7

I've been playing around with a lot of grammars that are not LL(1) recently, and many of them can be transformed into grammars that are LL(1).

However, I have never seen an example of an unambiguous language that is not LL(1). In other words, a language for which any unambiguous grammar for the language is not LL(1)), nor do I have any idea how I would go about proving that I had found one if I accidentally stumbled across one.

Does anyone know how to prove that a particular unambiguous language is not LL(1)?

DVK
  • 126,886
  • 32
  • 213
  • 327
templatetypedef
  • 362,284
  • 104
  • 897
  • 1,065

1 Answers1

5

I was thinking about the question a while and then found this language at Wikipedia:

S -> A | B
A -> 'a' A 'b' | ε
B -> 'a' B 'b' 'b' | ε

They claim the language described by the grammar above cannot be described by LL(k) grammar. You asked about LL(1) only and this is pretty straightforward. Having first symbol only, you don't know if the sequence is 'ab' or 'aab' (or any more recursive one) and therefore you cannot choose the right rule. So the language is definitely not LL(1).

Also for every sequence generated by this grammar there is only one derivation tree. So the language is unambiguous.

The second part of your question is a little harder. It is much easier to prove the language is LL(1), than the opposite (there is no LL(1) grammar describing the language). I think you just create a grammar describing the language, then you try to make it LL(1). After discovering a conflict which cannot be resolved you somehow have to take advantage of it and create a proof.

Binus
  • 1,065
  • 7
  • 14