2 votes 2 votes Q if any Grammar is LL(1) definitely LALR(1) ? It is true or false Please ans explain in detail. Compiler Design compiler-design parsing ll-parser lr-parser true-false + – kallu singh asked Sep 5, 2017 edited Jun 21, 2022 by Lakshman Bhaiya kallu singh 647 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
1 votes 1 votes if any Grammer is LL(1) then it compulsory BUP that means must be CLR(1) , powerful one of BUP definitely satisfy but there is no guarantee of LALR(1). bcoz BUP is more powerful than TDP.that means atleast most powerful person must satisfy rajoramanoj answered Sep 6, 2017 rajoramanoj comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes False! each LL(1) is CLR(1) but each CLR(1) may or may not be LALR(1). I think there is no relation b/w LL(1) and LALR(1) but LL(1) are proper subet of CLR(1). Note that, every epsilon free LL(1) grammar is LALR(1). Manu Thakur answered Sep 5, 2017 edited Sep 5, 2017 Manu Thakur comment Share Follow See all 3 Comments See all 3 3 Comments reply Shubhanshu commented Sep 5, 2017 reply Follow Share LALR is a LR parser uses LALR(1) language, and LL are proper subset of LR languages then, it should be true. ryt. 0 votes 0 votes just_bhavana commented Sep 6, 2017 reply Follow Share Among the various LR parsers, only CLR(1) is the strongest of all. So every LL(1) grammar is CLR(1). However, only a proper subset of LALR(1) grammars is LL(1). 0 votes 0 votes kallu singh commented Sep 6, 2017 reply Follow Share Can you explain explain with example which LL(1) but not LALR(1) 0 votes 0 votes Please log in or register to add a comment.