Regex for bit string containing at least 2 zeros but no consecutive zeros.












0












$begingroup$


This is what I have:



(1*011*011*)*



But I don't think this is accounting for an odd number of zeros, like "10101010101111". I think I have the right expression that satisfies no 2 consecutive zeros and even number of zeros.










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$endgroup$

















    0












    $begingroup$


    This is what I have:



    (1*011*011*)*



    But I don't think this is accounting for an odd number of zeros, like "10101010101111". I think I have the right expression that satisfies no 2 consecutive zeros and even number of zeros.










    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      This is what I have:



      (1*011*011*)*



      But I don't think this is accounting for an odd number of zeros, like "10101010101111". I think I have the right expression that satisfies no 2 consecutive zeros and even number of zeros.










      share|cite|improve this question









      $endgroup$




      This is what I have:



      (1*011*011*)*



      But I don't think this is accounting for an odd number of zeros, like "10101010101111". I think I have the right expression that satisfies no 2 consecutive zeros and even number of zeros.







      regular-expressions






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 3 '18 at 19:19









      pinklemonadepinklemonade

      827




      827






















          1 Answer
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          $begingroup$

          I'll get you started.



          You need to have two zeroes, so start by putting those in your regex, leaving space between, before and after:



          $$
          ______0______0______
          $$

          Now, consider the space on the left. Every zero in this region is followed by a $1$. Therefore, this portion can be broken into sums of $(01)$ and $1$, so it is $(01|1)^*$:
          $$
          (01|1)^*0______0______
          $$

          What about the space in the middle? It has similar rules, but we cannot quite use the same regex $(01|1)^*$ we used on the left. The problem is that this can begin with a zero, which would cause a $00$ with the first $0$ we placed. How can you fix this?






          share|cite|improve this answer









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            $begingroup$

            I'll get you started.



            You need to have two zeroes, so start by putting those in your regex, leaving space between, before and after:



            $$
            ______0______0______
            $$

            Now, consider the space on the left. Every zero in this region is followed by a $1$. Therefore, this portion can be broken into sums of $(01)$ and $1$, so it is $(01|1)^*$:
            $$
            (01|1)^*0______0______
            $$

            What about the space in the middle? It has similar rules, but we cannot quite use the same regex $(01|1)^*$ we used on the left. The problem is that this can begin with a zero, which would cause a $00$ with the first $0$ we placed. How can you fix this?






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              I'll get you started.



              You need to have two zeroes, so start by putting those in your regex, leaving space between, before and after:



              $$
              ______0______0______
              $$

              Now, consider the space on the left. Every zero in this region is followed by a $1$. Therefore, this portion can be broken into sums of $(01)$ and $1$, so it is $(01|1)^*$:
              $$
              (01|1)^*0______0______
              $$

              What about the space in the middle? It has similar rules, but we cannot quite use the same regex $(01|1)^*$ we used on the left. The problem is that this can begin with a zero, which would cause a $00$ with the first $0$ we placed. How can you fix this?






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                I'll get you started.



                You need to have two zeroes, so start by putting those in your regex, leaving space between, before and after:



                $$
                ______0______0______
                $$

                Now, consider the space on the left. Every zero in this region is followed by a $1$. Therefore, this portion can be broken into sums of $(01)$ and $1$, so it is $(01|1)^*$:
                $$
                (01|1)^*0______0______
                $$

                What about the space in the middle? It has similar rules, but we cannot quite use the same regex $(01|1)^*$ we used on the left. The problem is that this can begin with a zero, which would cause a $00$ with the first $0$ we placed. How can you fix this?






                share|cite|improve this answer









                $endgroup$



                I'll get you started.



                You need to have two zeroes, so start by putting those in your regex, leaving space between, before and after:



                $$
                ______0______0______
                $$

                Now, consider the space on the left. Every zero in this region is followed by a $1$. Therefore, this portion can be broken into sums of $(01)$ and $1$, so it is $(01|1)^*$:
                $$
                (01|1)^*0______0______
                $$

                What about the space in the middle? It has similar rules, but we cannot quite use the same regex $(01|1)^*$ we used on the left. The problem is that this can begin with a zero, which would cause a $00$ with the first $0$ we placed. How can you fix this?







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 3 '18 at 21:07









                Mike EarnestMike Earnest

                22k12051




                22k12051






























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