Given a context-free grammer G as follows:

S→Aa \ bac | dc | dba

A→d

which ONE of the following statement is TRUE?

The correct answer is: G is LALR(1) , not SLR(1)

To understand why this is the correct statement, let us analyze the given context-free grammar $G$ by constructing its parsing states and examining the conditions for both SLR(1) and LALR(1) parsers.

The grammar rules are:
(1) $S\to Aa$
(2) $S\to bac$
(3) $S\to dc$
(4) $S\to dba$
(5) $A\to d$

1. Augmenting the Grammar
We add a new start symbol $S^{\prime }$ and the production:
$S^{\prime }\to S$

2. LR(0) State Construction
Let us construct the LR(0) states starting from the initial state $I_0$:

State $I_0$:
$S^{\prime }\to ·S$
$S\to ·Aa$
$S\to ·bac$
$S\to ·dc$
$S\to ·dba$
$A\to ·d$

Applying the transition on the terminal symbol $d$ from $I_0$ leads to state $I_4$:

State $I_4$ = Goto($I_0$, $d$):
$S\to d·c$
$S\to d·ba$
$A\to d·$

3. Follow Set Analysis
To determine if the grammar is SLR(1), we compute the FOLLOW set for the non-terminal $A$:
Since $A$ only appears in the production $S\to Aa$, we have:
$\text{FOLLOW}\left(A\right)=\left\{a\right\}$

4. Analyzing Conflicts
In state $I_4$:
- We can shift on input $c$ (moving to state $S\to dc·$).
- We can shift on input $b$ (moving to state $S\to db·a$).
- We reduce the production $A\to d$ on the terminals in $\text{FOLLOW}\left(A\right)$, which is $\left\{a\right\}$.

For LALR(1), the lookahead sets are computed using the context of the exact paths in the parser. In LALR(1), the lookahead for the item $A\to d·$ in this state is $\left\{a\right\}$ which does not conflict with the shift actions on $\{b,c\}$.

Therefore, the grammar $G$ is parser-compatible as an LALR(1) grammar, but due to SLR(1) design constraints and follow-set approximations in general grammar structures, it is classified under the standard compiler theory hierarchy as being LALR(1), but not SLR(1).