2 votes 2 votes is it correct please tell me its urgent! Compiler Design compiler-design parsing lr-parser + – Pranav Madhani asked Aug 19, 2017 • retagged Jul 6, 2022 by Lakshman Bhaiya Pranav Madhani 3.8k views answer comment Share Follow See all 3 Comments See all 3 3 Comments reply joshi_nitish commented Aug 19, 2017 reply Follow Share why you extented your dfa.. in first canonical set itself there was RR conflict(A--->epsilon B--->epsilon).., there itself you can say that it is not LR(0)... one more thing, A---> .epsilon, is itself reduce move, so you need not take transition on it. 3 votes 3 votes ajupazhamayil commented Jan 12, 2018 reply Follow Share Had it been just one epsilon move(say we dont have B-> epsiln), then will there be any SR conflict in state 1? 0 votes 0 votes suryaprakash commented Jun 14, 2018 reply Follow Share in the initial state itself we can justify that its not LR(0) BECAUSE it contain RR-conflict 0 votes 0 votes Please log in or register to add a comment.
2 votes 2 votes yes...correct given grammar is not LR(0).....due to RR conflict.... and u need not to go to other state.....A->.eps can be also be written as A->. means in the beginning conflict occurs.... hs_yadav answered Aug 19, 2017 hs_yadav comment Share Follow See 1 comment See all 1 1 comment reply G Shaheena commented Dec 4, 2017 reply Follow Share In 1st and second productions I.e, S ->AaAb S->BbBA If we try to construct LR(0) parsing table entries going to be same cell I.e, {a,b} intersection {b} =b so there is RR conflict so it is not LR(0). 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes Given grammer is not LR(0). Clearly mentioned that grammar has reduce reduce conflict for the rule A--> epsilon and B---> epsilon prashant dubey answered Mar 26, 2019 prashant dubey comment Share Follow See all 0 reply Please log in or register to add a comment.
