0 votes 0 votes LL grammar for the language: $L = \{a^mb^nC^{n+m} \mid m\geq 0, n\geq 0\}$ is a. S->aSc | S1 ; S1 -> bS1c | lambda b. none Compiler Design ll-parser + – gate_forum asked Nov 17, 2018 • edited Nov 24, 2018 by gate_forum gate_forum 695 views answer comment Share Follow See all 8 Comments See all 8 8 Comments reply Show 5 previous comments gate_forum commented Nov 24, 2018 reply Follow Share Please note L = {a^m b^n c^(m+n) | m>= 0, n>=0} 0 votes 0 votes vishalshrm539 commented Nov 24, 2018 reply Follow Share It was a^nb^n earlier right? 0 votes 0 votes gate_forum commented Nov 24, 2018 reply Follow Share The question is whether option a. is LL grammar or not? 0 votes 0 votes Please log in or register to add a comment.