1 votes 1 votes Consider the following right linear grammar. S->aA/abc A->aA/bB/a B->bB/cC/b C->cC/c Find left linear grammar is equivalent to the above right linear grammar? Theory of Computation theory-of-computation + – saurav04 asked Jan 29, 2016 saurav04 2.2k views answer comment Share Follow See all 3 Comments See all 3 3 Comments reply saurav04 commented Jan 29, 2016 reply Follow Share Plz use DFA construction method so that i can understand my mistake 0 votes 0 votes S Ram commented Jan 3, 2019 reply Follow Share @Shaik Masthan any idea?? 0 votes 0 votes Shaik Masthan commented Jan 3, 2019 reply Follow Share RLG ---> FA ----> reverse the FA ---> RLG of FA$^R$ ----> Reverse the productions ===> LLG 2 votes 2 votes Please log in or register to add a comment.
0 votes 0 votes IS it, S->Aa A->Aa/Bb B->Bb/Cc C->Cc Dibyo jyoti Koleh answered Jan 29, 2016 Dibyo jyoti Koleh comment Share Follow See all 4 Comments See all 4 4 Comments reply saurav04 commented Jan 29, 2016 reply Follow Share given ans is S->Ac/abc A->Ac/Bb/c B->Bb/Ca/a C->Ca/a like to know ur approach..... 0 votes 0 votes shivanisrivarshini commented Jan 30, 2016 reply Follow Share Since the given grammer results in strings that start with a and ends with c L={abc,aabc,abbc,abcc ........ } S->aA/abc ===> as we are considering left linear S->Ac/abc ===> strings ends with c A->Bb/Ac/c ===> to have more than 1 c A->Ac/c and A->Bb to have b's B->Bb/b/Ca ===> same as for A C->Ca/a ===> to have a's Since we replace Left most Non terminal we need to have start terminal with leftmost symbol S->A->B->C->a so that we get 1st terminal 0 votes 0 votes saurav04 commented Jan 30, 2016 reply Follow Share there are more string then just ending with c , this is the regular expression for above grammar : { (aa^+) +(a^+bb^+) +(a^+b^+cc^+) } now explain each production how u got ...DFA constuction and then reversal of production and interchanging final and non final state wont work. 0 votes 0 votes shivanisrivarshini commented Jan 30, 2016 reply Follow Share a*b*c* ?? 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes $S \rightarrow Aa \ | \ Bb \ | \ Cc \ | \ abc$ $A \rightarrow Aa \ | \ a$ $B \rightarrow Bb \ | \ Ab$ $C \rightarrow Cc \ | \ Bc$ Mk Utkarsh answered Dec 30, 2019 Mk Utkarsh comment Share Follow See all 0 reply Please log in or register to add a comment.