Final Answer:
(a) This grammar is not LR(0) because it has shift-reduce conflicts due to the presence of "StmtList → Stmt" and "StmtList → StmtList semi Stmt" productions.
(b) This grammar is LR(0) as it does not have any shift-reduce or reduce-reduce conflicts. The productions are structured in a way that the parser can deterministically decide between shifting or reducing.
(c) This grammar is not LR(0) due to the ambiguity in the production "StmtList → StmtList semi StmtList", causing shift-reduce conflicts.
(d) This grammar is LR(0) since it's a right-recursive grammar with no conflicts, allowing for deterministic parsing using LR(0) parser.
Step-by-step explanation:
(a) The grammar has conflicts due to the presence of both "StmtList → StmtList semi Stmt" and "StmtList → Stmt" productions. This causes ambiguity when the parser encounters a 'semi' token after 'Stmt', leading to shift-reduce conflicts. It can't be parsed using LR(0) due to these conflicts.
(b) This grammar doesn't contain any ambiguity, as the production rules are well-structured. The 'StmtList' production is separated by 'semi' and 'Stmt', ensuring a clear distinction between the different parts of the language. Hence, it's LR(0) as it can be parsed unambiguously using an LR(0) parser.
(c) The grammar suffers from ambiguity introduced by the production "StmtList → StmtList semi StmtList," which creates shift-reduce conflicts. The parser can't decide whether to shift the 'semi' or reduce 'StmtList' to 'StmtList semi StmtList,' making it not LR(0).
(d) This grammar is LR(0) compatible because it is a right-recursive grammar without any conflicts. The 'StTail' production creates a sequence of 'semi' followed by itself or λ (empty string), maintaining the determinism required for LR(0) parsing.
This is incomplete question, the complete question is: Which of following grammars are LR(0)? Explain why.
(a)1 S → StmtList $
2 StmtList→ StmtList semi Stmt
3 | Stmt
4 Stmt → s
(b)1 S → StmtList $
2 StmtList→ Stmt semi StmtList
3 | Stmt
4 Stmt → s
(c)1 S → StmtList $
2 StmtList→ StmtList semi StmtList
3 | Stmt
4 Stmt → s
(d)1 S → StmtList $
2 StmtList→ s StTail
3 StTail → semi StTail
4 | λ