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?













1












$begingroup$



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.










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New contributor




Matan Halfon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$

















    1












    $begingroup$



    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.










    share|cite|improve this question









    New contributor




    Matan Halfon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$















      1












      1








      1





      $begingroup$



      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.










      share|cite|improve this question









      New contributor




      Matan Halfon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$





      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






      share|cite|improve this question









      New contributor




      Matan Halfon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|cite|improve this question









      New contributor




      Matan Halfon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|cite|improve this question




      share|cite|improve this question








      edited 17 hours ago









      Yuval Filmus

      196k15184349




      196k15184349






      New contributor




      Matan Halfon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      asked 17 hours ago









      Matan HalfonMatan Halfon

      91




      91




      New contributor




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      New contributor





      Matan Halfon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      Matan Halfon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






















          1 Answer
          1






          active

          oldest

          votes


















          2












          $begingroup$

          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.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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












          Your Answer





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          1 Answer
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          1 Answer
          1






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          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.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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
















          2












          $begingroup$

          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.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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














          2












          2








          2





          $begingroup$

          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.






          share|cite|improve this answer











          $endgroup$



          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.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          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


















          • $begingroup$
            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










          Matan Halfon is a new contributor. Be nice, and check out our Code of Conduct.










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          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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