Correct : a

The correct answer is Option A - G is neither LALR(1) nor SLR(1).
The grammar is:
S → Aa | bAc | dc | bda
A → d
This is a classic example used to demonstrate the difference between SLR(1)/LALR(1) and CLR(1) parsers. Let''s understand why.
Why G is not SLR(1) or LALR(1):
A → d appears in two different production contexts - S → Aa (where A is followed by ''a'') and S → bAc (where A is followed by ''c''). So FOLLOW(A) = {a, c}.
In the SLR(1) parsing table, when the parser has ''d'' on top of the stack and sees a lookahead of ''a'' or ''c'', it must decide whether to reduce A → d or shift. Additionally, ''d'' also appears directly in S → dc and S → bda, creating a reduce-reduce conflict - the parser can''t determine with just one lookahead whether ''d'' should be reduced as A → d or kept as part of S → dc or S → bda. This conflict makes G not SLR(1).
LALR(1) merges states with the same LR(0) core, but since the conflict originates from overlapping FOLLOW sets and not from state merging, LALR(1) also fails to resolve it. G is not LALR(1) either.
Why G is CLR(1):
CLR(1) uses full LR(1) items with precise lookahead sets per state. In CLR(1), the state where A → d• is the item will have the lookahead precisely as {a} in one state and {c} in another - they are separate states and never merged. This precise distinction resolves the conflict completely, making G a valid CLR(1) grammar.
Key takeaway - whenever a grammar is CLR(1) but not LALR(1), it means LALR(1) state merging causes a conflict that CLR(1) avoids by keeping those states separate.