I think for this PDA, q1 must be a final state, as language also accepts that strings for n=0. And also to accept a string, both conditions must be true: 1. Final state must be reached. 2. Stack must be empty. If any of these conditions come false, then string will not be accepted. Apart this, @NesoAcademy, you are providing a great content. Thanks a lot.
@@lifeofsreeh no, because the initial state is reached only by reading the empty string. No way every string will be accepted if you make q1 a final state
it's crazy, how you could explain that such brief and knowble, you save my computin theory example, thank you (its quiet annoying to read through teachers power point and got nothing)
In the PDA , transaction from q2 and q3 should be E,E-->E(E= empty transaction) because it should accept the case where n=0and the given PDA will work when n>=1
I been wanting to say this, the people behind Neso thanks alot for providing such quality content. I often dont listen to class so its because of you guys I have able to pass subjects like EEE and digital system last year and I still use this to learn for my classes even now. Much appreciated :D
but this will also be true for 000011 as he said one of the condition should satisfy but i feel both should satisfy reaching Z and final states what say
Such a great line that two conditions for acceptance of any string in pushdown automata is : First: reaching the final state Second:stack should be empty
hi there, thanks a ton on the amazing lecturee, they saved my life. Just a little feedback on something than can make it easier for the viewers. If you number the lectures for each topic, it would be easier to track which video follows from which as youtube doesn't always bring them in sequence.
@@brahamaggarwal1800 - They're not in the correct order. The "Context-Free Grammar" video references "previous videos" about PDAs when defining what a Context-Free Grammar is. But in the playlist those are #65-84, where this discussion of PDAs starts at #85. ... It is a bit circular anyway, as the Intro PDA video describes them as a way to describe Context-Free Grammar/Language.
For whom might be confused! PDA accepts 1. The final state 2. When the stack is EMPTY. Hence, this is the accurate PDA recognize just epsilon where n=0. 🙂
a small piece of advice, can you also add the title with the number of the course like the main picture of the video. then it would be much easier for me to know which tutor I am at now. thanks!
What an excellent video. Incidentally, the diagram accounts for n> 0. In order to account for n = 0, do we need an epsilon transition from q2 to q3? -- Best regards and thank you again!
You are doing the work of God. Whatever God is or however many Gods there are ... you are doing the work of the good. Thank you kindly, sir, from a dumb American.
would it be correct to adjust the content of the last transition function to be 1,z0, e? ( in case I got the naming wrong, I'm refering to the writing on the arrow from q3 to q4). I tried to redraw the automata from my understanding of the topic thus far and I noticed the discrepency between my drawing and the professor's. My justification is that my design is intended to transition to the final state when we reach the final input in the stack when the professor's design transitions after we've reached the final input and there are no more inputs. Is this justification correct?
What will be happened in case of null string ? As the condition is also true for n=0 then the state q1 should also be accepted if i don't make mistake.
Then the first transition will accept no input and push z_0 to the stack. To make any progress any other transition will need to accept no input, otherwise the empty string is sure to be rejected.
I'm sorry, but in the example of the language L = { 0^n 1^n : n >= 0 } I believe the first state (q1) should be an accepting state. This is because of the basic case where the input string is an empty word, represented as an epsilon (ε). This can be seen as 1 and 0 raised to the power of 0, resulting in an epsilon (ε) empty word. Therefore, I think the first state should be an accepting state.
The example is NOT correct because 'n' is greater or equal to zero. Meaning if n = 0, the string will be empty. In this case, you cannot reach the accepting state (final state) if the string is empty. Correct me if I'm wrong.
@Nesco Academy I made only two states both are accept states state1 will have loop transition 0,€,0 and transition going state 2 which is 1,0,€ and state 2 will alse have loop transition 1,0,€ is it correct or not
but this will also be true for 000011 as he said one of the condition should satisfy but i feel both should satisfy reaching Z and final states what say
I made only two states both are accept states state1 will have loop transition 0,€,0 and transition going state 2 which is 1,0,€ and state 2 will alse have loop transition 1,0,€ is it correct or not
Thanks a lot for the video. What happens when the string "1" is the input? At state q2, it will read 1 but the stack does not have a 0. Will this get rejected?
Believe me, with the very same language, your explanation is a way better than most people in here. Thank you!
I think for this PDA, q1 must be a final state, as language also accepts that strings for n=0.
And also to accept a string, both conditions must be true:
1. Final state must be reached.
2. Stack must be empty.
If any of these conditions come false, then string will not be accepted.
Apart this, @NesoAcademy, you are providing a great content.
Thanks a lot.
No , that case every string will be accepted
Edit :
The right way is to add a transition from q1 to q4
It's a NFA so what's why there isn't that transition.
there are two types of acceptance by pda.
1. acceptance by final state
2. acceptance by empty stack
correct but I think the mistake is in the question, where the condition should be n > 0.
@@lifeofsreeh no, because the initial state is reached only by reading the empty string. No way every string will be accepted if you make q1 a final state
You saved my life, I just wanted you to know !
legend
ye pyaar nahi toh aur kya hai
@@sagesy9774 I do not understand your language
@@Jerdz nothing it was a silly joke
@@Jerdz he meant "if this isn't love then what else can it be"
Such a great lesson. Clear, patient, concise and to the point. Good work you guys!
Netflix: Ughh.
Neso: Aah
Real heroes don't wear capes
1:07 Meaning of a, b -> c
3:24 Example
LEGENDARY Explanation...💯
SEHR HILFREICH UND TOLL VIDEO. DU BIST WUNDERBAR MENSCHEN!!!!
Best teaching channel
This guy is a hero.
these videos never gets old😅😉
Do u have any exam tomorrow
it's crazy, how you could explain that such brief and knowble, you save my computin theory example, thank you
(its quiet annoying to read through teachers power point and got nothing)
In the PDA , transaction from q2 and q3 should be E,E-->E(E= empty transaction) because it should accept the case where n=0and the given PDA will work when n>=1
where n = 0 the stack will be empty so it will also be accepted
Yeah much people coment that
omg you cleared PDA concept in one shot amazing man
Always the best ,keep up sir!
dude your name took over my screen i thought i got fault in my monitor
I been wanting to say this, the people behind Neso thanks alot for providing such quality content. I often dont listen to class so its because of you guys I have able to pass subjects like EEE and digital system last year and I still use this to learn for my classes even now. Much appreciated :D
I swear you're such a life saver! Amazing job
Great job on explaining this. Very explicit and clear for understanding.
Awesome Videos, He explains everything slowly and clearly.
i wish i see your photo one day, what a nice guy
After watching this video for over 15 times consecutively...I have understood PDA succesfully...:)
U are a lifesaver man!
Much Love
best video on the topic out there
Perhaps I misread it, but it seems like this PDA makes {0^n1^n | n>0}, not {0^n1^n | n>=0}
I think you're right
his work is correct if he put accept state on "q1"
yes.
but this will also be true for 000011 as he said one of the condition should satisfy but i feel both should satisfy reaching Z and final states what say
Yeah u r right
As always nailed it.
a real hero just that...
Have my university exam tomorrow
Learnt a lot from your lectures
Thank you
this is just exactly what I needed, thank you so much
Such a great line that two conditions for acceptance of any string in pushdown automata is :
First: reaching the final state
Second:stack should be empty
Hope every university can hire such a teacher to give lectures so that his students don't go to youtube university🙂
thanks for clearing confusion
You saved me..!!!! Thank you so much for making such a great content..❤️👍
E, E -> Zo
(E: input symbol, E: to be popped -> Zo: to be pushed)
Good explanation sir
Thank you so much for your videos!!! You are wonderful at delivering this material!!
ustad kamalll😍🥰🤩
hi there, thanks a ton on the amazing lecturee, they saved my life. Just a little feedback on something than can make it easier for the viewers. If you number the lectures for each topic, it would be easier to track which video follows from which as youtube doesn't always bring them in sequence.
you can always watch the whole playlist..... you will get all 112 videos in sequence.
@@brahamaggarwal1800 - They're not in the correct order. The "Context-Free Grammar" video references "previous videos" about PDAs when defining what a Context-Free Grammar is. But in the playlist those are #65-84, where this discussion of PDAs starts at #85. ... It is a bit circular anyway, as the Intro PDA video describes them as a way to describe Context-Free Grammar/Language.
Wonderful explanation!
Q1 should be a final state as well for the n=0 (case ε)
Wolff or replace the transition from q2 to q3 with e, e->e.
@@lukaspovilonis210 Then 00111 will also be accepted na ?
Awesome explanation I m lovin it
McFlurry Sir
@@ricaspinto 😭🤣
For whom might be confused!
PDA accepts
1. The final state
2. When the stack is EMPTY.
Hence, this is the accurate PDA recognize just epsilon where n=0.
🙂
transaction from q2 and q3 should be E,E-->E(E= empty transaction) because it should accept the case where n=0
7:09 - can q2 also be the initial state?
super explanations sir i like very much ....keep going on
Thank you so much, great teacher.
Thank you 🙏
Wonderful explanation sir jiiiii
thank you, good lesson.
sending soooooooooooo much love your way man your videos are awesome :)
Finally PDA Concept is clear
Thank you
i can not afford to buy paid your course. i am in 3rd sem now btech. pleaseeeeee dont remove these lectures. its is extremely helpfull
This was very helpful!
I love your videos. I am subscribing.
Thanks a lot, very helpful tutorial!
a small piece of advice, can you also add the title with the number of the course like the main picture of the video. then it would be much easier for me to know which tutor I am at now. thanks!
Perfectoooo.I love you guys :)
What an excellent video. Incidentally, the diagram accounts for n> 0. In order to account for n = 0, do we need an epsilon transition from q2 to q3? -- Best regards and thank you again!
You are doing the work of God. Whatever God is or however many Gods there are ... you are doing the work of the good. Thank you kindly, sir, from a dumb American.
🥰🥰🥰
that's too much of a compliment, God gave him this knowledge
Amazingly helpful
sir in this lecture how to accept empty string? because n>=0 so, it must be accepted. kindly explain in above PDA.
I comment on this video in 2024..my college professors use only your's video and ppt to teach us
really useful content thank you very much
Nice explanation
sir you are super
yes
would it be correct to adjust the content of the last transition function to be 1,z0, e? ( in case I got the naming wrong, I'm refering to the writing on the arrow from q3 to q4). I tried to redraw the automata from my understanding of the topic thus far and I noticed the discrepency between my drawing and the professor's. My justification is that my design is intended to transition to the final state when we reach the final input in the stack when the professor's design transitions after we've reached the final input and there are no more inputs. Is this justification correct?
your goatted 💯💯💯
What will be happened in case of null string ? As the condition is also true for n=0 then the state q1 should also be accepted if i don't make mistake.
Then the first transition will accept no input and push z_0 to the stack. To make any progress any other transition will need to accept no input, otherwise the empty string is sure to be rejected.
The base case for the diagram for 0^n 1^n is not right because it should be n >= 1. If it was n = 0, then it would take the empty string.
Superb
damn you are the first brown boi that i can listen to when it come to youtube learning. keep doing!
for q2 to q3 why don't we use ∑ , ∑ -> ∑ that way 00111 wouldn't be accepted?
I'm sorry, but in the example of the language L = { 0^n 1^n : n >= 0 } I believe the first state (q1) should be an accepting state. This is because of the basic case where the input string is an empty word, represented as an epsilon (ε). This can be seen as 1 and 0 raised to the power of 0, resulting in an epsilon (ε) empty word. Therefore, I think the first state should be an accepting state.
Yes you are right
thanks a lot
I don't understand what was the use of stack in this case? Why all the pushes and pops?
Thank you sir
Amazing
Thankyou sir
Great videos! ..but maybe shorten them? 4:30 - 6:30 a 2 min explanation of Zo.
Thanks a lot!
thanks
Awesome
The example is NOT correct because 'n' is greater or equal to zero. Meaning if n = 0, the string will be empty. In this case, you cannot reach the accepting state (final state) if the string is empty. Correct me if I'm wrong.
You have solved it for n>=1 as final state is at the end, but in the question n>=0 has been mentioned.
@Nesco Academy I made only two states both are accept states state1 will have loop transition 0,€,0 and transition going state 2 which is 1,0,€ and state 2 will alse have loop transition 1,0,€ is it correct or not
Why did we pushed down Os into stack but not 1 ?
Does the 1s given as inputs get read in the string even if they don't push and pop out of the stack??????.
if n=0 then epsilon in is L which means q1 also needs to be an accept state.
you set n equal 0, so epsilon should be accepted too, right? How does the automata also accept only the epsilon?
How about if n = 0 in this example?
What about when q3 reads 0, can we do 0,E->0 and move to q2 state? For example in 001011 while reading 4th input symbol
This grammar implies that the language is strictly n number of 0's followed by n number of 1's. There is no mixing of symbols.
but this will also be true for 000011 as he said one of the condition should satisfy but i feel both should satisfy reaching Z and final states what say
w = 000011 is not accepted by the PDA of the example. Why? After inserting the chain w, the stack contains 00z_0.
thx it helps
I made only two states both are accept states state1 will have loop transition 0,€,0 and transition going state 2 which is 1,0,€ and state 2 will alse have loop transition 1,0,€ is it correct or not
How null string will be satisfied in this example . Please explain
i dont understand the condition(n>=0) here what if n=0 pda not going to accept it
wht about if we get 1 first instead of 0 in q2
Thanks a lot for the video. What happens when the string "1" is the input? At state q2, it will read 1 but the stack does not have a 0. Will this get rejected?
yes! as the number of 0s is not equal to the number of 1s
sir won't this accept 00111 as well?