sumit goyal 1 please explain the difference
It is drawn like this
@ sumit goyal 1 You forgot dead state.$4$ states will be there moreover first state is final
NO,
we can write (a+b)*=(a*+b*)*
=(a*+b)*
=(a+b*)*
=(a*b*)*
=a*(ba*)*
=b*(ab*) *
or
make minimal dfa for both the RE and check LHS != RHS ,infact language both are different
$(a+b)^* \neq (ab^*)^*$
And last one should be b*(ab*)*
yes..by mistake correct one is
(a+b)*=(a*b*)*
=b*(ab*)*
$1^{st}$ one is RHS. $2^{nd}$ one is LHS. LHS $\neq$ RHS
@MRINMOY_HALDER check this
https://gateoverflow.in/169116/regular-expression
No ,because both are not same as u can see by the 1st REstring produce like ab,aabb.a,b,aa,bbetc but u can't get string like aba,aaabbabab,ababa etc .
whereas by the 2nt RE u will get all string produce by a and b.