In WXW^R, if w contains a 3 length string then let, it can be start with a and end with a in W^R but what about the other char of the W are they reversed or not in W^R because this expression cannot able to tell about the a(a+b)*a + b(a+b)*b other char........
Why this is not considered as Regular ? a^n . b^n | n>=0 we can write its Regular grammar then why is it not considered ? Regular grammar = S = a.S.b | epsilon
thats not regular grammar because its neither right linear nor left linear what you have written is context free grammar..... see the definition of regular grammar
L=a^m b^n | m+n=even ..... It is said to be regular. There exists a functional dependency. If m=1(odd) then n= (odd) so the value of n depends on the m value ,then why it is said to be regular. ❗❗ This is not in the video. I have a doubt on this question. To clarify this i came across the video ...could somebody explain me this.
You mentioned a functional dependency where if m is odd, then n must also be odd for m+n to be even. This is true, but it does not simply means that the language is non-regular. Regular languages can have dependencies, but they must be expressible in a way that can be captured by a finite automatation , Hence the give equation can be captured by FA so this is Regular Language
Excellent sir..no one explained clearly like you
Please make a video on Pumping Lemma for regular expression
Really very nice sir u have a bright future 👍👍👍👍. On teaching side , your explanation without any break r confusig r hesitating, very well keep it up
You deserve more subscriber ❤
In WXW^R, if w contains a 3 length string then let, it can be start with a and end with a in W^R but what about the other char of the W are they reversed or not in W^R because this expression cannot able to tell about the a(a+b)*a + b(a+b)*b other char........
was thinking the exact same thing, it's not regular grammar, he was wrong. A stack is required here.
Thanks sir, great video!!
very helpful content. Thank you sir
Very nice explanation Sir👍👍💯
Thank you so much sir😊
Why this is not considered as Regular ?
a^n . b^n | n>=0
we can write its Regular grammar then why is it not considered ?
Regular grammar = S = a.S.b | epsilon
thats not regular grammar because its neither right linear nor left linear
what you have written is context free grammar.....
see the definition of regular grammar
@@akash_assistExactly
@@akash_assistbut we can write regular expression corresponding to it as a*b* then why isn't it a regular language??
Sorry, I'm wrong, I got my answer..
use pumping lemma for this concept
Thank you sir ❤❤
Nice
Tku sir🙏🙏
Sir ismein answer mein star ki jagah plus aayega as x and y have to take something
yes bro u r right
Nice Video Sir
❤❤❤❤❤❤❤❤❤❤❤
L=a^m b^n | m+n=even ..... It is said to be regular.
There exists a functional dependency. If m=1(odd) then n= (odd) so the value of n depends on the m value ,then why it is said to be regular.
❗❗ This is not in the video. I have a doubt on this question. To clarify this i came across the video ...could somebody explain me this.
You mentioned a functional dependency where if m is odd, then n must also be odd for m+n to be even. This is true, but it does not simply means that the language is non-regular. Regular languages can have dependencies, but they must be expressible in a way that can be captured by a finite automatation , Hence the give equation can be captured by FA so this is Regular Language
Great 👍
Nice video sir g
13: sir why it is not regular..can you say it in english
Must watch
Sir, wcw^r strings regular expression
Generates strings not in L which violates the property that regular expression shldnot generate strings not in L
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No offense but
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He nailed TOC
👌👌🤘
Nice video Sir