I have discussed about intro to context free grammar examples. drive.google.com/drive/folder...
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Пікірлер: 55
@Ria-3119 күн бұрын
Tomorrow is my toc exam ur videos are very helpful sir🙏🏻🙌🏻👏🏻
@because2022
19 күн бұрын
All the best
@Shadyx7
19 күн бұрын
@@because2022sir indha sum ku alternative ah endha sum pakkalam coz indha sum enaku puriyala
@arunsuresh85472 ай бұрын
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 ?
@because2022
2 ай бұрын
No epsilon varalam, In that case S1=> AS and if you substitute a for A and epsilon for S. it will satisfy m>n
@user-vc5qk2ef6d3 ай бұрын
Sir , 6th sum answer la (epsilon E) varakoodadhu dhana sir instead ,s--->asb/A A-->aA/a indha ans crct dhana sir?
@because2022
3 ай бұрын
m,n should be >=0 right. So epsilon will be there.
@jiminshiiahjumma6470Ай бұрын
Sir as for productions . Namma own production edukalama which also satisfies the input string or Neenga edutha same than edukanuma?
@because2022
Ай бұрын
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.
@trending_video00372 ай бұрын
Thanks sir
@because2022
2 ай бұрын
Welcome
@deepalakshmi4371 Жыл бұрын
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
@because2022
Жыл бұрын
Yes un 3rd queston minimum bb will be there. SO epsilon is not requird.
@deepalakshmi4371
Жыл бұрын
@@because2022 thank you sir
@arunsuresh85472 ай бұрын
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
@because2022
2 ай бұрын
Yes we can write in many forms
@marliyabegam-sn6eu Жыл бұрын
Mam normal cfg sum nadathuga
@because2022
Жыл бұрын
Wt do u mean by normal cfg??
@nihethane8 ай бұрын
For 6 eg exact ans is S1->AS Ah? Sir
@because2022
8 ай бұрын
Yes. You need to write all three productions together.
@BayanFahimАй бұрын
Epsilon is not required because of the n=0 has the valve of bb Am I right sir?
@because2022
Ай бұрын
In which question?
@Monika-jc3wl Жыл бұрын
Sir, a^m b^n condition m>n thana sir eruku but m,n 0 kuduthaa condition fail aiduthee sir.
@because2022
Жыл бұрын
Why monika it is failing? For m,n=0, we get S->epsilon directly right?
@Monika-jc3wl
Жыл бұрын
Sir condition m>n thanaa sir but we r assigning both to 0
@because2022
Жыл бұрын
@@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.
@Monika-jc3wl
Жыл бұрын
TQ for the reply sir✨
@Tharik8456
Жыл бұрын
@@because2022 so there is no epsilon in this prblm right sir?
@UsadreamerokАй бұрын
Sir i know 1st and 2nd unit completlely can i go to 4th unit sir??
@because2022
Ай бұрын
No.
@Usadreamerok
Ай бұрын
@@because2022 ok sir
@vsvicky544820 күн бұрын
5:08 sir why u put ab after "/"
@because2022
20 күн бұрын
Because we are moving to next possiblity
@DDHereАй бұрын
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?
@because2022
Ай бұрын
Epsilon wont be part of language. It will have atleast a when n =0
@DDHere
Ай бұрын
@@because2022 thank you sir
@_MIRANSS Жыл бұрын
Sir, 6th sum a^mb^n A->aA/epsillon ( is possible )
@because2022
Жыл бұрын
No then m wont be greater than n when A is epsilon.
@wssgaming7050Ай бұрын
12:52 n>=1 ku answer ena varum sir ??? Orea confused ahh eiruku 😵
@because2022
Ай бұрын
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.
@user-vh8sm9xu3oАй бұрын
Sir for the 6th sum We can write production rule as S->aaSb/€ Is this possible ?
@because2022
Ай бұрын
It wont accept aaab.
@wssgaming7050
Ай бұрын
@@because2022 S->aSb/aS/a
@SyntaxsagoАй бұрын
in 5th sum s---->aaaaaSb s---->aaaaab s----->epsilon its correct answera sir...
@because2022
Ай бұрын
No when n=0, the language should accept aaa. But it doesnt.
@Usadreamerok
20 күн бұрын
Sir if you put n =1 then its aaaaab sir
@iccreations253420 күн бұрын
S->aaaSb/aa is wrong? Then why sir?
@because2022
20 күн бұрын
For which question?
@Mahi-de3cl
19 күн бұрын
@@because2022 5th question sir
@Santhosh_0042 Жыл бұрын
Sir, 6th sum answer ippudi varuma sir S-->aaSb/ab
@because2022
Жыл бұрын
No because it wont accept aab.
@SyedRasheed-rc6zp3 ай бұрын
4th purila sir
@because2022
3 ай бұрын
Plz watch few more examples and check 4th. Then you might understand. If you dont let me know.
Пікірлер: 55
Tomorrow is my toc exam ur videos are very helpful sir🙏🏻🙌🏻👏🏻
@because2022
19 күн бұрын
All the best
@Shadyx7
19 күн бұрын
@@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 ?
@because2022
2 ай бұрын
No epsilon varalam, In that case S1=> AS and if you substitute a for A and epsilon for S. it will satisfy m>n
Sir , 6th sum answer la (epsilon E) varakoodadhu dhana sir instead ,s--->asb/A A-->aA/a indha ans crct dhana sir?
@because2022
3 ай бұрын
m,n should be >=0 right. So epsilon will be there.
Sir as for productions . Namma own production edukalama which also satisfies the input string or Neenga edutha same than edukanuma?
@because2022
Ай бұрын
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.
Thanks sir
@because2022
2 ай бұрын
Welcome
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
@because2022
Жыл бұрын
Yes un 3rd queston minimum bb will be there. SO epsilon is not requird.
@deepalakshmi4371
Жыл бұрын
@@because2022 thank you 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
@because2022
2 ай бұрын
Yes we can write in many forms
Mam normal cfg sum nadathuga
@because2022
Жыл бұрын
Wt do u mean by normal cfg??
For 6 eg exact ans is S1->AS Ah? Sir
@because2022
8 ай бұрын
Yes. You need to write all three productions together.
Epsilon is not required because of the n=0 has the valve of bb Am I right sir?
@because2022
Ай бұрын
In which question?
Sir, a^m b^n condition m>n thana sir eruku but m,n 0 kuduthaa condition fail aiduthee sir.
@because2022
Жыл бұрын
Why monika it is failing? For m,n=0, we get S->epsilon directly right?
@Monika-jc3wl
Жыл бұрын
Sir condition m>n thanaa sir but we r assigning both to 0
@because2022
Жыл бұрын
@@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.
@Monika-jc3wl
Жыл бұрын
TQ for the reply sir✨
@Tharik8456
Жыл бұрын
@@because2022 so there is no epsilon in this prblm right sir?
Sir i know 1st and 2nd unit completlely can i go to 4th unit sir??
@because2022
Ай бұрын
No.
@Usadreamerok
Ай бұрын
@@because2022 ok sir
5:08 sir why u put ab after "/"
@because2022
20 күн бұрын
Because we are moving to next possiblity
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?
@because2022
Ай бұрын
Epsilon wont be part of language. It will have atleast a when n =0
@DDHere
Ай бұрын
@@because2022 thank you sir
Sir, 6th sum a^mb^n A->aA/epsillon ( is possible )
@because2022
Жыл бұрын
No then m wont be greater than n when A is epsilon.
12:52 n>=1 ku answer ena varum sir ??? Orea confused ahh eiruku 😵
@because2022
Ай бұрын
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 for the 6th sum We can write production rule as S->aaSb/€ Is this possible ?
@because2022
Ай бұрын
It wont accept aaab.
@wssgaming7050
Ай бұрын
@@because2022 S->aSb/aS/a
in 5th sum s---->aaaaaSb s---->aaaaab s----->epsilon its correct answera sir...
@because2022
Ай бұрын
No when n=0, the language should accept aaa. But it doesnt.
@Usadreamerok
20 күн бұрын
Sir if you put n =1 then its aaaaab sir
S->aaaSb/aa is wrong? Then why sir?
@because2022
20 күн бұрын
For which question?
@Mahi-de3cl
19 күн бұрын
@@because2022 5th question sir
Sir, 6th sum answer ippudi varuma sir S-->aaSb/ab
@because2022
Жыл бұрын
No because it wont accept aab.
4th purila sir
@because2022
3 ай бұрын
Plz watch few more examples and check 4th. Then you might understand. If you dont let me know.