madeeasy work book

Question

L={a^n b^n a^n |n=1,2,3.........} is an example of a language that is

a)context free

b)not context free

c)not context free but whose complement is CF

d)context free but whose complement is not CF

Comments

it is not CFL but its complement is CFL

$\bar{L}$ = b(a+b)* ∪ {a^mbⁿa*|m$\neq$n, m,n>=0} ∪ {a*b^maⁿ|m≠n, m,n>=0} ∪ {a^mb*aⁿ|m$\neq$n, m,n>=0} ∪ {a^mbⁿa(a+b)*|m$\neq$n,m,n>=1} + eps

above language is union of CFL's and regular, therefore it is CFL

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what u did am not getting. how to find the complement of {a^n b^n a^n }

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just put opposite conditions of (a^n b^n a^n)

like first-: starting with 'b' = b(a+b)*

then-: first a's not equal to first b's = (a^m b^n a*; m,n > 0; m != n)

then-: first b's and second a's not equal = (a* b^n a^m; n,m > 0; m != n)

then-: first a's and second a's not equal = (a^n b* a^m; n,m >0; m != n)

then-: epsilon

this will result in Union of Regular and CFL's which is CFL only