Why is this basic language not a regular language?












1












$begingroup$


L = {x in {0,1}* | x has an equal number of 0s & 1s}



Based on the recursive definition of regular languages, isn't it possible to form a single regular language set over the binary alphabet {0,1} by doing the following?




  1. concatenating 0's and 1's to form each of the binary strings satisfying the condition, resulting in a regular language consisting of that single string

  2. union all the single-string regular languages together into one single regular language






regular-languages







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  • $begingroup$
    There must be something wrong with your proof attempt, since it would apply to any language and there are surely some non-regular languages. The answers explain what.
    $endgroup$
    – David Richerby
    6 hours ago


















1












$begingroup$


L = {x in {0,1}* | x has an equal number of 0s & 1s}



Based on the recursive definition of regular languages, isn't it possible to form a single regular language set over the binary alphabet {0,1} by doing the following?




  1. concatenating 0's and 1's to form each of the binary strings satisfying the condition, resulting in a regular language consisting of that single string

  2. union all the single-string regular languages together into one single regular language






regular-languages







share|cite|improve this question







New contributor




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







$endgroup$












  • $begingroup$
    There must be something wrong with your proof attempt, since it would apply to any language and there are surely some non-regular languages. The answers explain what.
    $endgroup$
    – David Richerby
    6 hours ago
















1












1








1





$begingroup$


L = {x in {0,1}* | x has an equal number of 0s & 1s}



Based on the recursive definition of regular languages, isn't it possible to form a single regular language set over the binary alphabet {0,1} by doing the following?




  1. concatenating 0's and 1's to form each of the binary strings satisfying the condition, resulting in a regular language consisting of that single string

  2. union all the single-string regular languages together into one single regular language






regular-languages







share|cite|improve this question







New contributor




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







$endgroup$




L = {x in {0,1}* | x has an equal number of 0s & 1s}



Based on the recursive definition of regular languages, isn't it possible to form a single regular language set over the binary alphabet {0,1} by doing the following?




  1. concatenating 0's and 1's to form each of the binary strings satisfying the condition, resulting in a regular language consisting of that single string

  2. union all the single-string regular languages together into one single regular language






regular-languages




regular-languages






share|cite|improve this question







New contributor




Shukie 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




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









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asked 15 hours ago









ShukieShukie

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Shukie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • $begingroup$
    There must be something wrong with your proof attempt, since it would apply to any language and there are surely some non-regular languages. The answers explain what.
    $endgroup$
    – David Richerby
    6 hours ago




















  • $begingroup$
    There must be something wrong with your proof attempt, since it would apply to any language and there are surely some non-regular languages. The answers explain what.
    $endgroup$
    – David Richerby
    6 hours ago


















$begingroup$
There must be something wrong with your proof attempt, since it would apply to any language and there are surely some non-regular languages. The answers explain what.
$endgroup$
– David Richerby
6 hours ago






$begingroup$
There must be something wrong with your proof attempt, since it would apply to any language and there are surely some non-regular languages. The answers explain what.
$endgroup$
– David Richerby
6 hours ago












2 Answers
2






active

oldest

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3












$begingroup$

Every language is the union of regular languages:
$$
L = bigcup_{x in L} { x }.
$$
However, the set of regular languages is not closed under infinite unions. It isn't even closed under countably infinite unions, as your example demonstrates.



You can prove that your language is not regular in many ways. For example, if your language were regular, then so would its intersection with $0^*1^*$ be; yet this language is ${ 0^n 1^n : n geq 0 }$, the classical example of a non-regular language. You can also prove non-regularity of your language directly, using either the pumping lemma or Myhill–Nerode theory.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    You're on the right track! There's just one thing you're missing.



    The "build all the strings and union them together" approach works great—if you have a finite number of strings. There's a theorem that says "a union of finitely many regular languages is regular", but the key is the finitely many.



    In this case, there are infinitely many strings in the language, so the union-them-all trick no longer works.



    If you want to prove that the language is not regular (as opposed to just failing to prove that it is regular), try a fooling set proof:




    • Let $F$ be the language $1^*$. It's clearly infinite.

    • Let $x$ and $y$ be two distinct strings in $F$. These can be written as $1^i$ and $1^j$ with $i neq j$.

    • Now, let $z$ be $0^i$. From the definition of your language, we can see that $xz$ is in the language, but $yz$ is not.

    • Therefore, all the different strings in $F$ are distinguishable as prefixes. Since the language has infinitely many distinguishable prefixes, it cannot be regular.






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

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      active

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      active

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      3












      $begingroup$

      Every language is the union of regular languages:
      $$
      L = bigcup_{x in L} { x }.
      $$
      However, the set of regular languages is not closed under infinite unions. It isn't even closed under countably infinite unions, as your example demonstrates.



      You can prove that your language is not regular in many ways. For example, if your language were regular, then so would its intersection with $0^*1^*$ be; yet this language is ${ 0^n 1^n : n geq 0 }$, the classical example of a non-regular language. You can also prove non-regularity of your language directly, using either the pumping lemma or Myhill–Nerode theory.






      share|cite|improve this answer









      $endgroup$


















        3












        $begingroup$

        Every language is the union of regular languages:
        $$
        L = bigcup_{x in L} { x }.
        $$
        However, the set of regular languages is not closed under infinite unions. It isn't even closed under countably infinite unions, as your example demonstrates.



        You can prove that your language is not regular in many ways. For example, if your language were regular, then so would its intersection with $0^*1^*$ be; yet this language is ${ 0^n 1^n : n geq 0 }$, the classical example of a non-regular language. You can also prove non-regularity of your language directly, using either the pumping lemma or Myhill–Nerode theory.






        share|cite|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          Every language is the union of regular languages:
          $$
          L = bigcup_{x in L} { x }.
          $$
          However, the set of regular languages is not closed under infinite unions. It isn't even closed under countably infinite unions, as your example demonstrates.



          You can prove that your language is not regular in many ways. For example, if your language were regular, then so would its intersection with $0^*1^*$ be; yet this language is ${ 0^n 1^n : n geq 0 }$, the classical example of a non-regular language. You can also prove non-regularity of your language directly, using either the pumping lemma or Myhill–Nerode theory.






          share|cite|improve this answer









          $endgroup$



          Every language is the union of regular languages:
          $$
          L = bigcup_{x in L} { x }.
          $$
          However, the set of regular languages is not closed under infinite unions. It isn't even closed under countably infinite unions, as your example demonstrates.



          You can prove that your language is not regular in many ways. For example, if your language were regular, then so would its intersection with $0^*1^*$ be; yet this language is ${ 0^n 1^n : n geq 0 }$, the classical example of a non-regular language. You can also prove non-regularity of your language directly, using either the pumping lemma or Myhill–Nerode theory.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 13 hours ago









          Yuval FilmusYuval Filmus

          196k15184349




          196k15184349























              1












              $begingroup$

              You're on the right track! There's just one thing you're missing.



              The "build all the strings and union them together" approach works great—if you have a finite number of strings. There's a theorem that says "a union of finitely many regular languages is regular", but the key is the finitely many.



              In this case, there are infinitely many strings in the language, so the union-them-all trick no longer works.



              If you want to prove that the language is not regular (as opposed to just failing to prove that it is regular), try a fooling set proof:




              • Let $F$ be the language $1^*$. It's clearly infinite.

              • Let $x$ and $y$ be two distinct strings in $F$. These can be written as $1^i$ and $1^j$ with $i neq j$.

              • Now, let $z$ be $0^i$. From the definition of your language, we can see that $xz$ is in the language, but $yz$ is not.

              • Therefore, all the different strings in $F$ are distinguishable as prefixes. Since the language has infinitely many distinguishable prefixes, it cannot be regular.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                You're on the right track! There's just one thing you're missing.



                The "build all the strings and union them together" approach works great—if you have a finite number of strings. There's a theorem that says "a union of finitely many regular languages is regular", but the key is the finitely many.



                In this case, there are infinitely many strings in the language, so the union-them-all trick no longer works.



                If you want to prove that the language is not regular (as opposed to just failing to prove that it is regular), try a fooling set proof:




                • Let $F$ be the language $1^*$. It's clearly infinite.

                • Let $x$ and $y$ be two distinct strings in $F$. These can be written as $1^i$ and $1^j$ with $i neq j$.

                • Now, let $z$ be $0^i$. From the definition of your language, we can see that $xz$ is in the language, but $yz$ is not.

                • Therefore, all the different strings in $F$ are distinguishable as prefixes. Since the language has infinitely many distinguishable prefixes, it cannot be regular.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  You're on the right track! There's just one thing you're missing.



                  The "build all the strings and union them together" approach works great—if you have a finite number of strings. There's a theorem that says "a union of finitely many regular languages is regular", but the key is the finitely many.



                  In this case, there are infinitely many strings in the language, so the union-them-all trick no longer works.



                  If you want to prove that the language is not regular (as opposed to just failing to prove that it is regular), try a fooling set proof:




                  • Let $F$ be the language $1^*$. It's clearly infinite.

                  • Let $x$ and $y$ be two distinct strings in $F$. These can be written as $1^i$ and $1^j$ with $i neq j$.

                  • Now, let $z$ be $0^i$. From the definition of your language, we can see that $xz$ is in the language, but $yz$ is not.

                  • Therefore, all the different strings in $F$ are distinguishable as prefixes. Since the language has infinitely many distinguishable prefixes, it cannot be regular.






                  share|cite|improve this answer









                  $endgroup$



                  You're on the right track! There's just one thing you're missing.



                  The "build all the strings and union them together" approach works great—if you have a finite number of strings. There's a theorem that says "a union of finitely many regular languages is regular", but the key is the finitely many.



                  In this case, there are infinitely many strings in the language, so the union-them-all trick no longer works.



                  If you want to prove that the language is not regular (as opposed to just failing to prove that it is regular), try a fooling set proof:




                  • Let $F$ be the language $1^*$. It's clearly infinite.

                  • Let $x$ and $y$ be two distinct strings in $F$. These can be written as $1^i$ and $1^j$ with $i neq j$.

                  • Now, let $z$ be $0^i$. From the definition of your language, we can see that $xz$ is in the language, but $yz$ is not.

                  • Therefore, all the different strings in $F$ are distinguishable as prefixes. Since the language has infinitely many distinguishable prefixes, it cannot be regular.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 9 hours ago









                  DraconisDraconis

                  5,762921




                  5,762921






















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