Let Σ = {a, b, c}. Which of the following statements is true ?
a)For any A ⊆ Σ*, if A is regular, then so is {xx | x ∊ A}
b)For any A ⊆ Σ*, if A is regular, then so is {x | xx ∊ A}
c)For any A ⊆ Σ*, if A is context-free, then so is {xx | x ∊ A}
d)For any A ⊆ Σ*, if A is context-free, then so is {x | xx ∊ A}
We can get a DFA for L as follows: Take DFA for A. Let S be its start state. For each state D of A, consider 2 separate DFAs, one with S as the start state and D as the final state and another with D as the start state and final state in final states of A. If both these DFAs accept same language make D as final state.
This procedure works as the equivalence of 2 DFAs is decidable.
$\{a^nb^nc^*a^*b^nc^n|n>0\} $
This is CFL. But if we make L from A as per (d), it'll be L = {a^nb^nc^n|n>0} which is not context free..
Let Σ = {a, b, c}. Which of the following statements is true ?
a)For any A ⊆ Σ*, if A is regular, then so is {xx | x ∊ A}
b)For any A ⊆ Σ*, if A is regular, then so is {x | xx ∊ A}
c)For any A ⊆ Σ*, if A is context-free, then so is {xx | x ∊ A}
d)For any A ⊆ Σ*, if A is context-free, then so is {x | xx ∊ A}
We can get a DFA for L as follows: Take DFA for A. Let S be its start state. For each state D of A, consider 2 separate DFAs, one with S as the start state and D as the final state and another with D as the start state and final state in final states of A. If both these DFAs accept same language make D as final state.
This procedure works as the equivalence of 2 DFAs is decidable.
$\{a^nb^nc^*a^*b^nc^n|n>0\} $
This is CFL. But if we make L from A as per (d), it'll be L = {a^nb^nc^n|n>0} which is not context free..
GATECSE