0 votes 0 votes I think that the given grammar is LL(1) . please explain me if I'm wrong. Compiler Design compiler-design parsing lr-parser ll-parser ace-test-series + – atul_21 asked Nov 28, 2017 • retagged Jun 22, 2022 by Lakshman Bhaiya atul_21 647 views answer comment Share Follow See 1 comment See all 1 1 comment reply Shivam Chauhan commented Nov 28, 2017 reply Follow Share Grammar is not LL(1). Productions B -->1B and B-->$\epsilon$ will be under terminal 1 conflict 1 votes 1 votes Please log in or register to add a comment.
Best answer 2 votes 2 votes Bro answer is its not LL(1) grammAr... Look at this production B->1B|episolon.........here first of 1B is {1}...and as production contain episolon we have to see follow of B... Follow of B is First of C ...first of C={1}.....so we can say IN THE ROW CORRESPONDING TO "B" AND COLUMN NAMED "1" WE HAVE 2 ENTRIES ...SO GRAMMAR IS NOT LL(1) SHUBHAM SHASTRI answered Nov 28, 2017 • selected Nov 28, 2017 by atul_21 SHUBHAM SHASTRI comment Share Follow See 1 comment See all 1 1 comment reply atul_21 commented Nov 28, 2017 reply Follow Share I missed to see the follow of LHS when there is B -> eps @SHUBHAM SHASTRI thnx alot 1 votes 1 votes Please log in or register to add a comment.
1 votes 1 votes The Grammar is not LL1 A->0A1/ null first( 0,null) follow( 1,0, $) so A->0A1 will be in column (0) A->null will be in column( 0,1,$) hence it's not LL1 gari answered Nov 28, 2017 gari comment Share Follow See 1 comment See all 1 1 comment reply atul_21 commented Nov 28, 2017 reply Follow Share Yes this is correct. .thnku 1 votes 1 votes Please log in or register to add a comment.
