Thank you! I have my exams in a few days and was having a hard time wrapping my head around this because every other guide uses random unnecessary symbols and all. Thank you for making it so simple and to-the-point!
thank you very much for having subtitles enabled. i have a lot easier of a time processing speech when i have a visual aspect to go along with it, so the subtitles help a lot. the tutorial's very clear and to the point, and i'm able to use it to get ahead on my homework early. thank you.
In the transition table for DFA it seems you should not put the {...} because the image of the transition function is Q, not P(Q) for DFA. Doing that makes more intuitive why you need to create the combined state AB \in Q instead of using {A, B} \in P(Q) from the NFA.
It doesn't simply "know" when to move to the final state but rather when it faces a situation where it must chose to go to 2 different states, it branches out into 2 branches and moves forward in both directions and so on, but when one branch reaches Final state, the rest will be abandoned and it will be considered as "accepted". That's why in NFA it is said that "If there is any way to run the machine that ends in any set of states out of which at least one state is a final state, then the NFA 'accepts' ". Although this is the theoretical explanation behind it, it cannot be implemented as is without the use of some other ways and finally representing the same in DFA form.
Please take some complex example with states more than 4 and explain those so that we can clearly understand the concepts with such complex example since simple NFA are easier to convert to DFA than that of complex NFA
Like several commenters, I am confused as to why the DFA table features "AB" instead of "B?" Is this just a naming convention or is there a deeper principle here? In what way would it change the meaning of the DFA table if I used "B" instead of "AB?"
Suppose in NFA, X is the final state. Now, after converting to DFA, from state transition table, all state that has X in them will be final state in DFA. Eg: In DFA Transition table, states are: A, B, AX, CX, ACX Then AX, CX, ACX all will be final states.
A getting input 0 will stay in A and A getting input 1 goes to B. B getting input 0 gets back to A and B getting input 1 stays in B. Can't this be done for DFA.
AB on input 0 - AF AB on input 1 should give you ABF (AB and a failure state because has no other transitions on state B) Then AF on 0 - AF AF on 1 - ABF ABF on 0 - AF ABF on 1 - ABF However, I will admit that including the failure states may not be necessary and some professors do not include them. It comes down to preference and acceptable practices for your class.
sir If you loop on both "0" and "1" in q0, the automaton will always stay in q0 and never accept strings like "1", "01", etc., which should be accepted because they end in "1.".. So which one of is right ????
if something is supposed to end in 1 like the first example, it should accept something like "011111" as well as just "01" no? like the wording of these examples in almost all of these videos is so... wrong and vague at times that it makes me do it "incorrectly" based on his interpretation of the example. it would be much more clear if he said it accepts "the set of all strings over (0, 1) that ends in ONE '1'" but he doesn't say that, it just says ending in 1, which i interpret to mean any 0*1*, or any number of 0s and 1s assuming that it still ends in 1. also doesn't this NFA not work if it gets a string like '0101'? THAT ends in a 1, but it isn't accepted because when it gets to the 2nd 0, it goes to phi and then can't get back to the accept state. whereas the DFA WILL accept '0101'.
The reason why B is not in the DFA table is that there is no way B is connected to A or AB. Had B been connected to AB or A, it would be shown in the state table. Don't make the assumption that all states from NFA are to be shown in the DFA State transition Table and Diagram!!! That assumption of yours is wrong. I agree he should have explained it properly, but he didn't!
A bit late but for anyone new: in the NFA, B is the final state. We say that a state or set of states is final if ONE of the members is final. AB contains the B, hence it is final.
You are providing service to the students, which no one can provide.
Congratulations for 1M subscribers do 1m golden play button , whenever it comes,you deserve it.
It was very helpful. Your calm manner of your instructing is very effective.
Best explanation ever.. keep going, when I encounter any difficulty, I immediately go for your videos
Thank you! I have my exams in a few days and was having a hard time wrapping my head around this because every other guide uses random unnecessary symbols and all. Thank you for making it so simple and to-the-point!
@volt jax Yes and yes, are you someone I know? Hit me up on Telegram, @fushinari
My lectures and notes are a mind numbing mess of symbols. Thank you for explaining this in simple, regular language!
thank you very much for having subtitles enabled. i have a lot easier of a time processing speech when i have a visual aspect to go along with it, so the subtitles help a lot. the tutorial's very clear and to the point, and i'm able to use it to get ahead on my homework early. thank you.
In the transition table for DFA it seems you should not put the {...} because the image of the transition function is Q, not P(Q) for DFA. Doing that makes more intuitive why you need to create the combined state AB \in Q instead of using {A, B} \in P(Q) from the NFA.
This is the best lecture I've heard so far.Thank you so much sir🙏🙏🙏
I could not sleep for 2 days watched 5min of this video and fell asleep... Thanks for this... nice explaination.. Thnks for my sleep 💟
who needs ASMR
It´s correct. I have this exact example in my book. Point is NFA simply "know" when to move to the final state. Book however noted AB as B again.
It doesn't simply "know" when to move to the final state but rather when it faces a situation where it must chose to go to 2 different states, it branches out into 2 branches and moves forward in both directions and so on, but when one branch reaches Final state, the rest will be abandoned and it will be considered as "accepted".
That's why in NFA it is said that "If there is any way to run the machine that ends in any set of states out of which at least one state is a final state, then the NFA 'accepts' ".
Although this is the theoretical explanation behind it, it cannot be implemented as is without the use of some other ways and finally representing the same in DFA form.
Thank you, Neso Academy!
6:50 why does he say "ab will be my final state" twice instead of actually explaining why it is the final state?
After the transformation, every state that contains the initial final state is a final state I think.
How did you determine that AB is a final state?
For future Reference, In original NFA, B was final state, therfore any states with B in dfa is final state
5:37 please share that video of that union operation video.
Please take some complex example with states more than 4 and explain those so that we can clearly understand the concepts with such complex example since simple NFA are easier to convert to DFA than that of complex NFA
I have absent for this tutorial,but I have understood the concept.Tq for helping me,the best way of teaching tqsm🙏
same here mate
Sir I hope your the great teacher .. thank you sir ...
😍😍😍😍I am from Yemen
thank you for helping me 😍😍😍 thank you very much
15 Yemen Road ,Yemen??
these are killer tutorials :)
25 minutes before exam, thank you
6:57 how can you say that 'AB' is our final state.
Because in AB, we have the final state B so AB is final state ..
How is AB decided as final state?
As we can see B is the final state in NFA , so all the combinations with B in it , becomes the Final State in the DFA
bhai kitna easly samjhaya hai yaar
Like several commenters, I am confused as to why the DFA table features "AB" instead of "B?" Is this just a naming convention or is there a deeper principle here? In what way would it change the meaning of the DFA table if I used "B" instead of "AB?"
On input 1, it can either choose to stay at A or move on to B as shown in the diagram.
@@yoyo999-g4k it would have been help if u could mention it here
Happy Teachers day sir
but in the DFA we can take the state C for the null state .......?
need reply pls
Can there be many NFA diagrams for one question ?? If i take another type of NFA and I am getting different DFA answer ? Is it correct ??
Thank You sir
Why will we take AB as a final State in case of dfa?
Suppose in NFA, X is the final state.
Now, after converting to DFA, from state transition table, all state that has X in them will be final state in DFA.
Eg: In DFA Transition table, states are: A, B, AX, CX, ACX
Then AX, CX, ACX all will be final states.
@@supersakib62thanks for your valuable explanation ❤️
meeku naa hrudaya poorvaka namaskaralu guru gaaru
One day before the exam here. Thank you so much for saving us(students)
this guy is Superman
@@nahomaraya2377 definitely😭he is
how do we know that AB is the final state
there can be more states in the transition table
Why have u not taken dead configuration as dead state here as in previous example ?
Plz explain sir !
Thank you. Excellent presentation.
A getting input 0 will stay in A and A getting input 1 goes to B. B getting input 0 gets back to A and B getting input 1 stays in B. Can't this be done for DFA.
Sir why should we don't put 0,1 in the final stage
AB on input 0 - AF
AB on input 1 should give you ABF (AB and a failure state because has no other transitions on state B)
Then AF on 0 - AF
AF on 1 - ABF
ABF on 0 - AF
ABF on 1 - ABF
However, I will admit that including the failure states may not be necessary and some professors do not include them. It comes down to preference and acceptable practices for your class.
Love from Nepal sir
Why didn't we mention B transation state in the transaction table
beacuse state B is not reachable from previous states
All my college professors should be fired
My collage one should be first one
😂😂😂
This guy is a real democrat for creating so many educational videos for free
sir If you loop on both "0" and "1" in q0, the automaton will always stay in q0 and never accept strings like "1", "01", etc., which should be accepted because they end in "1.".. So which one of is right ????
sir, why ab is finish state?
Hello! So when we need to design a DFA do we always first design NFA and then convert it to DFA?
No, you can design DFA directly.
Thanks a lot sir...
Dislikers are the teachers who don't know how to teach
You hits the truth😂😂
in NFA B on getting input 1 can go to B?
Same Doubt
No, as for A we have considered that case, that is if any 0 or 1 comes before the last 1 it stays in state A itself.
@@prateeksengar6634 yeah, point
Thank you sir😊
sir I have a small doubt why was dead state not introduced here?
please upload each topic of engineering specially from RGPV, you are damn good, you can succeed more that that Khan academy
sirr plzzz make videos on design and analysis of algorithms plzzz
wait, if state B in NFA get's 0 again, then it goes back to A right? why is it empty state?
Syntactically, is {AB} the same as {A∪B}?
thank you
sir
thank u professor.
Brilliant video
why AB ? can't it be B ?
Thank you, sir!!
good lecture
What does AB state represents
if something is supposed to end in 1 like the first example, it should accept something like "011111" as well as just "01" no? like the wording of these examples in almost all of these videos is so... wrong and vague at times that it makes me do it "incorrectly" based on his interpretation of the example.
it would be much more clear if he said it accepts "the set of all strings over (0, 1) that ends in ONE '1'" but he doesn't say that, it just says ending in 1, which i interpret to mean any 0*1*, or any number of 0s and 1s assuming that it still ends in 1.
also doesn't this NFA not work if it gets a string like '0101'? THAT ends in a 1, but it isn't accepted because when it gets to the 2nd 0, it goes to phi and then can't get back to the accept state. whereas the DFA WILL accept '0101'.
what does phi Φ mean the context of union and concatenation? is it equivalent to the null set Ø in set theory?
Yes, it’s the null state which is an element of the superset of Sigma
The union of A and B from the NFA table for input 1 should have been {A,B} in the DFA table.... Instead, you wrote {AB}. But, why?
wont AB also be the initial state ?
in the NFA, doesnt B have to go to B on 1? it would still be an NFA wouldnt it?
If A has no self loop what can we do
I like how the videos are bite sized
Just chunks of 5-10 minute videos
thanks😭😭😭
I have a doubt I took three states for NFA..
(A)--1,0--(B)--1---((C)) (Here C is my final state )..
How to I convert it to DFA...!
if u cant directly design the DFA so first design NFA then convert it/ *SMORT*/
In the example in dfa can we write state AB as B
It could lead to confusion
Thankyou🙏🙏🙏🙏🙏
Fun fact my professor did a quiz to us and the question was same question in the video 😂😂
Thank you!
Very well explained. But repetitive examples again and again. It would be nicer if you have different problems.
sir,why can't we give 1 to b its also true know then why can't we do so pls reply sir
those cases will be covered since we also input 1 to A
Thankyou sir
Before these videos student's grades are F, after that videos all of A s.
Thanks man!
Big Thank U
which accepts string that ends with b length less than 2
why there is no dead trap in converted dfa?
Thanks
what happened to b?
you nailed it
Thank u sr
subset contruction method
Nice
❤❤❤
Good
In NFA when B get's an input of 1 doesn't it have to stay in B instead of going into the null state.
No, since B checks for the last character. There's nothing after the last character, so anything after that goes to death configuration.
great
how to determine a final state in converted dfa when we got new states
Every state which contains final state of NFA is stated as final state for dfa
مفهمتش مش عارفة ليش
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Please tell the acceptance of these NFA and DFA .. as you asked in assignment.. I can't find the answer
The reason why B is not in the DFA table is that there is no way B is connected to A or AB. Had B been connected to AB or A, it would be shown in the state table. Don't make the assumption that all states from NFA are to be shown in the DFA State transition Table and Diagram!!! That assumption of yours is wrong. I agree he should have explained it properly, but he didn't!
chill, he explained everything perfect.
I actually didn’t get why until @JOSEPH Blessingh said something. Thanks
nnice bro just for it
how AB will be final state?How do i know?
A bit late but for anyone new: in the NFA, B is the final state. We say that a state or set of states is final if ONE of the members is final. AB contains the B, hence it is final.
Can I have three State?Does it matter?
@@Broekmanium thanks🤜🤛
I wish I could see you earlier, I found you only on my final day! 😢