58 views
L={a^n b^n :n>=1} and R = (a+b)^*

L union R is going to be regular or not regular

plzz give reason L is not regular if N leads to infinity then how it can be regular ..........
retagged | 58 views

$L$ = {$ab,aabb,aaabbb,aaaabbbb.......$}     $\Rightarrow DCFL$
$R$ = {$\epsilon, a ,b , aa,bb,ab,ba,aaa,bbb ,aabb,aaaabbbb,,,,,,,,$}  $\Rightarrow Regular$
$L\cup R$ = {$\epsilon, a ,b , aa,bb,ab,ba,aaa,bbb ,aabb,aaaabbbb,,,,,,,,$} = $R$  $\Rightarrow Regular$

The thing is $R$ is a set of all strings over {$a,b$}. So $union$ of $R$ and $L$ is $R$ only. So it's regular.
$DCFL \cup Regular$ = $DCFL$ and Regular languages are subset of DCFL .
selected
+1 vote

it should be regular.

• (a+b)* U any language = (a+b)* -----> which is regular.