In 17:51 6th sum, epsilon vara koodathu thana sir ? It violates m>n .. Can i write productions like this for 6th sum : S -> aSb | aA A -> aA | epsilon Is it correct sir ?
Sir , oru cfl ku neraiya production iruka lama ? Eg: L={a^m*b^n , m>n , n>=0} CFG production 1 : S -> aSb|a CFG production 2 : S -> aSb | aA A -> aA | epsilon
Sir ..15:30 la ..m should be greater than ... N nu soltanga...aprm epdi sir ..m kum n kum ..same value 0 nu potu ... Epsilon poduringa ? .... N 0 va iridha ... M 1ah thana sir irukanum? ....epdi rendum same nu vachu Epsilon podringa sir .. please clarify sir?
Tomorrow is my toc exam ur videos are very helpful sir🙏🏻🙌🏻👏🏻
All the best
@@because2022sir indha sum ku alternative ah endha sum pakkalam coz indha sum enaku puriyala
In 17:51 6th sum, epsilon vara koodathu thana sir ? It violates m>n ..
Can i write productions like this for 6th sum :
S -> aSb | aA
A -> aA | epsilon
Is it correct sir ?
No epsilon varalam, In that case S1=> AS and if you substitute a for A and epsilon for S. it will satisfy m>n
Thanks sir
Welcome
Sir as for productions . Namma own production edukalama which also satisfies the input string or Neenga edutha same than edukanuma?
You can take other productions too but it should accept all strings of the language and should reject those which are not part of language.
Sir ,
6th sum answer la (epsilon E) varakoodadhu dhana sir
instead ,s--->asb/A
A-->aA/a
indha ans crct dhana sir?
m,n should be >=0 right. So epsilon will be there.
Episilon is not required because there should be condition starts with a ,but in 3 problem there is no starts with a..it's bb
Yes un 3rd queston minimum bb will be there. SO epsilon is not requird.
@@because2022 thank you sir
Mam normal cfg sum nadathuga
Wt do u mean by normal cfg??
Epsilon is not required because of the n=0 has the valve of bb
Am I right sir?
In which question?
For 6 eg exact ans is S1->AS Ah? Sir
Yes. You need to write all three productions together.
Sir , oru cfl ku neraiya production iruka lama ?
Eg:
L={a^m*b^n , m>n , n>=0}
CFG production 1 : S -> aSb|a
CFG production 2 : S -> aSb | aA
A -> aA | epsilon
Yes we can write in many forms
12:52 n>=1 ku answer ena varum sir ??? Orea confused ahh eiruku 😵
I want you to try and if you are struck, you can mail to venkat.kvhapp@gmail.com. Because in exam you wont get same questions.
Sir i know 1st and 2nd unit completlely can i go to 4th unit sir??
No.
@@because2022 ok sir
Sir, a^m b^n condition m>n thana sir eruku but m,n 0 kuduthaa condition fail aiduthee sir.
Why monika it is failing? For m,n=0, we get S->epsilon directly right?
Sir condition m>n thanaa sir but we r assigning both to 0
@@Monika-jc3wl sorry monika. It was my mistake. Basically m>n and n>=0 than condition. So basically when n is 0 we should have atleast one a.
TQ for the reply sir✨
@@because2022 so there is no epsilon in this prblm right sir?
Sir ..15:30 la ..m should be greater than ... N nu soltanga...aprm epdi sir ..m kum n kum ..same value 0 nu potu ... Epsilon poduringa ? ....
N 0 va iridha ... M 1ah thana sir irukanum? ....epdi rendum same nu vachu Epsilon podringa sir .. please clarify sir?
Epsilon wont be part of language. It will have atleast a when n =0
@@because2022 thank you sir
5:08 sir why u put ab after "/"
Because we are moving to next possiblity
Sir for the 6th sum
We can write production rule as S->aaSb/€
Is this possible ?
It wont accept aaab.
@@because2022 S->aSb/aS/a
Sir, 6th sum a^mb^n
A->aA/epsillon ( is possible )
No then m wont be greater than n when A is epsilon.
in 5th sum
s---->aaaaaSb
s---->aaaaab
s----->epsilon its correct answera sir...
No when n=0, the language should accept aaa. But it doesnt.
Sir if you put n =1 then its aaaaab sir
Sir,
6th sum answer ippudi varuma sir
S-->aaSb/ab
No because it wont accept aab.
S->aaaSb/aa is wrong? Then why sir?
For which question?
@@because2022 5th question sir
4th purila sir
Plz watch few more examples and check 4th. Then you might understand. If you dont let me know.