Regular Language

How is this a regular language?

{a^n b^2m| n>=0, m>=0}

Nishant Vijayvergiya
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

Aparajita Mehta
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..

Anjan
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 ?