0 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 Theory of Computation + – sanju77767 946 views answer comment Share Follow Print See all 5 Comments 5 5 Comments reply Show 2 previous comments Angkit commented May 3, 2018 reply Follow flag It is a DFA we need a dead state .Total 3 states.Please see this : 0 0 replyShare sanju77767 commented May 3, 2018 reply Follow flag 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 0 replyShare Angkit commented May 3, 2018 reply Follow flag L(a,aa,aaa,aaaaa,aaaaa,,aaaaa,,,,,,,,,,,,) in the above question this language is getting accepted. Because M can but null. 0 0 replyShare Please log in or register to add a comment.