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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. $\text{FIRST}(A) = \{a, b\} = \text{FIRST}(B)$
$\text{FOLLOW}(A) = \{a, b\}$
$\text{FOLLOW}(B) =\{a, b\}$
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• $\text{First}(S) = \text{First}(A) = \text{First}(B) = \{a,b,\epsilon\}$
• $\text{Follow}(A) = \{a,b\}$
• $\text{Follow}(B) = \text{Follow}(S) = \{a,b,\$\}$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?
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.

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