Consider two grammars $G_1$ and $G_2$ with the production rules given below: $G_1: S \to if\ E\ then\ S \mid if\ E\ then\ S\ else\ S \mid a$ $E \to b$ $G_2: S \to if\ E\ then\ S \mid M$ $M \to if\ E\ then\ M\ else\ S \mid c$ $E \to b$ where $if, then, else, a, b, c$ are the terminals. Which of the following option(s) is/are CORRECT?

### Analysis of Grammar $G_1$:

• $G_1$ is the classic "dangling-else" grammar.

• For the string if b then if b then a else a, there are two distinct leftmost derivations / parse trees:

  1. The else associates with the inner if.

  2. The else associates with the outer if.

• Since $G_1$ generates a string with more than one parse tree, $G_1$ is ambiguous [46].

• Any ambiguous grammar cannot be $LL(1)$, so $G_1$ is not $LL(1)$.

### Analysis of Grammar $G_2$:

• Consider the string if b then if b then c else if b then c:

  • Parse Tree 1:

    $$S \implies if\ E\ then\ S \implies if\ b\ then\ M \implies if\ b\ then\ if\ E\ then\ M\ else\ S \implies if\ b\ then\ if\ b\ then\ c\ else\ (if\ E\ then\ S) \dots$$

  • Parse Tree 2:

    $$S \implies if\ E\ then\ S \implies if\ b\ then\ (if\ E\ then\ S) \implies if\ b\ then\ if\ b\ then\ M \implies if\ b\ then\ if\ b\ then\ if\ E\ then\ M\ else\ S \dots$$

• Alternatively, consider the non-terminal $S \to if\ E\ then\ S \mid M$. The non-terminal $M$ has production $M \to if\ E\ then\ M\ else\ S \mid c$. Both productions from $S$ can generate sentences starting with if, causing First/First and Follow conflicts.

• More directly, $G_2$ does not unambiguously restrict the dangling-else structure because the else-branch in $M$ produces $S$, which allows un-matched if statements inside it to cause ambiguity in nested conditionals.

• Hence, $G_2$ is also ambiguous and consequently not $LL(1)$ [46].

Thus:

• Option (C) is CORRECT ($G_1$ and $G_2$ are not $LL(1)$) [46].

• Option (D) is CORRECT ($G_1$ and $G_2$ are ambiguous) [46].