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This procedure works as the equivalence of 2 DFAs is decidable.
 
This procedure works as the equivalence of 2 DFAs is decidable.
  
$\{a^nb^nc^*a^*b^nc^n|n>0\} $
+
'''Contradictions for other choices'''
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a) Consider A =  Σ*. Now for $w \in A L = {x | xx ∊ A}$ which is context sensitive
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c) Same example as for (a)
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d) $\{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  
 
This is CFL. But if we make L from A as per (d), it'll be  

Revision as of 02:06, 31 January 2014

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}

Solution

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.

Contradictions for other choices

a) Consider A = Σ*. Now for $w \in A L = {x | xx ∊ A}$ which is context sensitive

c) Same example as for (a)

d) $\{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..








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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}

Solution[edit]

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..








blog comments powered by Disqus