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











up vote
2
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










share|cite|improve this question


























    up vote
    2
    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










    share|cite|improve this question
























      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






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 12 at 6:48









      Tree Garen

      33419




      33419






















          1 Answer
          1






          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





















            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: "69"
            };
            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',
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            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
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














             

            draft saved


            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2994957%2fl-4-x-1x-2-cdot-10x-find-cfg-given-hints%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            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 $1A_{01}1$.
                $$
                A_{11}longrightarrow 1A_{00}1, text{ } 1A_{01}1, text{ } A_{10}A_{11}
                $$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 16 hours ago









                Joey Kilpatrick

                65418




                65418






























                     

                    draft saved


                    draft discarded



















































                     


                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2994957%2fl-4-x-1x-2-cdot-10x-find-cfg-given-hints%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

                    AnyDesk - Fatal Program Failure

                    QoS: MAC-Priority for clients behind a repeater

                    Актюбинская область