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For the grammar below, a partial $LL(1)$ parsing table is also presented along with the grammar. Entries that need to be filled are indicated as $E1, E2,$ and $E3$. $\varepsilon$ is the empty string, \$ indicates end of input, and, $\mid$ separates alternate right hand sides of productions.

  • $ S\rightarrow a A b B \mid b A a B \mid \varepsilon $
  • $ A\rightarrow S $
  • $ B\rightarrow S $

$$\begin{array}{|l|l|}\hline \text{}  &  \textbf{a} & \textbf{b}  &  \textbf{\$} \\\hline  \text{$S$} & \text{E1} &  \text{E2}  &  \text{$S\rightarrow \varepsilon $} \\\hline  \text{$A$} & \text{$A\rightarrow S$} &  \text{$A\rightarrow S$}  &  \text{error} \\\hline  \text{$B$} & \text{$B\rightarrow S$} & \text{$B\rightarrow S$}  &  \text{$E3$} \\\hline \end{array}$$

The FIRST and FOLLOW sets for the non-terminals $A$ and $B$ are

  1. $ \text{FIRST}(A) = \{a, b, \varepsilon\} =\text{FIRST}(B) $
    $ \text{FOLLOW}(A) = \{a, b\} $
    $ \text{FOLLOW}(B) = \{a, b, \$\} $
     
  2. $ \text{FIRST}(A) = \{a, b, \$\} $
    $ \text{FIRST}(B) = \{a, b, \varepsilon\} $
    $ \text{FOLLOW}(A) = \{a, b\} $
    $ \text{FOLLOW}(B) = \{\$\} $
     
  3. $ \text{FIRST}(A) = \{a, b, \varepsilon\} =\text{FIRST}(B) $
    $ \text{FOLLOW}(A) =\{a, b\} $
    $ \text{FOLLOW}(B) = \varnothing $
     
  4. $ \text{FIRST}(A) = \{a, b\} = \text{FIRST}(B) $
    $ \text{FOLLOW}(A) = \{a, b\} $
    $ \text{FOLLOW}(B) =\{a, b\} $
asked in Compiler Design by Boss (16k points)
edited by | 1.5k views

3 Answers

+13 votes
Best answer
  • $\text{First}(S) = \text{First}(A) = \text{First}(B) = \{a,b,\epsilon\}$
  • $\text{Follow}(A) = \{a,b\}$
  • $\text{Follow}(B) = \text{Follow}(S) = \{a,b,\$\}$

So, the answer to question 52 is option A.

answered by Boss (30.9k points)
edited by
+1
How can we now that the production to entered in E3 is B=>S and not B=>∊ ? Precisely, how do we decide to be inser B=>∊ versus some other production. For S at M[S, $], the production entered is S=>∊. What is the difference?
+5 votes
Q 52 : ans is A.

first we can easly find.

follow of A = {a,b}, all symbols appeared after A in RHS

follow of B = {a,b,\$}, as B is at the end in RHS, follow(B) = follow (S) and S is also at the end in RHS so follow(S) = follow(A) = {a,b} and S is start symbol, so its follow contains \$ also.

hence, follow(B) = {a,b,\$}

Q 53.

ans is C.
answered by Loyal (8.1k points)
–2 votes
Q52 answer - D

Q53 answer - A
answered by Loyal (8.7k points)
Answer:

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