so much books but nothing get help but as I have seen this tutorial it can get help me a lot thanks 🙏🙏
5:200 In the last video you mentioned that for completion of DFA for each state we have to give some input...
Suppose here in this example is it necessary to give= A takes input ->0,1 & goes to A, same for state B, then & only then it's complete DFA?
you're a great teacher, thank you so much for this !
thank you so much completely upset how I found your channel after my exam --;
I freaken love you and this channel for this content!
Thank you sir. You are teaching us in very nice way. Thank you so much.
Awesome videos, loving the playlist
I skipped the wix ad just at the moment when he told wait before skipping ad . 😂😂😂😂😂
May you live a long prosperous life!
finally understood automata ,thank you very much🙇♀🙇♀🙇♀🙇♀
But why we can't take a self looo on state C, as we complete the length 2 of the string?
You cannot self loop at C because it would be contradictory. C is the final state but looping again means you add another symbol, which would exceed the required length of 2. When you exceed that required length it means it would be invalid, which would not be possible as C is the final state. (Final state cannot be invalid).
very impressive,explanation is so clear this videos are so useful at exam point of view thanks !!
sir love you and appreciate all your efforts,,,best wishes from bangladesh
Wow you really helped me understand it better> Thank you very much.
Sir your teaching is really good
1000x better than my lecturer
Thanks a lot, man! Great examples and great explanation!
thankyou so much please complete the COMPILER DESIGN COA CN
tutorial and DBMS as soon as possible
At the first example why didn't we loop 0,1 at the final state C instead of going to the dead state
Can we put a self loop for b containg 0,1 and mention it as final state?
Can we?
Have you found out?
@@harshkumar7857 .
You are a great teacher ✌️
Why we've drawn D ? If we let this after C, it will also satisfy question?
Thanks bro I am searching for this ....
❤Well explained.. Thank you very much..😊🔥🔥🔥👍
i have a doubt in 5:22 ..why we extended from final state c to d ? can we not just do self loop in final state itself?can someone expalain?
That would make the length of final string 3
Sir also provide practice questions with solutions or just answer if possible. Thank you so much 🖋️📖📚📚📚
really, fantastic and dynamic videos.very helpful
I have a doubt sir is it necessary to have 3 states or we can use self loop for B state
Ya same doubt
Bro in question they have mentioned length as 2. So it is necessary to have 3 states
If you give self loop on b so it create 1 length as well as 2 length.
We we want exactly 2 length string.
I have one question sir...why 0,1 is taken as input after reaching state B.is it depends on the length of the string?
Me too😔
Really good explanation
Excellent explanation 👌💐
so if the question wasn't limited to the length of 2, we would have made "A" self loop with 0,1 right ?
Sir please tell how to know how many states we will get? How to check if the drawn diagram is correct?
To know whether the diagram is correct or not only he is checking with the string at the end
why can't "c" having self loop sir. pls clarify my doubt sir pls🙏
If it C has a self loop an input of 3 alphabets will be accepted because the third input would end at C and C is an accept state. He extended the diagram to have a 4th state D so as that the third input can end in a dead state(D), understood?
@@geolegacy2 yep dude
@@geolegacy2 Thank You
@Sunaan S we can't make a self loop on B. Beacause in case of third input or many next inputs, it will accept which is not possible
It also violates the rule of having length 2, if C contains self loop.
Thanks for the effort
Good Explantion..Sir...Thanks..
during your last example there is string of length 1 which is not a final state so tell me why it isnt a dead state.
Thank u man this was so helpful
A terminal symbol is needed, perhaps Sigma = {0,1,T}
A man jumps off a 5 story building,
as he passes the 3 floor (final state) he says so far so good
Why would it also be fine if on the example 5:00 there is no dead/trap state and just loots back to c?
when a string can reach the final state, that string is accepted
awsome way of teaching.........
Sir you are amazing!..plzz make videos on python,compiler design,design and analysis of algorithm sir✨✨
Why can't we give s self loop to state C and finish it there. Whats the point of making another state D? at 5:10
If we self loop state C the string length becomes 3 and it also accepts the string of length 3 as C is the final state.which is contrary to the given question.
Explaination is realy good....
really clear examples
Rather pay my tuition to this channel
What if we eliminate D and put the loop at State C?
the purpose is to eliminate any strings above two to dead state so a self loop on c would not send it to dead state but to final state meaning the you want any 10 combination of infinite length
Love from karnataka❤❤❤❤❤❤
why can't we put A to B (0,1) then from B to B self loop(0,1)?
did you learn the answer??
length 2 is mentioned in question so that's why we taken
You are life saver DUDe
thank u sir!
But is your solution the only right answer? or there are other possible DFA for this same questions.
To NESO Academy
Why can't the self loop be implemented in state C itself rather extending it to state D and then implementing a self loop and making it a length of 3.
If self loop is implemented in state C itself it stays in state C wether or not its '0' or '1'. As done in the previous lecture DFA (Example 1)
if we apply self loop on the state C only it will accept strings of longer lengths like(00000011111) which we dont want to do, we only want to make accept strings having length 2
I have exams tomorrow ,I was like dying then YOU.......... JUST SAVED MY LIFE......
nice explaination
Excellent 👍
At 5:00 , why don't u apply the self loop on C itself ?
because we are to construct a dfa that accepts sets of all strings over the length of "2" making c loop means sets larger than 2 would be accepted because they will then be in the final state
Sir,forgive me but in DFA
Rule 1:For each state in the DFA, there must be
exactly one transition defined for each symbol
in the alphabet
So:How 0,1 going to same state ? There is an undeterministic situation here i think ?
However thanks for your clear explanation.
> For each state in the DFA, there must be exactly one transition defined for each symbol in the alphabet
There was exactly one transition defined for each symbol. There was a transition defined for symbol 0, it goes from A to B. Then there was a separate transition defined for symbol 1, it goes from A to B. Transitions for both symbols go to the same state. He could've used two arrows, one on top going from A to B and labeled 0 and another one at the bottom also going from A to B and labeled 1. Instead he reused the same arrow and simply labeled it 0,1 to mark that they're in fact two separate transitions.
Please help me.. I have doubt is DFA contain only one input where NFA contain two or more that's a difference between them.. Then how is this they use two inputs?
It was very much useful
This is too easy... I'm suspicious...
1. why can the length of input be 3 and 1? 2. Can we define that after C get input, it goes to itself?
Is it necessary to draw dead state to every DFA?
Yes, if dead state isn't present your machine would simply truncate the remaining inputs . For eg: if the input is 000 or 001( these are strings of length 3 ,and hence should be rejected by the machine, but if the dfa doesn't have a dead end every input would simply stay in the final state at the end thereby accepting all string values and our designed dfa will be wrong ) . I just hope you understand whatever I tried to explain.
wonderful explanation. Thank you so much. But i am having one doubt that why you have not put self loop to state C. In first example of DFA construction as the state ended at B you given self loop then why not in C in example 2
Because state C is a final state and a string larger than 2 should not be accepted. A self loop in C would mean that strings larger than 2 would become accepted by the automaton.
Thank-you so much sir
is there any way to run video on higher than 2X.
why you have not put self loop in state c
please reply asap tomorrow is my exam
Hi I know this is late but for anyone else wondering. If you put a self loop on state C the string 001 would end on state C, which is the final state. This means it would be accepted. We need to ensure an invalid string ends in a dead state so that it is not accepted.
@@luke8983 Thank you so much
@Luke Nolan but our question was to accept strings of length then how it is possible to accept a string of length 3 or more
I would get the reply after 3 months😂😂
sir,is there any problem if i keep self loop to C which stores 0,1 in itself without sending to D
yeah definitely there is a problem cuz if u keep a self loop for C then it means that it accepts strings of length greater than 2 also, so to avoid this we'll send the input to a dead state
i just got the situation, thank you very much sir
Bro this is right because if we take loop over 'C' then problem at time 9:00 L=001 will accept the string and became final state on 'C' but lenght of is two so this is not possible to loop over 'C' in this question...
@@ckumar2028 perfect bro
@@SaiKumar-zw9eh tysm
Sir, here we can also put a self loop to state b and make that only as final state right?
Thank you so much sir
You are awesome thank you very much for explaining so easily, looking forward for other videos
I thought a finite automaton is only deterministic when the transition from one state and another contains only one symbol... you went from A to B with both 0 and 1 on the same transition. This makes this NOT A DFA...
it is not '0 and 1' it is '0or 1'. see carefully , you are taking any one input(either 0 or 1) at a time. since the input at a time is deterministic it is a DFA
Nice lecture
I dint understand one thing !! Is it compulsory to mention Dead state for all DFA problems ... my teacher he dint gave dead state for some problems ... is it optional ..or its mistake if we dont mention it ..
What if I give a self loop to B??
If you give a self loop to B you are allowing inputs greater than length 2 so we need 2 states A and B and a final state C
Then it occurred 1 length
@@aadarshmishra1488 i agree that we should go back to state b with a lenght of one on any input at the final state
@@bmwmaniac3 c is not final state
My teacher 😎
Why don't we create a loop for B for second conditions and make it as final state and create a dead state at c.
Since we can reduce one element.
Is it necessary to write dead state?
why he does not stop at the final state? I think this is enough as a solution to the issue
Yes I have also doubt in this can anyone please tell me@ Neso academy
What if i take only two states A and B performing self loop on A and on B also giving both of them 0,1 on the self loop..... please make me understand...quickly
If you will give self loop in state A then it can accept more than 2symbol which is not a valid input for this language. Cross check me if I am wrong.
@@rakhimahanta3357 question is also same bro that accepts only sings of length 2
good lecture
why aren't we adding a self loop in state C ?
why we make state D we can again come back to state C for any input 0 or 1
sir in these example if we give no input then also it goes to an different trap state
Thank you ❤
Do we have to mention each time which state we've chosen dead state?
No
hiii
hello
how r u?