1.A ε-free LL(1) grammar is also a SLR(1) grammar and thus LALR(1) too.
Since the grammar is LL(1), we can conclude that it is unambiguous, non-left recursive and left factored.
Since it is unambiguous, therefore we will never reach a point while parsing where we’ll have 2 choices for reduction, therefore RR conflict of SLR(1) not possible.
Now we have to think that how will ε-free help us in avoiding SR conflict.
Since it is ε-free we’ll never reach a point where we are confused that whether we need to shift the terminal symbol or reduce ε to the non-terminal.
2. A LL(1) grammar with symbols that have both empty and non-empty derivations is also a LALR(1) grammar.
All logic stays the same as done in the 1st statement. Just one additional point it that here we can look ahead and make a decision whether to shift or reduce. Hence LALR(1).
3. A LL(1) grammar with symbols that have only the empty derivation may or may not be LALR(1).
Can anyone please give an example of such grammar. Will such grammar make any sense?