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 = {x | xx ∊ A} as follows: Take DFA for A $(Q, \delta, \Sigma, S, F)$ with everything same except initially making F = {}. Now for each state $D \in Q$, 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 set of final states ⊆ F. If both these DFAs accept same language make D as final state.
This procedure works as the equivalence of 2 DFAs is decidable.
Contradictions for other choices
a) Consider A = Σ*. Now for $w \in A, L = \{xx | x \in A\} = \{ww | w \in Σ^*\} $ which is context sensitive
c) Same example as for (a)
d)Consider $A = \{a^nb^nc^*a^*b^nc^n|n\ge0\} $ This is CFL. But if we make L from A as per (d), it'll be $L = \{a^nb^nc^n|n\ge0\}$ 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 = {x | xx ∊ A} as follows: Take DFA for A $(Q, \delta, \Sigma, S, F)$ with everything same except initially making F = {}. Now for each state $D \in Q$, 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 set of final states ⊆ F. If both these DFAs accept same language make D as final state.
This procedure works as the equivalence of 2 DFAs is decidable.
Contradictions for other choices
a) Consider A = Σ*. Now for $w \in A, L = \{xx | x \in A\} = \{ww | w \in Σ^*\} $ which is context sensitive
c) Same example as for (a)
d)Consider $A = \{a^nb^nc^*a^*b^nc^n|n\ge0\} $ This is CFL. But if we make L from A as per (d), it'll be $L = \{a^nb^nc^n|n\ge0\}$ which is not context free..
GATECSE