Consider the following grammar $G$ $$\begin{aligned} S &\rightarrow F \mid H F &\rightarrow p \mid c H &\rightarrow d \mid c \end{aligned}$$ where $S, F$, and $H$ are non-terminal symbols, $p, d$, and $c$ are terminal symbols. Which of the following statement(s) is/are correct? S1. LL(1) can parse all strings that are generated using grammar $G$ S2. LR(1) can parse all strings that are generated using grammar $G$

Consider the given grammar $G$:

$$\begin{aligned} S &\rightarrow F \mid H F &\rightarrow p \mid c H &\rightarrow d \mid c \end{aligned}$$

1. Ambiguity Analysis:

   • For the terminal string $c$, there are two distinct leftmost derivations:

     1. $S \Rightarrow F \Rightarrow c$

     2. $S \Rightarrow H \Rightarrow c$

   • Since the string $c$ has two distinct parse trees (and two distinct leftmost/rightmost derivations), the grammar $G$ is ambiguous.

2. Parsing with LL(1) and LR(1):

   • No ambiguous grammar can be LL(1) because there will be multiple entries in the LL(1) parsing table (FIRST/FIRST conflict: $\text{FIRST}(F) \cap \text{FIRST}(H) = \{c\}$). Thus, an LL(1) parser cannot deterministically parse all strings generated by $G$.

   • Similarly, no ambiguous grammar can be LR(1) (or LR($k$) for any $k$) because it leads to reduce/reduce conflicts in the LR state representing the reductions $F \rightarrow c$ and $H \rightarrow c$. Thus, an LR(1) parser cannot parse all strings generated by $G$.

Hence, neither S1 nor S2 is correct (Option D).