Question

Given a Context-Free Grammar \( G \) as follows: \[ S \to A a \ |\ b A c \ |\ d c \ |\ b d a \] \[ A \to d \] Which ONE of the following statements is TRUE?

• A. {\( G \) is neither LALR(1) nor SLR(1)}
• B. {\( G \) is CLR(1), not LALR(1)}
• C. {\( G \) is LALR(1), not SLR(1)}
• D. {\( G \) is LALR(1), also SLR(1)}

Hint:

LALR(1) parsers are a compromise between SLR(1) and CLR(1), balancing efficiency and power. They are commonly used in compiler design.

Answer 1

The Correct Option is C

- The given grammar has no conflicts in its LR(1) parsing table, making it CLR(1). A grammar is CLR(1) (Canonical LR(1)) if its LR(1) parsing table has no conflicts, meaning that every state has a unique reduction or shift for each lookahead symbol. Since the grammar has no conflicts, it qualifies as CLR(1).
- It can be reduced to an LALR(1) grammar because the lookahead sets do not change the parsing table significantly. LALR(1) (Look-Ahead LR(1)) grammars are a subset of LR(1) grammars where the parsing table is simplified by merging states with identical cores (ignoring lookahead symbols). Since the lookahead sets in this grammar do not significantly affect the table, it can be reduced to an LALR(1) grammar.
- However, it does not satisfy the SLR(1) property, as there exist conflicts in the follow sets when reducing states. SLR(1) (Simple LR(1)) grammars are a more restrictive subset of LR(1) grammars where reductions are based on follow sets of non-terminals. If there are conflicts in the follow sets during reductions, the grammar does not satisfy the SLR(1) property.
Thus, the correct option is (C).