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Context-free languages and regular languages are both closed under the operations(s) of
(i) Union
(ii) Intersection
(iii) Concatenation
(i) Union
(ii) Intersection
(iii) Concatenation
- a)(i) and (ii) only
- b)(ii) and (iii) only
- c)(i) and (iii) only
- d)All of these
Correct answer is option 'C'. Can you explain this answer?
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Context-free languages and regular languages are both closed under the...
Regular and context free languages are both closed under union and concatenation while only regular is closed under intersection.
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Context-free languages and regular languages are both closed under the...
Union of Context-Free Languages
- The union of two context-free languages is also a context-free language.
- Let's say we have two context-free languages L1 and L2. We can construct a new context-free grammar that generates the union of L1 and L2.
- The new grammar can have a start symbol S, and the productions can be a combination of the productions of L1 and L2.
- For example, if L1 has a production S1 → α and L2 has a production S2 → β, then the new grammar can have a production S → S1 | S2, where | represents the union operation.
- Therefore, the union of two context-free languages is still a context-free language.
Intersection of Context-Free Languages
- The intersection of two context-free languages is not necessarily a context-free language.
- Let's say we have two context-free languages L1 and L2. If we take the intersection of L1 and L2, the resulting language may not satisfy the closure properties of context-free languages.
- For example, if L1 is the language of all strings consisting of an equal number of 'a's and 'b's, and L2 is the language of all strings consisting of an equal number of 'a's, 'b's, and 'c's, then the intersection of L1 and L2 would be the language of all strings consisting of an equal number of 'a's and 'b's, but also containing 'c's.
- This language is not context-free because it cannot be generated by a context-free grammar.
- Therefore, the intersection of two context-free languages is not necessarily a context-free language.
Concatenation of Context-Free Languages
- The concatenation of two context-free languages is also a context-free language.
- Let's say we have two context-free languages L1 and L2. We can construct a new context-free grammar that generates the concatenation of L1 and L2.
- The new grammar can have a start symbol S, and the productions can be a combination of the productions of L1 and L2, ensuring that the end of L1 is connected to the beginning of L2.
- For example, if L1 has a production S1 → α and L2 has a production S2 → β, then the new grammar can have a production S → S1S2, where S1S2 represents the concatenation of S1 and S2.
- Therefore, the concatenation of two context-free languages is still a context-free language.
Conclusion
- From the above explanations, we can conclude that context-free languages are closed under the operations of union and concatenation, but not under intersection.
- Regular languages, on the other hand, are closed under all three operations of union, intersection, and concatenation.
- Therefore, the correct answer is option C) (i) and (iii) only.
- The union of two context-free languages is also a context-free language.
- Let's say we have two context-free languages L1 and L2. We can construct a new context-free grammar that generates the union of L1 and L2.
- The new grammar can have a start symbol S, and the productions can be a combination of the productions of L1 and L2.
- For example, if L1 has a production S1 → α and L2 has a production S2 → β, then the new grammar can have a production S → S1 | S2, where | represents the union operation.
- Therefore, the union of two context-free languages is still a context-free language.
Intersection of Context-Free Languages
- The intersection of two context-free languages is not necessarily a context-free language.
- Let's say we have two context-free languages L1 and L2. If we take the intersection of L1 and L2, the resulting language may not satisfy the closure properties of context-free languages.
- For example, if L1 is the language of all strings consisting of an equal number of 'a's and 'b's, and L2 is the language of all strings consisting of an equal number of 'a's, 'b's, and 'c's, then the intersection of L1 and L2 would be the language of all strings consisting of an equal number of 'a's and 'b's, but also containing 'c's.
- This language is not context-free because it cannot be generated by a context-free grammar.
- Therefore, the intersection of two context-free languages is not necessarily a context-free language.
Concatenation of Context-Free Languages
- The concatenation of two context-free languages is also a context-free language.
- Let's say we have two context-free languages L1 and L2. We can construct a new context-free grammar that generates the concatenation of L1 and L2.
- The new grammar can have a start symbol S, and the productions can be a combination of the productions of L1 and L2, ensuring that the end of L1 is connected to the beginning of L2.
- For example, if L1 has a production S1 → α and L2 has a production S2 → β, then the new grammar can have a production S → S1S2, where S1S2 represents the concatenation of S1 and S2.
- Therefore, the concatenation of two context-free languages is still a context-free language.
Conclusion
- From the above explanations, we can conclude that context-free languages are closed under the operations of union and concatenation, but not under intersection.
- Regular languages, on the other hand, are closed under all three operations of union, intersection, and concatenation.
- Therefore, the correct answer is option C) (i) and (iii) only.
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