"ab" is also part of language cuz incase of n=m and m=1 then "ab" will be also part of the language
@navneetkumarmishra8411
4 жыл бұрын
single a is also possible since m can be 1 and n can be = m...please look after this maam
@liamjames4742
3 жыл бұрын
I noticed it too.
@infoiswealth1992
3 жыл бұрын
Yes condition shoud be n>m
@gwynethkay9341
3 жыл бұрын
i noticed that too
@navyareddy1349
3 жыл бұрын
Yes
@pavangamer663
2 жыл бұрын
Yes
@novicecoder5753
2 жыл бұрын
Mam you said that even if one element is left in stack that string is not accepted but how did you accept that 🙄🙄
@ChallengeAll
3 ай бұрын
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
@tech_and_memes
2 жыл бұрын
Mam it would be great if you could address the doubts of the students in comments 🙏😁
@siddhanttotade5702
2 жыл бұрын
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
7 ай бұрын
we can not take n=2 and m=2 as it is given n>m n!=m
@vrishnishreevb2003
Жыл бұрын
As 'a' is remaining in the stack, how did u make it acceptable then?
@tejashwakhare2561
4 жыл бұрын
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 !!
@dondlabhanuprakash1458
4 жыл бұрын
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
3 жыл бұрын
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
3 жыл бұрын
Yes bro n=m=1
@Brahmanand_129
2 жыл бұрын
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...
@jkavitha1527
2 жыл бұрын
The condition should be for this question is n>m,m>=1.
@RelaxingMusic-mo3px
2 жыл бұрын
The condition is not only greater than m, it can be equal. How can we add this condition?
@nitishnegi2449
2 жыл бұрын
Consider it like a^2n b^n
@wellingtoncunha852
2 жыл бұрын
muito bom. Recife-PE Brasil
@someshkumargupta256
3 жыл бұрын
aapka condition n >= m and or aap n > m le rhi ....????
@cst030santhoshm4
2 жыл бұрын
Can u say explain the production rules
@5d4pranith41
3 жыл бұрын
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
2 ай бұрын
@jonathanjundarino7784
4 жыл бұрын
I think the b,a|ε is doesn't mean to pop a? it means extra b? It should be a,Z0|aZ0
@srb5704
Жыл бұрын
Why we not take € in previous video to push the z⁰ .?
@baladimirputin708
2 жыл бұрын
in the question take n>m in place of n>=m don't drop any comment regarding this
@sohelshaikh2697
3 жыл бұрын
mam this language also take ab
@ShubhamKumar-pv8pj
10 ай бұрын
It is n>m
@vishalshankarmishra9984
4 жыл бұрын
If stack has more than one a then what?
@CSEJoshi
3 жыл бұрын
we should take loop on final state
@someshkumargupta256
3 жыл бұрын
aapka condition a >= m hai or aap a > m le rhi mam ???
@tarunsingh5615
3 жыл бұрын
Sahi bol raha hai bhai €, Z0/€ BHI chahiye q1----->q2 me
@sharvesh0369
7 ай бұрын
why is there difference in stack variables in first part and this second part?
@shreedhar..4970
5 ай бұрын
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 😊
@VaibhaviKamble-oq3yv
Жыл бұрын
Plz clear the concept...doubt raises that string is accepted or not in stack....and also try to good pronounce and not fumble
@itschurya4564
4 жыл бұрын
Best vedio, thankyou so much mam..no one could explain like you do
@infofacts4332
3 жыл бұрын
hmm
@saumyamishra5060
6 ай бұрын
Aaien a bcha h or accepted hain ..nyc. 😅
@shreedhar..4970
5 ай бұрын
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 😊
@shreedhar..4970
5 ай бұрын
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
2 ай бұрын
It is easy to understand whether string going to be accepted or not....
@tinashekadiki8885
4 жыл бұрын
at least 1b and at least 1a as well
@himanshuchauhan4440
4 жыл бұрын
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
3 жыл бұрын
then acc to me we should take epsilon,a/epsilon loop in final state..its not wrong
Пікірлер: 57