Hi I am trying to solve a LL(1) form question for first n follow rule The question is


What I have done to eliminate left recursion

C::= j

Is it correct?

Thanks guys, it has been solved.


closed as unclear what you're asking by D.W., FrankW, vonbrand, Juho, Artem Kaznatcheev Mar 27 '14 at 7:54

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    $\begingroup$ This question appears to be unsuited for this site because questions of the form: "This is the exercise problem, this is my solution. Please grade!" are not interesting for anyone but you. Please see this related meta discussion. If you want to ask a specific question about a specific part of your attempt, please edit the question accordingly and it may be reopened. Otherwise, you might want to visit Computer Science Chat and get some feedback there. $\endgroup$ – Raphael Mar 27 '14 at 8:38

The only left-recursive rule is the one for $B$. The nonterminal $A$ produces no problems, $\operatorname{first}(B) = \{ e \}$ while $\operatorname{first}(C) = \{ g, j \}$. So you need to eliminate left recursion for $B$:

\begin{align} B &::= e f B' \\ B' &::= d B' \mid \epsilon \end{align}

  • $\begingroup$ So the question is right then what would be first(A)?? It would be first(B)+first(C) ?? $\endgroup$ – Kunj Mar 13 '14 at 19:18
  • $\begingroup$ $\operatorname{first}(A) = \operatorname{first}(BC) \cup \operatorname{first}(C) = \operatorname{first}(B) \cup \operatorname{first}(C) = \{d, e, g, h \}$ $\endgroup$ – vonbrand Mar 13 '14 at 19:23
  • $\begingroup$ confused. as first(B)={e} while first(C)={g,j} then where does {d,h} comes from $\endgroup$ – Kunj Mar 13 '14 at 19:33
  • $\begingroup$ Messed up, sorry (can't edit a comment and see the question) $\endgroup$ – vonbrand Mar 13 '14 at 19:42
  • $\begingroup$ so it should be First(A) = {e,g,j} right? $\endgroup$ – Kunj Mar 13 '14 at 19:44

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