1 votes 1 votes Find the closure of S $\rightarrow$ .A ,dollar for the grammar S --> A A --> AB | $\varepsilon$ B --> aB | b // I'm getting S --> .A , dollar A --> .AB, dollar A --> . $\varepsilon$ , a,b Compiler Design compiler-design parsing + – just_bhavana asked Aug 8, 2017 • edited Aug 8, 2017 by just_bhavana just_bhavana 670 views answer comment Share Follow See all 5 Comments See all 5 5 Comments reply joshi_nitish commented Aug 8, 2017 reply Follow Share S--->.A, $ A--->.AB, $ /.ϵ, $ A--->.AB, a/b /.ϵ, a/b 1 votes 1 votes just_bhavana commented Aug 8, 2017 reply Follow Share So you're saying that in A--->.AB, lookaheads will be { $, a, b} ? 0 votes 0 votes joshi_nitish commented Aug 8, 2017 reply Follow Share yes.. 0 votes 0 votes just_bhavana commented Aug 8, 2017 reply Follow Share can you elaborate in detail please ? @joshi_nitish 0 votes 0 votes joshi_nitish commented Aug 8, 2017 reply Follow Share S--->.A, $ A--->.AB, $ /.ϵ, $ upto above you might be clear...now, in A--->.AB, \$ dot(.) is before A you have to write productions of 'A' and his time new lookaheads will be first(B$) i.e a,b... 3 votes 3 votes Please log in or register to add a comment.