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Q: Arguing that $L= \{a^n | n\geq 0\} \cup \{a^nb^n| n\geq 0\}$ is not $LL(k)$ for any $k$

Abhishek Ghosh $\text{Consider the language $L= \{a^n | n\geq 0\} \cup \{a^nb^n| n\geq 0\}$ }$ $\text{and the following statements.}$ $\quad\quad\text{I. $L$ is deterministic context-free.}$ $\quad\quad\text{II. $L$ is context-free but not deterministic context-free.}$ $\quad\quad\text{III. $L$ is not $LL(k)$...

 

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