Every deterministic context free grammar can be represented by a LR(1) grammar, so this question can be rephrased as: can I build an equivalent LL(k) grammar from every LR(k) grammar? Can I have an example of deterministic context free language that can not have an LL(k) grammar?

I’m not an expert on this topic, but I found these course notes (including some bibliographical references) which state that the language *L* = {*x ^{n}* :

*n*∈ ℕ} ∪ {

*x*:

^{n}y^{n}*n*∈ ℕ} has no LL(

*k*) parser, while being deterministic context-free (see pp. 24 and 27).

Edit: I found a better reference. The paper Two iteration theorems for the LL(*k*) languages by J.C. Beatty contains a proof that the LR language *L* = {*a ^{n}b^{n}*,

*a*:

^{n}c^{n}*n*≥ 1} is not LL (see Theorem 5.2).