MHB Context-Free Langs: S-L Check if Not CF?

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The discussion revolves around the relationship between context-free languages (CFLs) and regular languages (RLs), specifically focusing on the language difference S - L, where S is a CFL and L is an RL. It is established that the difference S - L can be expressed as S ∩ L^c, where L^c is the complement of L, which is also regular. This leads to the conclusion that S - L is context-free due to the closure properties of CFLs under intersection with RLs. Participants clarify that while CFLs are closed under intersection with RLs, they are not closed under complementation or intersection with other CFLs. This distinction is crucial when considering the language L - S, which is not guaranteed to be context-free. The conversation emphasizes the importance of understanding these closure properties and how they apply to various language operations. The discussion concludes with a mutual exchange of holiday greetings, maintaining a friendly tone throughout.
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Hello! :)
Let S be a context-free language and L a regular language.Is there a case that the language S-L is not context-free?
How can I check this?
 
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evinda said:
Hello! :)
Let S be a context-free language and L a regular language.Is there a case that the language S-L is not context-free?
How can I check this?

I think that it is not possible that the language S-L is not context-free,because of the closure properties.Am I right?
 
evinda said:
I think that it is not possible that the language S-L is not context-free,because of the closure properties.Am I right?
Yes, the class of context-free languages is closed under intersection with regular languages..
 
Evgeny.Makarov said:
Yes, the class of context-free languages is closed under intersection with regular languages..

Nice... :) and how can I prove this?Could you give me an example?
 
It is proved by combining a pushdown automaton and a deterministic finite automaton similarly to the proof that regular languages are closed under intersection. Namely, the set of states is the Cartesian product of the sets of the two automata, the accepting pairs are those where both states are accepting in their respective automata. Transitions are also done in the same way as in the given automata. That is, both automata are executed in parallel and accept only when both accept. This is described, for example, in Theorem 3.5.4 in "Elements of the Theory of Computation" by Lewis and Papadimitriou.

It is a good exercise to think why this construction does not work for combining two pushdown automata (which are inherently nondeterministic) and thus proving that CFL are closed under intersection (which is not the case).
 
Evgeny.Makarov said:
It is proved by combining a pushdown automaton and a deterministic finite automaton similarly to the proof that regular languages are closed under intersection. Namely, the set of states is the Cartesian product of the sets of the two automata, the accepting pairs are those where both states are accepting in their respective automata. Transitions are also done in the same way as in the given automata. That is, both automata are executed in parallel and accept only when both accept. This is described, for example, in Theorem 3.5.4 in "Elements of the Theory of Computation" by Lewis and Papadimitriou.

It is a good exercise to think why this construction does not work for combining two pushdown automata (which are inherently nondeterministic) and thus proving that CFL are closed under intersection (which is not the case).

To use the closure property of intersection,do I have to use the fact that S-L=S\cap L^{c} ,where L^{c} is the complement of L and is also regular?
 
evinda said:
To use the closure property of intersection,do I have to use the fact that S-L=S\cap L^{c} ,where L^{c} is the complement of L and is also regular?
That's right.
 
Evgeny.Makarov said:
That's right.

Nice,thank you! :) And what's about the language L-S?Is it also context-free and do I have to show it with the same way?
 
evinda said:
And what's about the language L-S?Is it also context-free and do I have to show it with the same way?
Context-free languages are not closed under complementation (nor intersection). And since the language of all words is regular, the difference of regular and CF is not CF in general.
 
  • #10
Evgeny.Makarov said:
Context-free languages are not closed under complementation (nor intersection). And since the language of all words is regular, the difference of regular and CF is not CF in general.

Nice,thank you very much! :) And...merry Christmas! ;)
 
  • #11
evinda said:
And...merry Christmas! ;)
Merci! And same to you!
 
  • #12
Evgeny.Makarov said:
It is proved by combining a pushdown automaton and a deterministic finite automaton similarly to the proof that regular languages are closed under intersection. Namely, the set of states is the Cartesian product of the sets of the two automata, the accepting pairs are those where both states are accepting in their respective automata. Transitions are also done in the same way as in the given automata. That is, both automata are executed in parallel and accept only when both accept. This is described, for example, in Theorem 3.5.4 in "Elements of the Theory of Computation" by Lewis and Papadimitriou.

It is a good exercise to think why this construction does not work for combining two pushdown automata (which are inherently nondeterministic) and thus proving that CFL are closed under intersection (which is not the case).

To show that S-L is context-free,could I also show that each regular language is context-free and that the set of all regular languages is a subset of the set of the context-free languages?So,the difference is only the set of the context-free languages.Or am I wrong?
 
  • #13
Please describe the logic of this argument in greater detail. That is, suppose that we know that regular languages are context-free. How do we use this fact to prove that S - L is context-free?
 
  • #14
Evgeny.Makarov said:
Please describe the logic of this argument in greater detail. That is, suppose that we know that regular languages are context-free. How do we use this fact to prove that S - L is context-free?

From the image:
View attachment 1850
we see that the set of the regular languages is a proper subset of the set of the context-free languages.So,if we subtract the regular language from the context-free one,the set that remains is context-free.Can I say it like that?
 

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  • #15
evinda said:
the set of the regular languages is a proper subset of the set of the context-free languages.
This is true.
evinda said:
So,if we subtract the regular language from the context-free one,the set that remains is context-free.
And this is a completely different thing. The first quote talks about the difference between the set of context-free languages and the set of regular languages, and the second quote talks about the difference between an individual context-free language and an individual regular language.
 
  • #16
Evgeny.Makarov said:
This is true.
And this is a completely different thing. The first quote talks about the difference between the set of context-free languages and the set of regular languages, and the second quote talks about the difference between an individual context-free language and an individual regular language.

I understand...Thanks a lot! :)
 

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