@@azcomputing sir ma nay pumping lemma for non language wali example may X=3, Y=4, Z=3 put kiya ha or jab Y ko pump kia ha to wo a^n and b^n equal aa gya ha, please guide me
@@tech4inspiration619the ending string will be aaaaabbaabbbbb which is not in the language as their is a sequence required in the question that number of b's must be followed by equal number of a's so it's non regular language
Dear Sir, Thank you for the explanation. I am watching your videos from Italy and My exam is on 26th January 2022. My professor taught this topic in a very different method. I just need to ask that how do we prove it in a mathematical fashion.
First of all thanks for watching. Yes, there are different and multiple methods to cover a single topic. Similarly we can prove it in mathematical fashiin in different ways. Basically, pumping lemma is used as a proof for irregularity of a language. If a language is regular, it always satisfies pumping lemma.
What if we divide S in such a way that X Y and Z still follow the Description even after Pumping Y? For example, instead of considering Y as 4 b's after the 1st b.. We consider Y as the middle, 3 a's and 3 b's? So when we pump, it will still satisfy the RE.. The point is, whaat is the criteria of division of S between X Y and Z?
I would love to hear from You if you could provide these mathematical proofs using pumping lemma: L1 = {a^p b a^q b a^q b a^r | p, q, r >= 1} L2 = {a^p b a^q b a^q b a^q+r | p, q, r >= 1} L3 = {a^p+q b a^q+r | p, q, r >= 1} State whether L1, L2, and L3 are regular languages, and provide mathematical proof.
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thanks, Your appreciation means a lot to me
Thanku so much sir jii kl mera paper hai apki videos sy mjhy bht mili Allah pk apko is ka ajar dain 🥰
Mera to abhi he 2 ghante bad😂
Mera bi kal h😂😂😂😂@@abujar.786
😀nice explanation sir best for pumping lemma
Excellent bhai very helpful video
Thank you sir , college me khuch smj me hi nhi aya tha ab smja
Sir kmall method❤❤❤
Wonderful Explanation❤❤❤
What an explanation ❤
such a great explanation😍😍
Thank you sir! excellent work
I like your efforts ❤❤❤
Thank you sir 😊
Thanks sir this video is very helpful and good
Welcome
@@azcomputing sir ma nay pumping lemma for non language wali example may X=3, Y=4, Z=3 put kiya ha or jab Y ko pump kia ha to wo a^n and b^n equal aa gya ha, please guide me
@@tech4inspiration619the ending string will be aaaaabbaabbbbb which is not in the language as their is a sequence required in the question that number of b's must be followed by equal number of a's so it's non regular language
Thank you sir its really help me
Very nice explanation
thank you sir
Thanks sir
Nice sir 🥰
Mza agya
What in the case if starting condition is ab and ending is bab, in that case I found pumping lemma is giving false result, is it?
Difference between pumping lemma for CFLs & pumping lemma for regular language?? Is these are different from one another? Answer me plz...
Good sir
Zabardast bro
Dear Sir, Thank you for the explanation. I am watching your videos from Italy and My exam is on 26th January 2022. My professor taught this topic in a very different method. I just need to ask that how do we prove it in a mathematical fashion.
First of all thanks for watching.
Yes, there are different and multiple methods to cover a single topic. Similarly we can prove it in mathematical fashiin in different ways.
Basically, pumping lemma is used as a proof for irregularity of a language. If a language is regular, it always satisfies pumping lemma.
❤❤❤❤❤❤
Jazakallah sir.
You're method is great
Sir agr a ki power 834
Or b ki power 733
Ya kasy slove krny gy
Kindly tell me
What if we divide S in such a way that X Y and Z still follow the Description even after Pumping Y?
For example, instead of considering Y as 4 b's after the 1st b.. We consider Y as the middle, 3 a's and 3 b's? So when we pump, it will still satisfy the RE..
The point is, whaat is the criteria of division of S between X Y and Z?
Sir plzz myhill nerode theorm ka bta dain..,possible ho tu kal tak ...perso sham ma paper h
Very nice explanation sir🤌👌
kindly map it on any regular expression please
I would love to hear from You if you could provide these mathematical proofs using pumping lemma:
L1 = {a^p b a^q b a^q b a^r | p, q, r >= 1}
L2 = {a^p b a^q b a^q b a^q+r | p, q, r >= 1}
L3 = {a^p+q b a^q+r | p, q, r >= 1}
State whether L1, L2, and L3 are regular languages, and provide mathematical proof.
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very good explanation
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