Proof that the grammar is LL(2)

Question

I am given the following grammar:

$ S \rightarrow AabAba \\ A \rightarrow a | \epsilon $

and I have to prove it is LL(2).

I know what LL(k) means - one can choose a production based on k characters from the input.

So consider the input $ababa$. Of course we start with $S$, so we apply the production to $AabAba$. Now we see $ab$ in the input, so we choose the production to $\epsilon$ for the first $A$. We get rid of the $ab$ and we have $aba$.

And what now? A while ago, we've chosen $\epsilon$ for $A$, seeing $ab$ in the input. Now we see the same thing, but are supposed to choose $A \rightarrow a$. But, as I understand, we cannot do this, because that would mean nondeterminism.

Is this a counterexample? How is this grammar LL(2)?

Answer 1

Sippu and Soisalon-Soininen (1982) carefully distinguish between two definitions of LL(k) grammars, one of which -- the one I think you are using -- they call strong LL(k):

A grammar $G$ is $\text{LL}(k)$ if $\text{FIRST}_k(\omega_1\delta)$ and $\text{FIRST}_k(\omega_2\delta)$ are disjoint whenever $xA\delta$ is a left sentential form of $G$ and $A \to \omega_1$ and $A \to \omega_2$ are distinct productions of $G$.

Grammar $G$ is $\text{strong LL}(k)$ if $\text{FIRST}_k(\omega_1\text{FOLLOW}_k(A))$ and $\text{FIRST}_k(\omega_2\text{FOLLOW}_k(A))$ are disjoint whenever $A \to \omega_1$ and $A \to \omega_2$ are distinct productions of $G$.

The difference has to do with the state machine. A grammar is $\text{strong LL}(k)$ if you can use the same procedure to decide which production for a non-terminal to use regardless of the contents of the parser stack. An $\text{LL}(k)$ grammar, on the other hand, can use the contents of the stack ($x$ in the above definition) as well as the lookahead. (For these purposes the contents of the stack can be condensed into a finite state automaton.)

While the two definitions are not equivalent -- the strong condition is much more restrictive -- it can be demonstrated that every $\text{LL}(k)$ grammar can be changed to a $\text{strong LL}(k)$ grammar simply by labelling each usage of a non-terminal. So in the case of your grammar, which is $\text{LL}(k)$ but not $\text{strong LL}(k)$, we can create the equivalent grammar:

$\begin{align} S &\to A_1abA_2ba \\ A_1 &\to a | \epsilon \\ A_2 &\to a | \epsilon \\ \end{align}$

which is $\text{strong LL}(k)$.

For more details, see the paper linked above.