LL(1) expressive power

Question

Does there exist LL(1) grammar for every Regular Language ?

                          OR

For Every Regular Language there exist atleast one LL(1) Grammar . True /False

Answer 1

True : For every Regular Language there exists a LL(1) grammar.
This is true because we can always get a right recursive grammar for any LL(1) that generates a Regular language.

Comments

@Amsar that should be a typo there. Because regular set does not have prefix property.

And why? It is just about allowing one more choice. But that is not given- I don't know the practical reason for this.

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Oky. thanks.. :)

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What about ambiguous grammar? How will it be LL(1)?

Answer 2

I THINK FOR EVERY RL ,LL(1) GRAMMAR NOT EXISTS BECAUSE

S->aS|a

IT IS NOT LL(1)BUT IT IS REGULAR GRAMMAR.

FINITE AUTOMATA POSSIBLE FOR THE ABOVE GRAMMAR

Comments

But for language of this Regular grammar , i.e. a⁺ 

      there exist a LL(1) grammar.

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it is not LL(1)

on input symbol 'a' it has two transitions i.e)s->as and s->a

so there is a conflict occurs

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After removing non-determinism, it will be LL(1).

S -> aS'

S' -> S/∈