How to guess the end of pushdown automata by emptying the stack?












1












$begingroup$



Let language $L={sigma_1w_1csigma_2w_2c...sigma_nw_nc}$ where:
$$
1le n\
sigma_iin {a,b}\
w_iin {a,b}^+\
exists i:ile|w_i|
$$

and at least one condition from the two conditions below holds:



the first $i$ letters of $w_i$ form a different word from $(sigma_1sigma_2...sigma_i)^R$ OR $|w_i|=n$.



Build a pushdown automaton which receives $L$ by emptying its stack.




The solution of these two conditions separately is here: https://imgur.com/a/T9PiPgT.



As far as I understand in the automata for both condition 1 and condion 2 ($|w_i|=n$) essentially:




  1. We build an automaton which simply describes reading $sigma, w_i, c$ first.

  2. Then we guess the $w_i$ which is according to condition 1 or 2.

  3. Then we read the $sigma, c$ surrounding the guessed $w_i$ and the $w_i$ itself.


I'm new to pushdown automata, but it looks like a lot of automata could be solved in such a generic way:




  1. Read the words according to some logic described in the problem.

  2. guess some "good" word (by using $epsilon$ connection) and delete it from the stack.


Am I missing something?



Also I don't understand why we need $2$ cases in the solution for each condition. The solutions for each condition look almost identical.



Lastly, I don't understand why in the automaton for the condition $1$ in state $q_5$ we need to read the end of $w_i$ first, then read the "good" $w_i$. Why can't we read the "good" $w_i$ straight after the guess?










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$endgroup$

















    1












    $begingroup$



    Let language $L={sigma_1w_1csigma_2w_2c...sigma_nw_nc}$ where:
    $$
    1le n\
    sigma_iin {a,b}\
    w_iin {a,b}^+\
    exists i:ile|w_i|
    $$

    and at least one condition from the two conditions below holds:



    the first $i$ letters of $w_i$ form a different word from $(sigma_1sigma_2...sigma_i)^R$ OR $|w_i|=n$.



    Build a pushdown automaton which receives $L$ by emptying its stack.




    The solution of these two conditions separately is here: https://imgur.com/a/T9PiPgT.



    As far as I understand in the automata for both condition 1 and condion 2 ($|w_i|=n$) essentially:




    1. We build an automaton which simply describes reading $sigma, w_i, c$ first.

    2. Then we guess the $w_i$ which is according to condition 1 or 2.

    3. Then we read the $sigma, c$ surrounding the guessed $w_i$ and the $w_i$ itself.


    I'm new to pushdown automata, but it looks like a lot of automata could be solved in such a generic way:




    1. Read the words according to some logic described in the problem.

    2. guess some "good" word (by using $epsilon$ connection) and delete it from the stack.


    Am I missing something?



    Also I don't understand why we need $2$ cases in the solution for each condition. The solutions for each condition look almost identical.



    Lastly, I don't understand why in the automaton for the condition $1$ in state $q_5$ we need to read the end of $w_i$ first, then read the "good" $w_i$. Why can't we read the "good" $w_i$ straight after the guess?










    share|cite|improve this question









    $endgroup$















      1












      1








      1


      0



      $begingroup$



      Let language $L={sigma_1w_1csigma_2w_2c...sigma_nw_nc}$ where:
      $$
      1le n\
      sigma_iin {a,b}\
      w_iin {a,b}^+\
      exists i:ile|w_i|
      $$

      and at least one condition from the two conditions below holds:



      the first $i$ letters of $w_i$ form a different word from $(sigma_1sigma_2...sigma_i)^R$ OR $|w_i|=n$.



      Build a pushdown automaton which receives $L$ by emptying its stack.




      The solution of these two conditions separately is here: https://imgur.com/a/T9PiPgT.



      As far as I understand in the automata for both condition 1 and condion 2 ($|w_i|=n$) essentially:




      1. We build an automaton which simply describes reading $sigma, w_i, c$ first.

      2. Then we guess the $w_i$ which is according to condition 1 or 2.

      3. Then we read the $sigma, c$ surrounding the guessed $w_i$ and the $w_i$ itself.


      I'm new to pushdown automata, but it looks like a lot of automata could be solved in such a generic way:




      1. Read the words according to some logic described in the problem.

      2. guess some "good" word (by using $epsilon$ connection) and delete it from the stack.


      Am I missing something?



      Also I don't understand why we need $2$ cases in the solution for each condition. The solutions for each condition look almost identical.



      Lastly, I don't understand why in the automaton for the condition $1$ in state $q_5$ we need to read the end of $w_i$ first, then read the "good" $w_i$. Why can't we read the "good" $w_i$ straight after the guess?










      share|cite|improve this question









      $endgroup$





      Let language $L={sigma_1w_1csigma_2w_2c...sigma_nw_nc}$ where:
      $$
      1le n\
      sigma_iin {a,b}\
      w_iin {a,b}^+\
      exists i:ile|w_i|
      $$

      and at least one condition from the two conditions below holds:



      the first $i$ letters of $w_i$ form a different word from $(sigma_1sigma_2...sigma_i)^R$ OR $|w_i|=n$.



      Build a pushdown automaton which receives $L$ by emptying its stack.




      The solution of these two conditions separately is here: https://imgur.com/a/T9PiPgT.



      As far as I understand in the automata for both condition 1 and condion 2 ($|w_i|=n$) essentially:




      1. We build an automaton which simply describes reading $sigma, w_i, c$ first.

      2. Then we guess the $w_i$ which is according to condition 1 or 2.

      3. Then we read the $sigma, c$ surrounding the guessed $w_i$ and the $w_i$ itself.


      I'm new to pushdown automata, but it looks like a lot of automata could be solved in such a generic way:




      1. Read the words according to some logic described in the problem.

      2. guess some "good" word (by using $epsilon$ connection) and delete it from the stack.


      Am I missing something?



      Also I don't understand why we need $2$ cases in the solution for each condition. The solutions for each condition look almost identical.



      Lastly, I don't understand why in the automaton for the condition $1$ in state $q_5$ we need to read the end of $w_i$ first, then read the "good" $w_i$. Why can't we read the "good" $w_i$ straight after the guess?







      proof-explanation formal-languages automata






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      asked Dec 15 '18 at 8:41









      YosYos

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