pushdown automata example | Part-2/2 | TOC | Lec-83 | Bhanu Priya

pushdown automata examples in english

[@navneetkumarmishra8411]

single a is also possible since m can be 1 and n can be = m...please look after this maam

[@liamjames4742]

I noticed it too.

[@infoiswealth1992]

Yes condition shoud be n>m

[@gwynethkay9341]

i noticed that too

[@navyareddy1349]

Yes

[@pavangamer663]

Yes

[@ocmannazirbriet]

"ab" is also part of language cuz incase of n=m and m=1 then "ab" will be also part of the language

[@novicecoder5753]

Mam you said that even if one element is left in stack that string is not accepted but how did you accept that 🙄🙄

[@ChallengeAll]

That example was because the number of a's is equal to the number of b's in the strings, but this example is different. It must leave an extra a or a's in the stack

[@siddhanttotade5702]

As U said , there must be a 'a' left in the stack then the string should be accepted. but here in this PDA example , the string that contain odd no. of "a's" and even no. "b's" can accepted not the string which has even no. of "a's" and even no. of "b's". as the question says , n>=m , m>=1 ...so if we take m=2 then n=2, which is even "a's" and even "b's" ....so after popping there is no 'a' left in the stack.

[@aabhastripathi1760]

we can not take n=2 and m=2 as it is given n>m n!=m

[@vrishnishreevb2003]

As 'a' is remaining in the stack, how did u make it acceptable then?

[@tech_and_memes]

Mam it would be great if you could address the doubts of the students in comments 🙏😁

[@dondlabhanuprakash1458]

n>=m can represents equal no of a's and b's but why you cant take this in language. eg: L={ ab,aabb,aaabbb,......}

[@CSEJoshi]

i think mam may be wrote wrongly acc to this sol the condition should be n>m then only it's gonna be correct

[@d.rudrama165]

Yes bro n=m=1

[@5d4pranith41]

If a is present in the stack then how can you say that there are no elements to read ( writing Epsilon empty ) Actually from q1 to q2 it should be b(E),a/E..........

[@lazyonigiri5665]

[@srb5704]

Why we not take € in previous video to push the z⁰ .?

[@shreedhar..4970]

If string is not completely read before that stack is empty then unacceptable, If string is completely read string is acceptable even stack is not null in this case state will changed because input is null but top of the stack is a. Little tricky but try to understand 😊

[@ujjwal5332]

It is easy to understand whether string going to be accepted or not....

[@Brahmanand_129]

Mam ab also possible because n>=m and m>=1 as we took m=1 then n may also be 1 so ab also possible...

[@tejashwakhare2561]

Mam b, a main E ki jagah a ayega cause E is the condition where we don't have to pop or push like you did before final stage !!! It should be b , a ---> a for (poping of a ) or b , a / a check into it !!

[@someshkumargupta256]

aapka condition n >= m and or aap n > m le rhi ....????

[@jkavitha1527]

The condition should be for this question is n>m,m>=1.

[@sharvesh0369]

why is there difference in stack variables in first part and this second part?

[@shreedhar..4970]

If string is not completely read before that stack is empty then unacceptable, If string is completely read string is acceptable even stack is not null in this case state will changed because input is null but top of the stack is a. Little tricky but try to understand 😊

[@baladimirputin708]

in the question take n>m in place of n>=m don't drop any comment regarding this

[@someshkumargupta256]

aapka condition a >= m hai or aap a > m le rhi mam ???

[@tarunsingh5615]

Sahi bol raha hai bhai €, Z0/€ BHI chahiye q1----->q2 me

[@jonathanjundarino7784]

I think the b,a|ε is doesn't mean to pop a? it means extra b? It should be a,Z0|aZ0

[@RelaxingMusic-mo3px]

The condition is not only greater than m, it can be equal. How can we add this condition?

[@VaibhaviKamble-oq3yv]

Plz clear the concept...doubt raises that string is accepted or not in stack....and also try to good pronounce and not fumble

[@himanshuchauhan4440]

Ye galat ha..... Lets say n=10 and m=1 ... It satisfies the condition n>=m but once b is popped we'll be left with 9 more a's and u only popped a single a at last... Taht makes 8 a's left ????

[@CSEJoshi]

then acc to me we should take epsilon,a/epsilon loop in final state..its not wrong

[@giduturisnehalatha8331]

exactly...same doubt...quiet controversal

[@nitishnegi2449]

Consider it like a^2n b^n

[@ShubhamKumar-pv8pj]

It is n>m

[@cst030santhoshm4]

Can u say explain the production rules

[@vishalshankarmishra9984]

If stack has more than one a then what?

[@CSEJoshi]

we should take loop on final state

[@saumyamishra5060]

Aaien a bcha h or accepted hain ..nyc. 😅

[@shreedhar..4970]

If string is not completely read before that stack is empty then unacceptable, If string is completely read string is acceptable even stack is not null in this case state will changed because input is null but top of the stack is a. Little tricky but try to understand 😊

[@sohelshaikh2697]

mam this language also take ab

[@itschurya4564]

Best vedio, thankyou so much mam..no one could explain like you do

[@infofacts4332]

hmm

[@wellingtoncunha852]

muito bom. Recife-PE Brasil

[@akshithchandra6177]

there is a flaw in this

[@tinashekadiki8885]

at least 1b and at least 1a as well

[@rockerboy4184]

Don't watch this video