Let $L_4$ be the language over alphabet ${0, 1}$ defined by $L_3 = {x : #_1(x) = 2 cdot #_{10}(x)}$ design...











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Let $L_4$ be the language over alphabet ${0, 1}$ defined by
$L_4 = {x : #_1(x) = 2 cdot #_{10}(x)}$



Here is a design for a PDA that accepts $L_4$. See diagram below, where we use e to donote $epsilon$



-For a string $x$, we use k to stand for the quantity $#_1(x) - 2 cdot#_{10}(x)$



-If $k = 0$, then the stack should be empty



-If $k neq 0$, then the stack should have a $Y$ at the bottom and $|k| - 1$ Xs on top.



-$q_0, q_1$ are for $x$ where $k = 0$. $q_0$ is for $x$ that does not end with 1. $q_1$ is for $x$ that ends with $1.$



-$q_2, q_3, q_4$ are for $x$ where $k > 0$. $q_2$ is for $x$ that ends with $1.$ $q_3$ is an extra state to sllow for popping 2 items off the stack. $q_4$ is for x that ends with $0$.



$q_5, q_6, q_7$ are for x where $k < 0$. $q_5$ is for x that ends with $0$. $q_6$ is an extra state to allow for pushing 2 items on the stack. $q_7$ is for x that ends with $1$.



Complete the PDA by adding appropiate transitions. To help you get started initial state is indicated some transitions are given, and accepting states are indicated by double parentheses.enter image description here



My attempt:



enter image description here



Not sure how to really go about this. I tried to satisfy very easy cases like the empty string is already accepted. 101 is accepted. But keep ending up stuck because I need to have k-1 x's










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    Let $L_4$ be the language over alphabet ${0, 1}$ defined by
    $L_4 = {x : #_1(x) = 2 cdot #_{10}(x)}$



    Here is a design for a PDA that accepts $L_4$. See diagram below, where we use e to donote $epsilon$



    -For a string $x$, we use k to stand for the quantity $#_1(x) - 2 cdot#_{10}(x)$



    -If $k = 0$, then the stack should be empty



    -If $k neq 0$, then the stack should have a $Y$ at the bottom and $|k| - 1$ Xs on top.



    -$q_0, q_1$ are for $x$ where $k = 0$. $q_0$ is for $x$ that does not end with 1. $q_1$ is for $x$ that ends with $1.$



    -$q_2, q_3, q_4$ are for $x$ where $k > 0$. $q_2$ is for $x$ that ends with $1.$ $q_3$ is an extra state to sllow for popping 2 items off the stack. $q_4$ is for x that ends with $0$.



    $q_5, q_6, q_7$ are for x where $k < 0$. $q_5$ is for x that ends with $0$. $q_6$ is an extra state to allow for pushing 2 items on the stack. $q_7$ is for x that ends with $1$.



    Complete the PDA by adding appropiate transitions. To help you get started initial state is indicated some transitions are given, and accepting states are indicated by double parentheses.enter image description here



    My attempt:



    enter image description here



    Not sure how to really go about this. I tried to satisfy very easy cases like the empty string is already accepted. 101 is accepted. But keep ending up stuck because I need to have k-1 x's










    share|cite|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      Let $L_4$ be the language over alphabet ${0, 1}$ defined by
      $L_4 = {x : #_1(x) = 2 cdot #_{10}(x)}$



      Here is a design for a PDA that accepts $L_4$. See diagram below, where we use e to donote $epsilon$



      -For a string $x$, we use k to stand for the quantity $#_1(x) - 2 cdot#_{10}(x)$



      -If $k = 0$, then the stack should be empty



      -If $k neq 0$, then the stack should have a $Y$ at the bottom and $|k| - 1$ Xs on top.



      -$q_0, q_1$ are for $x$ where $k = 0$. $q_0$ is for $x$ that does not end with 1. $q_1$ is for $x$ that ends with $1.$



      -$q_2, q_3, q_4$ are for $x$ where $k > 0$. $q_2$ is for $x$ that ends with $1.$ $q_3$ is an extra state to sllow for popping 2 items off the stack. $q_4$ is for x that ends with $0$.



      $q_5, q_6, q_7$ are for x where $k < 0$. $q_5$ is for x that ends with $0$. $q_6$ is an extra state to allow for pushing 2 items on the stack. $q_7$ is for x that ends with $1$.



      Complete the PDA by adding appropiate transitions. To help you get started initial state is indicated some transitions are given, and accepting states are indicated by double parentheses.enter image description here



      My attempt:



      enter image description here



      Not sure how to really go about this. I tried to satisfy very easy cases like the empty string is already accepted. 101 is accepted. But keep ending up stuck because I need to have k-1 x's










      share|cite|improve this question













      Let $L_4$ be the language over alphabet ${0, 1}$ defined by
      $L_4 = {x : #_1(x) = 2 cdot #_{10}(x)}$



      Here is a design for a PDA that accepts $L_4$. See diagram below, where we use e to donote $epsilon$



      -For a string $x$, we use k to stand for the quantity $#_1(x) - 2 cdot#_{10}(x)$



      -If $k = 0$, then the stack should be empty



      -If $k neq 0$, then the stack should have a $Y$ at the bottom and $|k| - 1$ Xs on top.



      -$q_0, q_1$ are for $x$ where $k = 0$. $q_0$ is for $x$ that does not end with 1. $q_1$ is for $x$ that ends with $1.$



      -$q_2, q_3, q_4$ are for $x$ where $k > 0$. $q_2$ is for $x$ that ends with $1.$ $q_3$ is an extra state to sllow for popping 2 items off the stack. $q_4$ is for x that ends with $0$.



      $q_5, q_6, q_7$ are for x where $k < 0$. $q_5$ is for x that ends with $0$. $q_6$ is an extra state to allow for pushing 2 items on the stack. $q_7$ is for x that ends with $1$.



      Complete the PDA by adding appropiate transitions. To help you get started initial state is indicated some transitions are given, and accepting states are indicated by double parentheses.enter image description here



      My attempt:



      enter image description here



      Not sure how to really go about this. I tried to satisfy very easy cases like the empty string is already accepted. 101 is accepted. But keep ending up stuck because I need to have k-1 x's







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