Strong LL(1) grammar check

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I need to find rules that don't satisfy strong LL(1) rule:




$First_1(alpha*Follow_1(A)) cap First_1(beta*Follow_1(A)) = varnothing$



For any rule $A implies alpha | beta$




I have the following grammar:



$S implies aBA | BCA | epsilon$



$B implies bB | epsilon$



$A implies bB$



$C implies aS | BBS$



I've calculated First and Follow sets:



$First_1(S) = a, epsilon, b $



$First_1(B) = b, epsilon $



$First_1(A) = b $



$First_1(C) = a, epsilon, b $



$Follow_1(S) = $, b $



$Follow_1(B) = $, b, a $



$Follow_1(A) = $, b $



$Follow_1(C) = b $



After that I've started to compute strong LL(1) formula, but have some difficulties with multiplication:



For the rules $S implies aBA, S implies BCA$



$First_1(aBA*Follow_1(S)) cap First_1(BCA * Follow_1(S)) = \ (First_1(aBA) * Follow_1(S)) cap (First_1(BCA) * Follow_1(S)) = \ (a * epsilon, b) cap (a, epsilon, b * epsilon, b). $



From this point I don't know what is the result of those sets multiplications, I've seen the example of usage of this formula with LL(2) grammar and in the end there was $aa, ba * b$ which gave the result $ba, bb$, but for the LL(1) I'm not sure how to correct calculate it.




formal-languages context-free-grammar




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    up vote
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    down vote

    favorite












    I need to find rules that don't satisfy strong LL(1) rule:




    $First_1(alpha*Follow_1(A)) cap First_1(beta*Follow_1(A)) = varnothing$



    For any rule $A implies alpha | beta$




    I have the following grammar:



    $S implies aBA | BCA | epsilon$



    $B implies bB | epsilon$



    $A implies bB$



    $C implies aS | BBS$



    I've calculated First and Follow sets:



    $First_1(S) = a, epsilon, b $



    $First_1(B) = b, epsilon $



    $First_1(A) = b $



    $First_1(C) = a, epsilon, b $



    $Follow_1(S) = $, b $



    $Follow_1(B) = $, b, a $



    $Follow_1(A) = $, b $



    $Follow_1(C) = b $



    After that I've started to compute strong LL(1) formula, but have some difficulties with multiplication:



    For the rules $S implies aBA, S implies BCA$



    $First_1(aBA*Follow_1(S)) cap First_1(BCA * Follow_1(S)) = \ (First_1(aBA) * Follow_1(S)) cap (First_1(BCA) * Follow_1(S)) = \ (a * epsilon, b) cap (a, epsilon, b * epsilon, b). $



    From this point I don't know what is the result of those sets multiplications, I've seen the example of usage of this formula with LL(2) grammar and in the end there was $aa, ba * b$ which gave the result $ba, bb$, but for the LL(1) I'm not sure how to correct calculate it.




    formal-languages context-free-grammar




    share|cite|improve this question





















      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      I need to find rules that don't satisfy strong LL(1) rule:




      $First_1(alpha*Follow_1(A)) cap First_1(beta*Follow_1(A)) = varnothing$



      For any rule $A implies alpha | beta$




      I have the following grammar:



      $S implies aBA | BCA | epsilon$



      $B implies bB | epsilon$



      $A implies bB$



      $C implies aS | BBS$



      I've calculated First and Follow sets:



      $First_1(S) = a, epsilon, b $



      $First_1(B) = b, epsilon $



      $First_1(A) = b $



      $First_1(C) = a, epsilon, b $



      $Follow_1(S) = $, b $



      $Follow_1(B) = $, b, a $



      $Follow_1(A) = $, b $



      $Follow_1(C) = b $



      After that I've started to compute strong LL(1) formula, but have some difficulties with multiplication:



      For the rules $S implies aBA, S implies BCA$



      $First_1(aBA*Follow_1(S)) cap First_1(BCA * Follow_1(S)) = \ (First_1(aBA) * Follow_1(S)) cap (First_1(BCA) * Follow_1(S)) = \ (a * epsilon, b) cap (a, epsilon, b * epsilon, b). $



      From this point I don't know what is the result of those sets multiplications, I've seen the example of usage of this formula with LL(2) grammar and in the end there was $aa, ba * b$ which gave the result $ba, bb$, but for the LL(1) I'm not sure how to correct calculate it.




      formal-languages context-free-grammar




      share|cite|improve this question











      I need to find rules that don't satisfy strong LL(1) rule:




      $First_1(alpha*Follow_1(A)) cap First_1(beta*Follow_1(A)) = varnothing$



      For any rule $A implies alpha | beta$




      I have the following grammar:



      $S implies aBA | BCA | epsilon$



      $B implies bB | epsilon$



      $A implies bB$



      $C implies aS | BBS$



      I've calculated First and Follow sets:



      $First_1(S) = a, epsilon, b $



      $First_1(B) = b, epsilon $



      $First_1(A) = b $



      $First_1(C) = a, epsilon, b $



      $Follow_1(S) = $, b $



      $Follow_1(B) = $, b, a $



      $Follow_1(A) = $, b $



      $Follow_1(C) = b $



      After that I've started to compute strong LL(1) formula, but have some difficulties with multiplication:



      For the rules $S implies aBA, S implies BCA$



      $First_1(aBA*Follow_1(S)) cap First_1(BCA * Follow_1(S)) = \ (First_1(aBA) * Follow_1(S)) cap (First_1(BCA) * Follow_1(S)) = \ (a * epsilon, b) cap (a, epsilon, b * epsilon, b). $



      From this point I don't know what is the result of those sets multiplications, I've seen the example of usage of this formula with LL(2) grammar and in the end there was $aa, ba * b$ which gave the result $ba, bb$, but for the LL(1) I'm not sure how to correct calculate it.





      formal-languages context-free-grammar





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      asked Jul 30 at 11:33









      Kevin

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