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?



Can I have an LL grammar for every deterministic context free language?
I’m not an expert on this topic, but I found these [course notes](http://www.gdi.uni-bamberg.de/teaching/SS10/GdI-GTI-B/tomlect8.pdf) (including some bibliographical references) which state that the language *L* = {*xn* : *n* ∈ ℕ} ∪ {*xnyn* : *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](http://www.cs.uwaterloo.ca/research/tr/1978/CS-78-24.pdf) by J.C. Beatty contains a proof that the LR language *L* = {*anbn*, *ancn* : *n* ≥ 1} is not LL (see Theorem 5.2).


