Aparajita Mehta @aparajita added a Question 19 Jan 2015 Regular Language How is this a regular language? {a^n b^2m| n>=0, m>=0} 3Comments thumbs up down up0 like Nishant Vijayvergiya @nishantv 19 Jan 2015 11:00 pm n,m are independent of each other. so no. of a's and no. of b's are independent of each other which typically makes this L={∈,a,b,abb,..} for which it is easy to construct a DFA. Hence given lang. is regular up0 like Log in or register to post comments Aparajita Mehta @aparajita 21 Jan 2015 02:42 am I was trying it through pumping lemma.. i took a string : aaabbbb (n=3, m=2) with pumping lemma : we have xy^i z I assumed x to be aa , y=ab, z=bbb for i=1 its cool for i=2 we will have: aa(ab)^2 bbb= aaababbbb which doesnot belong to the language. please tell if I am wrong.. up0 like Log in or register to post comments Anjan @prab_1 21 Nov 2015 10:54 am n,m are independent of each other so regular but here while proving with pumping lemma it failed above? can anyone say this ? up0 like Log in or register to post comments Pages1 2 3 4 5 6 7 8 9 … next › last »

n,m are independent of each other.

so no. of a's and no. of b's are independent of each other which typically makes this L={∈,a,b,abb,..} for which it is easy to construct a DFA.

Hence given lang. is regular

I was trying it through pumping lemma..

i took a string : aaabbbb (n=3, m=2)

with pumping lemma : we have xy^i z

I assumed x to be aa , y=ab, z=bbb

for i=1 its cool

for i=2 we will have: aa(ab)^2 bbb= aaababbbb which doesnot belong to the language.

please tell if I am wrong..

n,m are independent of each other so regular but here while proving with pumping lemma it failed above?

can anyone say this ?

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