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Task Build the Parse Table

Let $\mathcal{G}$ be the following context-free grammar: $$ \begin{aligned} E &\to TE' \\ E' &\to +TE'\; |\; \epsilon \\ T &\to F T' \; \\ T' &\to *FT'|\;\;\epsilon \\F &\to (E)\; | \;id \\ \end{aligned} $$

I calculated $ FIRST $ and $ FOLLOW $ sets \begin{aligned} FIRST(F) = FIRST(T) = FIRST(E) = \{ (,id \} \\ FIRST(T') = \{*,\epsilon\}\\ FIRST(E') = \{+,\epsilon\} \\FOLLOW(E) = FOLLOW(E') = \{ ),$ \} \\ FOLLOW(T)=FOLLOW(T') = \{+,),$\}\\ FOLLOW(F) = \{ +, * ,$ , )\} \end{aligned}

Here is the rule while constructing parse table\ $$For\; each\; terminal\; a\; in\; FIRST(A)\;,\; add \; A \to \alpha; to\; M[A,a]$$

When applying this rule to the production $$F\to(E)$$ I need to add this to $M[F,id]$ and $M[F,(]$ so when applying this rule again to $F\to id$, Should I need to add again to $M[F,id]$? Then ; It; is not LL(1) ]; grammar right? But it is given LL(1) . How , any mistake I have done?

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  • $\begingroup$ It's a long time ago, but I thought F -> (E) together with the rule that F is reachable means E can be followed by ). $\endgroup$
    – gnasher729
    Aug 9, 2022 at 11:13
  • $\begingroup$ Yeah, Thanks for pointing it out I've edited it, I made an error while typing! $\endgroup$
    – bigstreet
    Aug 9, 2022 at 11:26

1 Answer 1

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You have missed one important note. The rule is such

For each production $A→α$ of the grammar, do the following:
For each terminal $a$ in $FIRST(A)$, add $A→α;$ to $M[A,a]$

that is we only consider the elements in first set which is contributed by the production $A→α$ which here is $F→(E)$ ,

so for this rule ie $F→(E)$ the $FIRST(F) = \{(\}$ only

and for $F→id$ , $FIRST(F) = \{id\}$

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