$L_4 = {x: #_{1}(x) = 2 cdot #_{10}(x) }$ Find CFG given hints











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Attempt:



$S to A_{00}SA_{11}$



$A_{00} to 0, 0A_{00}, 0A_{10}$



$A_{01} to A_{00}1, A_{00}A_{11}, A_{01}1, A_{00}1$



$A_{10} to 1A_{10}, A_{10}0, 1A_{00}$



$A_{11} to 1, 1A_{11}, 1A_{01}$



Not sure what I'm doing. Any help is appreciated










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    up vote
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    down vote

    favorite
    1












    enter image description here



    Attempt:



    $S to A_{00}SA_{11}$



    $A_{00} to 0, 0A_{00}, 0A_{10}$



    $A_{01} to A_{00}1, A_{00}A_{11}, A_{01}1, A_{00}1$



    $A_{10} to 1A_{10}, A_{10}0, 1A_{00}$



    $A_{11} to 1, 1A_{11}, 1A_{01}$



    Not sure what I'm doing. Any help is appreciated










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      up vote
      2
      down vote

      favorite
      1









      up vote
      2
      down vote

      favorite
      1






      1





      enter image description here



      Attempt:



      $S to A_{00}SA_{11}$



      $A_{00} to 0, 0A_{00}, 0A_{10}$



      $A_{01} to A_{00}1, A_{00}A_{11}, A_{01}1, A_{00}1$



      $A_{10} to 1A_{10}, A_{10}0, 1A_{00}$



      $A_{11} to 1, 1A_{11}, 1A_{01}$



      Not sure what I'm doing. Any help is appreciated










      share|cite|improve this question













      enter image description here



      Attempt:



      $S to A_{00}SA_{11}$



      $A_{00} to 0, 0A_{00}, 0A_{10}$



      $A_{01} to A_{00}1, A_{00}A_{11}, A_{01}1, A_{00}1$



      $A_{10} to 1A_{10}, A_{10}0, 1A_{00}$



      $A_{11} to 1, 1A_{11}, 1A_{01}$



      Not sure what I'm doing. Any help is appreciated







      context-free-grammar






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      asked Nov 12 at 6:48









      Tree Garen

      33419




      33419






















          1 Answer
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          It's important to remember that each production rule must produce a string in $L_4$ i.e. a string that has exactly half of its $1$'s followed by a $0$.



          For $A_{00}$, we must produce every string that begins and ends with $0$ and that has exactly half of its $1$'s followed by a $0$. Clearly $0$ meets this criteria. Consider a string produced by $A_{00}$. What is it's second character? It could be $1$ or $0$. If it is $1$, then we can write the string as $0A_{10}$. If it is $0$, then we can write the string as $0A_{00}$.
          $$
          A_{00}longrightarrow 0, text{ } 0A_{00}, text{ } 0A_{10}
          $$



          For $A_{01}$, the same logic applies based off of the second character.
          $$
          A_{01}longrightarrow 0A_{01}, text{ } 0A_{11}
          $$



          It is more complicated for $A_{10}$ and $A_{11}$.



          For $A_{10}$, consider the second to last element of a string produced by $A_{10}$. If it is $0$, then our string is of the form $1...00$ which could then be rewritten $A_{10}0$. Otherwise it is of the form $1...10$. We will split this again into the case $11...10$ and $10...10$ and make arguments about each separately.



          A string in our language of the form $10...10$ must be of the form $A_{11}A_{10}$. The simplest way to show this (that I can think of) is through the Intermediate Value Theorem. I can provide further proof upon request. Our other case is strings of the form $11...10$ which we will further break down into $110...10$, which can be produced by $11A_{00}$, and $111...10$ which can be produced by $11A_{11}0$.
          $$
          A_{10}longrightarrow A_{10}0, text{ } 11A_{00}, text{ } 11A_{11}0, text{ }A_{11}A_{10}
          $$



          For $A_{11}$, we will break down into strings of the form $11...1$ and $10...1$. Nearly the same proof for strings of the form $10...10$ shows that strings of the form $11...1$ can be produced by $A_{10}A_{11}$. We can break down $10...1$ into $10...01$, which can be represented by $1A_{00}1$, and $10...11$, which can be produced by $1A_{01}1$.
          $$
          A_{11}longrightarrow 1A_{00}1, text{ } 1A_{01}1, text{ } A_{10}A_{11}
          $$






          share|cite|improve this answer





















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            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            0
            down vote



            accepted










            It's important to remember that each production rule must produce a string in $L_4$ i.e. a string that has exactly half of its $1$'s followed by a $0$.



            For $A_{00}$, we must produce every string that begins and ends with $0$ and that has exactly half of its $1$'s followed by a $0$. Clearly $0$ meets this criteria. Consider a string produced by $A_{00}$. What is it's second character? It could be $1$ or $0$. If it is $1$, then we can write the string as $0A_{10}$. If it is $0$, then we can write the string as $0A_{00}$.
            $$
            A_{00}longrightarrow 0, text{ } 0A_{00}, text{ } 0A_{10}
            $$



            For $A_{01}$, the same logic applies based off of the second character.
            $$
            A_{01}longrightarrow 0A_{01}, text{ } 0A_{11}
            $$



            It is more complicated for $A_{10}$ and $A_{11}$.



            For $A_{10}$, consider the second to last element of a string produced by $A_{10}$. If it is $0$, then our string is of the form $1...00$ which could then be rewritten $A_{10}0$. Otherwise it is of the form $1...10$. We will split this again into the case $11...10$ and $10...10$ and make arguments about each separately.



            A string in our language of the form $10...10$ must be of the form $A_{11}A_{10}$. The simplest way to show this (that I can think of) is through the Intermediate Value Theorem. I can provide further proof upon request. Our other case is strings of the form $11...10$ which we will further break down into $110...10$, which can be produced by $11A_{00}$, and $111...10$ which can be produced by $11A_{11}0$.
            $$
            A_{10}longrightarrow A_{10}0, text{ } 11A_{00}, text{ } 11A_{11}0, text{ }A_{11}A_{10}
            $$



            For $A_{11}$, we will break down into strings of the form $11...1$ and $10...1$. Nearly the same proof for strings of the form $10...10$ shows that strings of the form $11...1$ can be produced by $A_{10}A_{11}$. We can break down $10...1$ into $10...01$, which can be represented by $1A_{00}1$, and $10...11$, which can be produced by $1A_{01}1$.
            $$
            A_{11}longrightarrow 1A_{00}1, text{ } 1A_{01}1, text{ } A_{10}A_{11}
            $$






            share|cite|improve this answer

























              up vote
              0
              down vote



              accepted










              It's important to remember that each production rule must produce a string in $L_4$ i.e. a string that has exactly half of its $1$'s followed by a $0$.



              For $A_{00}$, we must produce every string that begins and ends with $0$ and that has exactly half of its $1$'s followed by a $0$. Clearly $0$ meets this criteria. Consider a string produced by $A_{00}$. What is it's second character? It could be $1$ or $0$. If it is $1$, then we can write the string as $0A_{10}$. If it is $0$, then we can write the string as $0A_{00}$.
              $$
              A_{00}longrightarrow 0, text{ } 0A_{00}, text{ } 0A_{10}
              $$



              For $A_{01}$, the same logic applies based off of the second character.
              $$
              A_{01}longrightarrow 0A_{01}, text{ } 0A_{11}
              $$



              It is more complicated for $A_{10}$ and $A_{11}$.



              For $A_{10}$, consider the second to last element of a string produced by $A_{10}$. If it is $0$, then our string is of the form $1...00$ which could then be rewritten $A_{10}0$. Otherwise it is of the form $1...10$. We will split this again into the case $11...10$ and $10...10$ and make arguments about each separately.



              A string in our language of the form $10...10$ must be of the form $A_{11}A_{10}$. The simplest way to show this (that I can think of) is through the Intermediate Value Theorem. I can provide further proof upon request. Our other case is strings of the form $11...10$ which we will further break down into $110...10$, which can be produced by $11A_{00}$, and $111...10$ which can be produced by $11A_{11}0$.
              $$
              A_{10}longrightarrow A_{10}0, text{ } 11A_{00}, text{ } 11A_{11}0, text{ }A_{11}A_{10}
              $$



              For $A_{11}$, we will break down into strings of the form $11...1$ and $10...1$. Nearly the same proof for strings of the form $10...10$ shows that strings of the form $11...1$ can be produced by $A_{10}A_{11}$. We can break down $10...1$ into $10...01$, which can be represented by $1A_{00}1$, and $10...11$, which can be produced by $1A_{01}1$.
              $$
              A_{11}longrightarrow 1A_{00}1, text{ } 1A_{01}1, text{ } A_{10}A_{11}
              $$






              share|cite|improve this answer























                up vote
                0
                down vote



                accepted







                up vote
                0
                down vote



                accepted






                It's important to remember that each production rule must produce a string in $L_4$ i.e. a string that has exactly half of its $1$'s followed by a $0$.



                For $A_{00}$, we must produce every string that begins and ends with $0$ and that has exactly half of its $1$'s followed by a $0$. Clearly $0$ meets this criteria. Consider a string produced by $A_{00}$. What is it's second character? It could be $1$ or $0$. If it is $1$, then we can write the string as $0A_{10}$. If it is $0$, then we can write the string as $0A_{00}$.
                $$
                A_{00}longrightarrow 0, text{ } 0A_{00}, text{ } 0A_{10}
                $$



                For $A_{01}$, the same logic applies based off of the second character.
                $$
                A_{01}longrightarrow 0A_{01}, text{ } 0A_{11}
                $$



                It is more complicated for $A_{10}$ and $A_{11}$.



                For $A_{10}$, consider the second to last element of a string produced by $A_{10}$. If it is $0$, then our string is of the form $1...00$ which could then be rewritten $A_{10}0$. Otherwise it is of the form $1...10$. We will split this again into the case $11...10$ and $10...10$ and make arguments about each separately.



                A string in our language of the form $10...10$ must be of the form $A_{11}A_{10}$. The simplest way to show this (that I can think of) is through the Intermediate Value Theorem. I can provide further proof upon request. Our other case is strings of the form $11...10$ which we will further break down into $110...10$, which can be produced by $11A_{00}$, and $111...10$ which can be produced by $11A_{11}0$.
                $$
                A_{10}longrightarrow A_{10}0, text{ } 11A_{00}, text{ } 11A_{11}0, text{ }A_{11}A_{10}
                $$



                For $A_{11}$, we will break down into strings of the form $11...1$ and $10...1$. Nearly the same proof for strings of the form $10...10$ shows that strings of the form $11...1$ can be produced by $A_{10}A_{11}$. We can break down $10...1$ into $10...01$, which can be represented by $1A_{00}1$, and $10...11$, which can be produced by $1A_{01}1$.
                $$
                A_{11}longrightarrow 1A_{00}1, text{ } 1A_{01}1, text{ } A_{10}A_{11}
                $$






                share|cite|improve this answer












                It's important to remember that each production rule must produce a string in $L_4$ i.e. a string that has exactly half of its $1$'s followed by a $0$.



                For $A_{00}$, we must produce every string that begins and ends with $0$ and that has exactly half of its $1$'s followed by a $0$. Clearly $0$ meets this criteria. Consider a string produced by $A_{00}$. What is it's second character? It could be $1$ or $0$. If it is $1$, then we can write the string as $0A_{10}$. If it is $0$, then we can write the string as $0A_{00}$.
                $$
                A_{00}longrightarrow 0, text{ } 0A_{00}, text{ } 0A_{10}
                $$



                For $A_{01}$, the same logic applies based off of the second character.
                $$
                A_{01}longrightarrow 0A_{01}, text{ } 0A_{11}
                $$



                It is more complicated for $A_{10}$ and $A_{11}$.



                For $A_{10}$, consider the second to last element of a string produced by $A_{10}$. If it is $0$, then our string is of the form $1...00$ which could then be rewritten $A_{10}0$. Otherwise it is of the form $1...10$. We will split this again into the case $11...10$ and $10...10$ and make arguments about each separately.



                A string in our language of the form $10...10$ must be of the form $A_{11}A_{10}$. The simplest way to show this (that I can think of) is through the Intermediate Value Theorem. I can provide further proof upon request. Our other case is strings of the form $11...10$ which we will further break down into $110...10$, which can be produced by $11A_{00}$, and $111...10$ which can be produced by $11A_{11}0$.
                $$
                A_{10}longrightarrow A_{10}0, text{ } 11A_{00}, text{ } 11A_{11}0, text{ }A_{11}A_{10}
                $$



                For $A_{11}$, we will break down into strings of the form $11...1$ and $10...1$. Nearly the same proof for strings of the form $10...10$ shows that strings of the form $11...1$ can be produced by $A_{10}A_{11}$. We can break down $10...1$ into $10...01$, which can be represented by $1A_{00}1$, and $10...11$, which can be produced by $1A_{01}1$.
                $$
                A_{11}longrightarrow 1A_{00}1, text{ } 1A_{01}1, text{ } A_{10}A_{11}
                $$







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                answered 16 hours ago









                Joey Kilpatrick

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