Given a Context-Free Grammar $G$ as follows: $S \rightarrow Aa \mid bAc \mid dc \mid bda$ $A \rightarrow d$ Which ONE of the following statements is TRUE?

Let us analyze the grammar $G$ by constructing the LR(0) and LR(1) items:

1. Augmented Grammar:

   1. $S^{\prime} \rightarrow S$

   2. $S \rightarrow Aa$

   3. $S \rightarrow bAc$

   4. $S \rightarrow dc$

   5. $S \rightarrow bda$

   6. $A \rightarrow d$

2. SLR(1) Analysis:

   • $\text{FOLLOW}(A)$: From $S \rightarrow Aa$, $a \in \text{FOLLOW}(A)$. From $S \rightarrow bAc$, $c \in \text{FOLLOW}(A)$. Hence, $\text{FOLLOW}(A) = \{a, c\}$.

   • In the initial canonical collection state $I_0 = \text{CLOSURE}(S^{\prime} \rightarrow \cdot S)$: $$\begin{aligned} S^{\prime} &\rightarrow \cdot S \\ S &\rightarrow \cdot Aa \\ S &\rightarrow \cdot bAc \\ S &\rightarrow \cdot dc \\ S &\rightarrow \cdot bda \\ A &\rightarrow \cdot d \end{aligned}$$

   • On input symbol $d$ from $I_0$, we transition to state $I_1$:

     $$I_1 = \text{GOTO}(I_0, d) = \{ S \rightarrow d \cdot c, \; A \rightarrow d \cdot \}$$

   • In state $I_1$, on terminal $c$:

     • There is a shift action due to $S \rightarrow d \cdot c$.

     • There is a reduce action due to $A \rightarrow d \cdot$ since $c \in \text{FOLLOW}(A)$.

   • This causes a Shift-Reduce (SR) conflict on symbol $c$. Therefore, the grammar $G$ is not SLR(1).

3. CLR(1) and LALR(1) Analysis:

   • In the LR(1) collection, the lookaheads are computed explicitly:

     • In $I_0$, the item $[A \rightarrow \cdot d, a]$ has lookahead $a$ because it is generated from [S \rightarrow \cdot Aa, \]$.

     • The item [S \rightarrow \cdot dc, \]$yields$[S \rightarrow d \cdot c, \$]$with a shift on$c$.

   • Thus, in the state for $\text{GOTO}(I_0, d)$:

     • Shift on $c$ for [S \rightarrow d \cdot c, \]$.

     • Reduce by $A \rightarrow d$ on lookahead $a$ for $[A \rightarrow d \cdot, a]$.

   • Since the lookaheads for shift and reduce are disjoint ($\{c\} \cap \{a\} = \emptyset$), there is no conflict.

   • Similarly, after transitioning on $b$, the state containing $[A \rightarrow \cdot d, c]$ has lookahead $c$ and [S \rightarrow bd \cdot a, \]$has shift on$a$, which are also conflict-free.

   • No merge conflicts arise when converting to LALR(1).

Hence, $G$ is LALR(1), not SLR(1), which matches option (C).