1 votes 1 votes $L=\{x^n:n \in N\} \cup\{x^ny^n|n \in N\}$ This language does not have LL(k) parser while being deterministic context free.Why? Compiler Design compiler-design context-free-language ll-parser descriptive + ā Ayush Upadhyaya asked Mar 3, 2019 • retagged Jun 21, 2022 by Lakshman Bhaiya Ayush Upadhyaya 744 views answer comment Share Follow See all 9 Comments See all 9 9 Comments reply Show 6 previous comments Arjun commented Mar 3, 2019 reply Follow Share yes, intuitively that is correct š Similar to why LR(k) is more powerful than LL(k). 0 votes 0 votes Shaik Masthan commented Mar 3, 2019 reply Follow Share sir, can you show in formal way, why it doesn't have any LL(k) parser? 0 votes 0 votes Arjun commented Mar 3, 2019 reply Follow Share I don't know a formal way. But the above reason by @Ayush Upadhyaya is sufficient. https://stackoverflow.com/questions/14674912/why-are-there-lr0-parsers-but-not-ll0-parsers 1 votes 1 votes Please log in or register to add a comment.
0 votes 0 votes Till DCFL there exist a LR(K) parser. abhishekmehta4u answered Mar 3, 2019 abhishekmehta4u comment Share Follow See all 0 reply Please log in or register to add a comment.
