Arjun Suresh (talk | contribs) (→Solution) |
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| − | Let Σ = {a, b, c}. Which of the following statements is true ? | + | Let <math>Σ = {a, b, c}</math>. Which of the following statements is true ? |
| − | a)For any A ⊆ Σ*, if A is regular, then so is {xx | x ∊ A} | + | a)For any <math>A ⊆ Σ^*</math>, if <math>A</math> is regular, then so is <math>\{xx | x ∊ A\}</math> |
| − | '''b)For any A ⊆ Σ*, if A is regular, then so is {x | xx ∊ A}''' | + | '''b)For any <math>A ⊆ Σ^*</math>, if <math>A</math> is regular, then so is <math>\{x | xx ∊ A\}</math>''' |
| − | c)For any A ⊆ Σ*, if A is context-free, then so is {xx | x ∊ A} | + | c)For any <math>A ⊆ Σ^*</math>, if <math>A</math> is context-free, then so is <math>\{xx | x ∊ A\}</math> |
| − | d)For any A ⊆ Σ*, if A is context-free, then so is {x | xx ∊ A} | + | d)For any <math>A ⊆ Σ^*</math>, if <math>A</math> is context-free, then so is <math>\{x | xx ∊ A\}</math> |
===Solution=== | ===Solution=== | ||
| − | We can get a DFA for L = {x | xx ∊ A} as follows: | + | We can get a DFA for <math>L = \{x | xx ∊ A\}</math> as follows: |
| − | Take DFA for A $(Q, \delta, \Sigma, S, F)$ with everything same except initially making F = | + | Take DFA for <math>A</math> $(Q, \delta, \Sigma, S, F)$ with everything same except initially making $F = \phi$. |
| − | 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. | + | Now for each state $D \in Q$, consider 2 separate DFAs, one with <math>S</math> as the start state and <math>D</math> as the final state and another with <math>D</math> as the start state and set of final states $⊆ F$. If both these DFAs accept same language make <math>D</math> as final state. |
This procedure works as the equivalence of 2 DFAs is decidable. | This procedure works as the equivalence of 2 DFAs is decidable. | ||
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'''Contradictions for other choices''' | '''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 | + | a) Consider <math>A = Σ^*</math>. Now for $w \in A, L = \{xx | x \in A\} = \{ww | w \in Σ^*\} $ which is context sensitive |
c) Same example as for (a) | c) Same example as for (a) | ||
Let <math>Σ = {a, b, c}</math>. Which of the following statements is true ?
a)For any <math>A ⊆ Σ^*</math>, if <math>A</math> is regular, then so is <math>\{xx | x ∊ A\}</math>
b)For any <math>A ⊆ Σ^*</math>, if <math>A</math> is regular, then so is <math>\{x | xx ∊ A\}</math>
c)For any <math>A ⊆ Σ^*</math>, if <math>A</math> is context-free, then so is <math>\{xx | x ∊ A\}</math>
d)For any <math>A ⊆ Σ^*</math>, if <math>A</math> is context-free, then so is <math>\{x | xx ∊ A\}</math>
We can get a DFA for <math>L = \{x | xx ∊ A\}</math> as follows: Take DFA for <math>A</math> $(Q, \delta, \Sigma, S, F)$ with everything same except initially making $F = \phi$. Now for each state $D \in Q$, consider 2 separate DFAs, one with <math>S</math> as the start state and <math>D</math> as the final state and another with <math>D</math> as the start state and set of final states $⊆ F$. If both these DFAs accept same language make <math>D</math> as final state.
This procedure works as the equivalence of 2 DFAs is decidable.
Contradictions for other choices
a) Consider <math>A = Σ^*</math>. 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 <math>Σ = {a, b, c}</math>. Which of the following statements is true ?
a)For any <math>A ⊆ Σ^*</math>, if <math>A</math> is regular, then so is <math>\{xx | x ∊ A\}</math>
b)For any <math>A ⊆ Σ^*</math>, if <math>A</math> is regular, then so is <math>\{x | xx ∊ A\}</math>
c)For any <math>A ⊆ Σ^*</math>, if <math>A</math> is context-free, then so is <math>\{xx | x ∊ A\}</math>
d)For any <math>A ⊆ Σ^*</math>, if <math>A</math> is context-free, then so is <math>\{x | xx ∊ A\}</math>
We can get a DFA for <math>L = \{x | xx ∊ A\}</math> as follows: Take DFA for <math>A</math> $(Q, \delta, \Sigma, S, F)$ with everything same except initially making $F = \phi$. Now for each state $D \in Q$, consider 2 separate DFAs, one with <math>S</math> as the start state and <math>D</math> as the final state and another with <math>D</math> as the start state and set of final states $⊆ F$. If both these DFAs accept same language make <math>D</math> as final state.
This procedure works as the equivalence of 2 DFAs is decidable.
Contradictions for other choices
a) Consider <math>A = Σ^*</math>. 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