Language whose intersection with a CFL is always a CFLIf $L_1$ is regular and $L_1 cap L_2$ context-free, is...
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Language whose intersection with a CFL is always a CFL
If $L_1$ is regular and $L_1 cap L_2$ context-free, is $L_2$ always context-free?Can every context free language written as a intersection of another context free language and a regular language?Intersection of a language with a regular language imply context freeProving/Disproving that language L is non-regular/CFLLower bound for number of nonterminals in a CFGIs intersection of regular language and context free language is “always” context free languageProof Idea: How to prove the intersection of regular language and CFL is a CFL?CFL Intersection with Regular Language proveGiven a CFL L and a regular language R, is $overline{L} cap R = emptyset$ decidable or undecidable?Is a kind of reverse Kleene star of a context-free language context-free?
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Prove or disprove: If the language $L$ is such that for every context-free language $L_0$, the language $L cap L_0$ is context-free, then $L$ is regular.
I haven't managed to prove this, but I'm pretty sure there is no counterexample.
formal-languages regular-languages context-free
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Prove or disprove: If the language $L$ is such that for every context-free language $L_0$, the language $L cap L_0$ is context-free, then $L$ is regular.
I haven't managed to prove this, but I'm pretty sure there is no counterexample.
formal-languages regular-languages context-free
New contributor
$endgroup$
add a comment |
$begingroup$
Prove or disprove: If the language $L$ is such that for every context-free language $L_0$, the language $L cap L_0$ is context-free, then $L$ is regular.
I haven't managed to prove this, but I'm pretty sure there is no counterexample.
formal-languages regular-languages context-free
New contributor
$endgroup$
Prove or disprove: If the language $L$ is such that for every context-free language $L_0$, the language $L cap L_0$ is context-free, then $L$ is regular.
I haven't managed to prove this, but I'm pretty sure there is no counterexample.
formal-languages regular-languages context-free
formal-languages regular-languages context-free
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New contributor
edited 17 hours ago
Yuval Filmus
196k15184349
196k15184349
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asked 17 hours ago
Matan HalfonMatan Halfon
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$begingroup$
Let $L = {a^n b^n : n geq 0}$, and let $L_0$ be an arbitrary context-free language. Define $L_1 = L_0 cap a^* b^*$, and note that $L_1$ is context-free and $L cap L_0 = L cap L_1$. Let $S = {(i,j) : a^i b^j in L_1}$.
According to Parikh's theorem, the set $S$ is semilinear. The set $D = {(n,n) geq 0}$ is also semilinear (in fact, it is linear). Since the semilinear sets are closed under intersection, $S cap D$ is also semilinear. Since $S cap D$ is (essentially) one-dimensional, it is eventually periodic. This shows that there is a finite language $F$, a modulus $m geq 1$ and a subset $A subseteq {0,ldots,m-1}$ such that
$$
L cap L_1 = F Delta { a^n b^n : n bmod m in A },
$$
where $Delta$ is symmetric difference. It is easy to check that ${a^nb^n : n bmod m in A}$ is context-free, and so $L cap L_1$ is context-free.
Summarizing, we have shown that $L$ is a non-regular language which satisfies your condition.
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$begingroup$
i dont know (or allowed )to use the Parikj's theorem there is any way you can show it without the linearity, thanks a lot
$endgroup$
– Matan Halfon
13 hours ago
$begingroup$
Unfortunately I don’t care about this sort of artificial constraint. In mathematics we can use everything we know.
$endgroup$
– Yuval Filmus
13 hours ago
$begingroup$
If you don't know something, why not look it up? Be curious.
$endgroup$
– Yuval Filmus
8 hours ago
add a comment |
Your Answer
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1 Answer
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$begingroup$
Let $L = {a^n b^n : n geq 0}$, and let $L_0$ be an arbitrary context-free language. Define $L_1 = L_0 cap a^* b^*$, and note that $L_1$ is context-free and $L cap L_0 = L cap L_1$. Let $S = {(i,j) : a^i b^j in L_1}$.
According to Parikh's theorem, the set $S$ is semilinear. The set $D = {(n,n) geq 0}$ is also semilinear (in fact, it is linear). Since the semilinear sets are closed under intersection, $S cap D$ is also semilinear. Since $S cap D$ is (essentially) one-dimensional, it is eventually periodic. This shows that there is a finite language $F$, a modulus $m geq 1$ and a subset $A subseteq {0,ldots,m-1}$ such that
$$
L cap L_1 = F Delta { a^n b^n : n bmod m in A },
$$
where $Delta$ is symmetric difference. It is easy to check that ${a^nb^n : n bmod m in A}$ is context-free, and so $L cap L_1$ is context-free.
Summarizing, we have shown that $L$ is a non-regular language which satisfies your condition.
$endgroup$
$begingroup$
i dont know (or allowed )to use the Parikj's theorem there is any way you can show it without the linearity, thanks a lot
$endgroup$
– Matan Halfon
13 hours ago
$begingroup$
Unfortunately I don’t care about this sort of artificial constraint. In mathematics we can use everything we know.
$endgroup$
– Yuval Filmus
13 hours ago
$begingroup$
If you don't know something, why not look it up? Be curious.
$endgroup$
– Yuval Filmus
8 hours ago
add a comment |
$begingroup$
Let $L = {a^n b^n : n geq 0}$, and let $L_0$ be an arbitrary context-free language. Define $L_1 = L_0 cap a^* b^*$, and note that $L_1$ is context-free and $L cap L_0 = L cap L_1$. Let $S = {(i,j) : a^i b^j in L_1}$.
According to Parikh's theorem, the set $S$ is semilinear. The set $D = {(n,n) geq 0}$ is also semilinear (in fact, it is linear). Since the semilinear sets are closed under intersection, $S cap D$ is also semilinear. Since $S cap D$ is (essentially) one-dimensional, it is eventually periodic. This shows that there is a finite language $F$, a modulus $m geq 1$ and a subset $A subseteq {0,ldots,m-1}$ such that
$$
L cap L_1 = F Delta { a^n b^n : n bmod m in A },
$$
where $Delta$ is symmetric difference. It is easy to check that ${a^nb^n : n bmod m in A}$ is context-free, and so $L cap L_1$ is context-free.
Summarizing, we have shown that $L$ is a non-regular language which satisfies your condition.
$endgroup$
$begingroup$
i dont know (or allowed )to use the Parikj's theorem there is any way you can show it without the linearity, thanks a lot
$endgroup$
– Matan Halfon
13 hours ago
$begingroup$
Unfortunately I don’t care about this sort of artificial constraint. In mathematics we can use everything we know.
$endgroup$
– Yuval Filmus
13 hours ago
$begingroup$
If you don't know something, why not look it up? Be curious.
$endgroup$
– Yuval Filmus
8 hours ago
add a comment |
$begingroup$
Let $L = {a^n b^n : n geq 0}$, and let $L_0$ be an arbitrary context-free language. Define $L_1 = L_0 cap a^* b^*$, and note that $L_1$ is context-free and $L cap L_0 = L cap L_1$. Let $S = {(i,j) : a^i b^j in L_1}$.
According to Parikh's theorem, the set $S$ is semilinear. The set $D = {(n,n) geq 0}$ is also semilinear (in fact, it is linear). Since the semilinear sets are closed under intersection, $S cap D$ is also semilinear. Since $S cap D$ is (essentially) one-dimensional, it is eventually periodic. This shows that there is a finite language $F$, a modulus $m geq 1$ and a subset $A subseteq {0,ldots,m-1}$ such that
$$
L cap L_1 = F Delta { a^n b^n : n bmod m in A },
$$
where $Delta$ is symmetric difference. It is easy to check that ${a^nb^n : n bmod m in A}$ is context-free, and so $L cap L_1$ is context-free.
Summarizing, we have shown that $L$ is a non-regular language which satisfies your condition.
$endgroup$
Let $L = {a^n b^n : n geq 0}$, and let $L_0$ be an arbitrary context-free language. Define $L_1 = L_0 cap a^* b^*$, and note that $L_1$ is context-free and $L cap L_0 = L cap L_1$. Let $S = {(i,j) : a^i b^j in L_1}$.
According to Parikh's theorem, the set $S$ is semilinear. The set $D = {(n,n) geq 0}$ is also semilinear (in fact, it is linear). Since the semilinear sets are closed under intersection, $S cap D$ is also semilinear. Since $S cap D$ is (essentially) one-dimensional, it is eventually periodic. This shows that there is a finite language $F$, a modulus $m geq 1$ and a subset $A subseteq {0,ldots,m-1}$ such that
$$
L cap L_1 = F Delta { a^n b^n : n bmod m in A },
$$
where $Delta$ is symmetric difference. It is easy to check that ${a^nb^n : n bmod m in A}$ is context-free, and so $L cap L_1$ is context-free.
Summarizing, we have shown that $L$ is a non-regular language which satisfies your condition.
edited 1 hour ago
answered 17 hours ago
Yuval FilmusYuval Filmus
196k15184349
196k15184349
$begingroup$
i dont know (or allowed )to use the Parikj's theorem there is any way you can show it without the linearity, thanks a lot
$endgroup$
– Matan Halfon
13 hours ago
$begingroup$
Unfortunately I don’t care about this sort of artificial constraint. In mathematics we can use everything we know.
$endgroup$
– Yuval Filmus
13 hours ago
$begingroup$
If you don't know something, why not look it up? Be curious.
$endgroup$
– Yuval Filmus
8 hours ago
add a comment |
$begingroup$
i dont know (or allowed )to use the Parikj's theorem there is any way you can show it without the linearity, thanks a lot
$endgroup$
– Matan Halfon
13 hours ago
$begingroup$
Unfortunately I don’t care about this sort of artificial constraint. In mathematics we can use everything we know.
$endgroup$
– Yuval Filmus
13 hours ago
$begingroup$
If you don't know something, why not look it up? Be curious.
$endgroup$
– Yuval Filmus
8 hours ago
$begingroup$
i dont know (or allowed )to use the Parikj's theorem there is any way you can show it without the linearity, thanks a lot
$endgroup$
– Matan Halfon
13 hours ago
$begingroup$
i dont know (or allowed )to use the Parikj's theorem there is any way you can show it without the linearity, thanks a lot
$endgroup$
– Matan Halfon
13 hours ago
$begingroup$
Unfortunately I don’t care about this sort of artificial constraint. In mathematics we can use everything we know.
$endgroup$
– Yuval Filmus
13 hours ago
$begingroup$
Unfortunately I don’t care about this sort of artificial constraint. In mathematics we can use everything we know.
$endgroup$
– Yuval Filmus
13 hours ago
$begingroup$
If you don't know something, why not look it up? Be curious.
$endgroup$
– Yuval Filmus
8 hours ago
$begingroup$
If you don't know something, why not look it up? Be curious.
$endgroup$
– Yuval Filmus
8 hours ago
add a comment |
Matan Halfon is a new contributor. Be nice, and check out our Code of Conduct.
Matan Halfon is a new contributor. Be nice, and check out our Code of Conduct.
Matan Halfon is a new contributor. Be nice, and check out our Code of Conduct.
Matan Halfon is a new contributor. Be nice, and check out our Code of Conduct.
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