Pushdown Automata (Graphical Notation)

[@sarba85528]

Believe me, with the very same language, your explanation is a way better than most people in here. Thank you!

[@Jerdz]

You saved my life, I just wanted you to know !

[@angshumansarma2836]

legend

[@sagesy9774]

ye pyaar nahi toh aur kya hai

[@Jerdz]

@@sagesy9774 I do not understand your language

[@sagesy9774]

@@Jerdz nothing it was a silly joke

[@sam_682]

@@Jerdz he meant "if this isn't love then what else can it be"

[@10_yogeshchandrapandey90]

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 , that case every string will be accepted Edit : The right way is to add a transition from q1 to q4

[@DFULTCDS]

It's a NFA so what's why there isn't that transition.

[@nitigyajoshi4658]

there are two types of acceptance by pda. 1. acceptance by final state 2. acceptance by empty stack

[@hemantdewpal1612]

correct but I think the mistake is in the question, where the condition should be n > 0.

[@kriskurti7497]

@@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

[@justinmorgan7495]

Such a great lesson. Clear, patient, concise and to the point. Good work you guys!

[@atifayaz3495]

Netflix: Ughh. Neso: Aah

[@piyushborse3085]

LEGENDARY Explanation...💯

[@gzhekoff]

Real heroes don't wear capes

[@egedurusoy7029]

SEHR HILFREICH UND TOLL VIDEO. DU BIST WUNDERBAR MENSCHEN!!!!

[@SHEETALSHARMA-tz7sm]

1:07 Meaning of a, b -> c 3:24 Example

[@ARSAGAMING69]

Best teaching channel

[@Shaam_Ark]

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

[@Sandeep-eb5sf]

these videos never gets old😅😉

[@harshinigogula1863]

Do u have any exam tomorrow

[@台閣]

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)

[@gracestewart544]

This guy is a hero.

[@taranrishith]

Always the best ,keep up sir!

[@yourdailyfails1]

dude your name took over my screen i thought i got fault in my monitor

[@kyathineeharika5670]

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

[@utsavpathak2122]

where n = 0 the stack will be empty so it will also be accepted

[@jriveros3901]

Yeah much people coment that

[@dibbyendukarmakar8351]

After watching this video for over 15 times consecutively...I have understood PDA succesfully...:)

[@zhouhang5330]

Great job on explaining this. Very explicit and clear for understanding.

[@brayanvizcaino]

Awesome Videos, He explains everything slowly and clearly.

[@kettleghost3721]

I swear you're such a life saver! Amazing job

[@TrustTheFund]

Perhaps I misread it, but it seems like this PDA makes {0^n1^n | n>0}, not {0^n1^n | n>=0}

[@frostbite585]

I think you're right

[@OmranAlHadad]

his work is correct if he put accept state on "q1"

[@GrimstboritH]

yes.

[@vaibhavmourya65]

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

[@sahilsaini2764]

Yeah u r right

[@rj3937]

U are a lifesaver man! Much Love

[@jatinbhardwaj3185]

Have my university exam tomorrow Learnt a lot from your lectures Thank you

[@samriddhigulati3109]

best video on the topic out there

[@tafveezahmad9692]

omg you cleared PDA concept in one shot amazing man

[@ahmedi_zakid]

i wish i see your photo one day, what a nice guy

[@rabeyabasri7070]

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]

you can always watch the whole playlist..... you will get all 112 videos in sequence.

[@sjhuz01]

@@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.

[@bushraw66]

this is just exactly what I needed, thank you so much

[@sidharthrajagarwal3025]

As always nailed it.

[@vatstech8725]

You saved me..!!!! Thank you so much for making such a great content..❤️👍

[@joyf7112]

Thank you so much for your videos!!! You are wonderful at delivering this material!!

[@mohammadasif6439]

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

[@chavaligangadevi8663]

Good explanation sir

[@subtitleslyrics7057]

E, E -> Zo (E: input symbol, E: to be popped -> Zo: to be pushed)

[@plsiok9626]

a real hero just that...

[@huanchenlee3356]

Hope every university can hire such a teacher to give lectures so that his students don't go to youtube university🙂

[@jeongwookim4130]

Wonderful explanation!

[@nandhucharan5858]

super explanations sir i like very much ....keep going on

[@FahimKarim-b6r]

thanks for clearing confusion

[@Meiliful]

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!

[@ARSAGAMING69]

7:09 - can q2 also be the initial state?

[@KinGxWolF]

Q1 should be a final state as well for the n=0 (case ε)

[@lukaspovilonis210]

Wolff or replace the transition from q2 to q3 with e, e->e.

[@sayantaniguha8519]

@@lukaspovilonis210 Then 00111 will also be accepted na ?

[@huzaifakhan4657]

ustad kamalll😍🥰🤩

[@saketkumar4972]

transaction from q2 and q3 should be E,E-->E(E= empty transaction) because it should accept the case where n=0

[@myname1484]

Thank you 🙏

[@KElehOS]

Thank you so much, great teacher.

[@parisaghanad8042]

sending soooooooooooo much love your way man your videos are awesome :)

[@Shivam-eh5fc]

Awesome explanation I m lovin it

[@ricaspinto]

McFlurry Sir

[@DarkOceanShark]

@@ricaspinto 😭🤣

[@Zeddy27182]

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. 🙂

[@jff711]

Thanks a lot, very helpful tutorial!

[@mazharrazmian]

I don't understand what was the use of stack in this case? Why all the pushes and pops?

[@lanblast9423]

Wonderful explanation sir jiiiii

[@dogunboundhounds9649]

I love your videos. I am subscribing.

[@S_Mist01]

Thank you

[@MuhammadNadeem-rc5bk]

sir in this lecture how to accept empty string? because n>=0 so, it must be accepted. kindly explain in above PDA.

[@DrRizzwan]

thank you, good lesson.

[@HosRo4161]

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!

[@pramodkoushiktr1895]

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

[@6Pope9]

This was very helpful!

[@mortezamahdavi2129]

Perfectoooo.I love you guys :)

[@qwertymama7344]

for q2 to q3 why don't we use ∑ , ∑ -> ∑ that way 00111 wouldn't be accepted?

[@Armando_Gutierrez]

Amazingly helpful

[@AnasTahira]

Why did we pushed down Os into stack but not 1 ?

[@mitul9638]

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.

[@lucasdarianschwendlervieir3714]

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.

[@calmingly8044]

you set n equal 0, so epsilon should be accepted too, right? How does the automata also accept only the epsilon?

[@okee]

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.

[@david_m157]

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.

[@VivekSingh-in6rq]

really useful content thank you very much

[@tuonghocat2310]

How about if n = 0 in this example?

[@sanju_ch_8782]

Does the 1s given as inputs get read in the string even if they don't push and pop out of the stack??????.

[@mohamadtabbakh9887]

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?

[@azkymohamed123]

Thank you sir

[@ytassignit]

sir won't this accept 00111 as well?

[@somu-p9p]

i dont understand the condition(n>=0) here what if n=0 pda not going to accept it

[@eswararao6617]

Nice explanation

[@neelb8653]

wht about if we get 1 first instead of 0 in q2

[@abdirahman848]

@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

[@jronyjrony2340]

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.

[@sghqz]

🥰🥰🥰

[@henoktademe1491]

that's too much of a compliment, God gave him this knowledge

[@jkssbjobtutorial.5061]

sir you are super

[@ricaspinto]

yes

[@venkatdurgasai2142]

For 000111 also have same diagram or it will change ? Please anyone say

[@vex8521]

if the string is 01 then it will not reach the final state...but it should...then what to do?

[@natrutto]

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.

[@mr.unknown6179]

Yes you are right

[@abdirahman848]

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

[@ליאורדבורה]

if n=0 then epsilon in is L which means q1 also needs to be an accept state.

[@skellep]

Great videos! ..but maybe shorten them? 4:30 - 6:30 a 2 min explanation of Zo.

[@devratagarwal5021]

How null string will be satisfied in this example . Please explain

[@efseesfesfsefse2124]

your goatted 💯💯💯

[@a.screations8693]

How to select transition states

[@manaveanni6646]

Superb

[@aayushmaurya1672]

how a null string will be accepted by this generated PDA?

[@ashstories6489]

thanks a lot

[@dhanushsivajaya1356]

Thankyou sir

[@shahe1183]

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

[@frostbite585]

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.

[@anubhavsingh5863]

You have solved it for n>=1 as final state is at the end, but in the question n>=0 has been mentioned.

[@Chandankumar-qw6hb]

thanks

[@jonatangall3715]

Thanks a lot!

[@pawansmy1844]

Finally PDA Concept is clear

[@gauravbhandari1184]

Amazing

[@saima9943]

why didn't we push 1 into the stack? why only zero?

[@theflash34529]

Exam ke liye pdh rha

[@adirakun9164]

what if it was 1010 what will we do after reaching q3 for 0 input