0 votes 0 votes L={a^n \ n>=0} M={b^n \ n>=0} L.M is a regular language and the DFA for this is going to be ending with b and epsilon and it will have two states Am I correct or not sanju77767 asked May 3, 2018 sanju77767 460 views answer comment Share Follow See all 5 Comments See all 5 5 Comments reply Angkit commented May 3, 2018 reply Follow Share It is a DFA we need a dead state .Total 3 states 0 votes 0 votes sanju77767 commented May 3, 2018 reply Follow Share plzzz upload ur DFA which u hav created and plzzz check the concatination properly this language does not going to dead state it is the starting state will also be the final state and the language is L={epsilon,b,bb,bbb,bbbb ab,abb,aaab,aaaab,aab,,,,,,,,,,} 0 votes 0 votes Angkit commented May 3, 2018 reply Follow Share It is a DFA we need a dead state .Total 3 states.Please see this : 0 votes 0 votes sanju77767 commented May 3, 2018 reply Follow Share the language which is getting accepted is L={epsilon,b,bb,bbb,bbbb ab,abb,aaab,aaaab,aab,,,,,,,,,,} ur language is also accepting this language L(a,aa,aaa,aaaaa,aaaaa,,aaaaa,,,,,,,,,,,,) in the above question this language is not getting accepted plzz see this DFA this will the DFA of the language concatination of Language L and language M it is not necessary that every DFA should always contain Dead state 0 votes 0 votes Angkit commented May 3, 2018 reply Follow Share L(a,aa,aaa,aaaaa,aaaaa,,aaaaa,,,,,,,,,,,,) in the above question this language is getting accepted. Because M can but null. 0 votes 0 votes Please log in or register to add a comment.