Why is this basic language not a regular language?Binary regular language?Is the language that accepts...

Modeling an IPv4 Address

Can an x86 CPU running in real mode be considered to be basically an 8086 CPU?

Have astronauts in space suits ever taken selfies? If so, how?

An academic/student plagiarism

How does one intimidate enemies without having the capacity for violence?

Why doesn't H₄O²⁺ exist?

Has the BBC provided arguments for saying Brexit being cancelled is unlikely?

What do the dots in this tr command do: tr .............A-Z A-ZA-Z <<< "JVPQBOV" (with 13 dots)

Is it possible to do 50 km distance without any previous training?

How could an uplifted falcon's brain work?

Fencing style for blades that can attack from a distance

Is this a crack on the carbon frame?

Is it legal for company to use my work email to pretend I still work there?

Why doesn't Newton's third law mean a person bounces back to where they started when they hit the ground?

Is it important to consider tone, melody, and musical form while writing a song?

Prove that NP is closed under karp reduction?

Writing rule stating superpower from different root cause is bad writing

How much RAM could one put in a typical 80386 setup?

Approximately how much travel time was saved by the opening of the Suez Canal in 1869?

Show that if two triangles built on parallel lines, with equal bases have the same perimeter only if they are congruent.

How is it possible to have an ability score that is less than 3?

Why are 150k or 200k jobs considered good when there are 300k+ births a month?

What defenses are there against being summoned by the Gate spell?

What does "Puller Prush Person" mean?



Why is this basic language not a regular language?


Binary regular language?Is the language that accepts strings concatenated with their reverse regular?Definition of a regular languageRegular language concatenation with supersetGenerators of families of langauges?Can the definition of regular languages be simplified?Prove language is regularProve a language is regular - Regular language of 0's and 1'sDoes this proof work for infinite regular languagesProving that language is regular or not













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










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
    7 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










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
    7 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










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






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.









share|cite|improve this question




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.









asked 16 hours ago









ShukieShukie

61




61




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.





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.






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












  • $begingroup$
    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
    7 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
    7 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
7 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
7 hours ago












2 Answers
2






active

oldest

votes


















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









    $endgroup$














      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "419"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: false,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: null,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });






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










      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcs.stackexchange.com%2fquestions%2f106556%2fwhy-is-this-basic-language-not-a-regular-language%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      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 15 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 10 hours ago









                  DraconisDraconis

                  5,762921




                  5,762921






















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










                      draft saved

                      draft discarded


















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













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












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
















                      Thanks for contributing an answer to Computer Science Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcs.stackexchange.com%2fquestions%2f106556%2fwhy-is-this-basic-language-not-a-regular-language%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      Popular posts from this blog

                      El tren de la libertad Índice Antecedentes "Porque yo decido" Desarrollo de la...

                      Castillo d'Acher Características Menú de navegación

                      Puerta de Hutt Referencias Enlaces externos Menú de navegación15°58′00″S 5°42′00″O /...