is this language regular and why pumping lemma doesn't work?

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I was told that this language is regular but as I can show below, pumping lemma is not working for it. What am I doing wrong? Is this language really regular? Why?
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formal-languages regular-languages pumping-lemma







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  • Note that you can use LaTeX here to typeset mathematics in a more readable way. See here for a short introduction.
    – Raphael♦
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up vote
1
down vote

favorite












I was told that this language is regular but as I can show below, pumping lemma is not working for it. What am I doing wrong? Is this language really regular? Why?
enter image description here






formal-languages regular-languages pumping-lemma







share|cite|improve this question























  • Note that you can use LaTeX here to typeset mathematics in a more readable way. See here for a short introduction.
    – Raphael♦
    12 mins ago












up vote
1
down vote

favorite









up vote
1
down vote

favorite











I was told that this language is regular but as I can show below, pumping lemma is not working for it. What am I doing wrong? Is this language really regular? Why?
enter image description here






formal-languages regular-languages pumping-lemma







share|cite|improve this question















I was told that this language is regular but as I can show below, pumping lemma is not working for it. What am I doing wrong? Is this language really regular? Why?
enter image description here






formal-languages regular-languages pumping-lemma




formal-languages regular-languages pumping-lemma






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edited 12 mins ago









Raphael♦

56.2k22138304




56.2k22138304










asked 2 hours ago









aky

212




212











  • Note that you can use LaTeX here to typeset mathematics in a more readable way. See here for a short introduction.
    – Raphael♦
    12 mins ago
















  • Note that you can use LaTeX here to typeset mathematics in a more readable way. See here for a short introduction.
    – Raphael♦
    12 mins ago















Note that you can use LaTeX here to typeset mathematics in a more readable way. See here for a short introduction.
– Raphael♦
12 mins ago




Note that you can use LaTeX here to typeset mathematics in a more readable way. See here for a short introduction.
– Raphael♦
12 mins ago










2 Answers
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Write the word $s'$ as



$$
s' = 0^(p-beta) left(1^p01^p0^beta right)0^(p -beta)
$$
to see that it is in fact in $L$.






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    up vote
    1
    down vote













    It's a "trick" question. The language is regular because
    beginalign*
    aba^mathrmRmid a,bin0,1^*
    &= bigvarepsilon bvarepsilon^mathrmRmid bin0,1^*big
    cup biga b a^mathrmRmid ain0,1^*, bin0,1^*big\
    &= 0,1^*
    cup biga b a^mathrmRmid ain0,1^*, bin0,1^*big\
    &= 0,1^*,.
    endalign*






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      2 Answers
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      2 Answers
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      up vote
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      Write the word $s'$ as



      $$
      s' = 0^(p-beta) left(1^p01^p0^beta right)0^(p -beta)
      $$
      to see that it is in fact in $L$.






      share|cite|improve this answer
























        up vote
        2
        down vote













        Write the word $s'$ as



        $$
        s' = 0^(p-beta) left(1^p01^p0^beta right)0^(p -beta)
        $$
        to see that it is in fact in $L$.






        share|cite|improve this answer






















          up vote
          2
          down vote










          up vote
          2
          down vote









          Write the word $s'$ as



          $$
          s' = 0^(p-beta) left(1^p01^p0^beta right)0^(p -beta)
          $$
          to see that it is in fact in $L$.






          share|cite|improve this answer












          Write the word $s'$ as



          $$
          s' = 0^(p-beta) left(1^p01^p0^beta right)0^(p -beta)
          $$
          to see that it is in fact in $L$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 1 hour ago









          Daniel Mroz

          3264




          3264




















              up vote
              1
              down vote













              It's a "trick" question. The language is regular because
              beginalign*
              aba^mathrmRmid a,bin0,1^*
              &= bigvarepsilon bvarepsilon^mathrmRmid bin0,1^*big
              cup biga b a^mathrmRmid ain0,1^*, bin0,1^*big\
              &= 0,1^*
              cup biga b a^mathrmRmid ain0,1^*, bin0,1^*big\
              &= 0,1^*,.
              endalign*






              share|cite|improve this answer
























                up vote
                1
                down vote













                It's a "trick" question. The language is regular because
                beginalign*
                aba^mathrmRmid a,bin0,1^*
                &= bigvarepsilon bvarepsilon^mathrmRmid bin0,1^*big
                cup biga b a^mathrmRmid ain0,1^*, bin0,1^*big\
                &= 0,1^*
                cup biga b a^mathrmRmid ain0,1^*, bin0,1^*big\
                &= 0,1^*,.
                endalign*






                share|cite|improve this answer






















                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  It's a "trick" question. The language is regular because
                  beginalign*
                  aba^mathrmRmid a,bin0,1^*
                  &= bigvarepsilon bvarepsilon^mathrmRmid bin0,1^*big
                  cup biga b a^mathrmRmid ain0,1^*, bin0,1^*big\
                  &= 0,1^*
                  cup biga b a^mathrmRmid ain0,1^*, bin0,1^*big\
                  &= 0,1^*,.
                  endalign*






                  share|cite|improve this answer












                  It's a "trick" question. The language is regular because
                  beginalign*
                  aba^mathrmRmid a,bin0,1^*
                  &= bigvarepsilon bvarepsilon^mathrmRmid bin0,1^*big
                  cup biga b a^mathrmRmid ain0,1^*, bin0,1^*big\
                  &= 0,1^*
                  cup biga b a^mathrmRmid ain0,1^*, bin0,1^*big\
                  &= 0,1^*,.
                  endalign*







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 30 mins ago









                  David Richerby

                  61.9k1595179




                  61.9k1595179



























                       

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