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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. •$S\rightarrow a A b B \mid b A a B \mid \varepsilon$•$A\rightarrow S$•$B\rightarrow S$a b$
$S$ E1 E2 $S\rightarrow \varepsilon$
$A$ $A\rightarrow S$ $A\rightarrow S$ error
$B$ $B\rightarrow S$ $B\rightarrow S$ E3

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\}$asked edited | 1.2k views 3 Answers +13 votes Best answer$\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.

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.

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