2 votes 2 votes Let L be the language accepted by the automaton $L = ${$(a^{n})b:n≥0$}. Find a dfa that accepts the language $L^{2} - L$. Theory of Computation theory-of-computation regular-language peter-linz peter-linz-edition4 finite-automata grammar + – Apoorva Jain asked Jun 30, 2018 • edited Mar 20, 2019 by Naveen Kumar 3 Apoorva Jain 765 views answer comment Share Follow See all 9 Comments See all 9 9 Comments reply joshi_nitish commented May 5, 2018 reply Follow Share yes exactly. 1 votes 1 votes srestha commented May 5, 2018 reply Follow Share here $L^{2}=L.L$ only concatenation operation is working here So, $L^{2}$will contain $a^{n}b.a^{n}b=\left \{ bb,abab,aabaab,aaabaaab,.... \right \}$ where $L=a^{n}b=\left \{ b,ab,aab,aaab,.... \right \}$ 0 votes 0 votes joshi_nitish commented May 5, 2018 reply Follow Share @srestha, the $L^2$ given by you is not correct. correct will be, $L^2=a^mb.a^nb\ |m,n>=0$ 2 votes 2 votes srestha commented May 5, 2018 reply Follow Share yes it will be $a^{*}b.a^{*}b$ and $L$ contains 1 b and $L^{2}$ contains 2 b that is why $L$ and $L^{2}$ are not same rt? 0 votes 0 votes srestha commented May 5, 2018 reply Follow Share thanks @nitish 0 votes 0 votes smsubham commented Jul 10, 2018 reply Follow Share L and L^2 have no string common. Why? L = a*b , L^2 = L. L = a*ba*b Note L has single b in all strings, L^2 has two b in all strings so no string will be common. So make DFA for a*ba*b regular expression. 1 votes 1 votes Apoorva Jain commented Jul 15, 2018 reply Follow Share Thank you ... I understood the solution... 0 votes 0 votes smsubham commented Jul 15, 2018 reply Follow Share You are welcome. Always write the initial strings in language to solve such questions. 1 votes 1 votes Apoorva Jain commented Jul 15, 2018 reply Follow Share Ok 0 votes 0 votes Please log in or register to add a comment.
1 votes 1 votes L and L2 both do not have common strings. Because L = anb and L2= (anb)(amb)|n,m>=0 L2 contains 2 b's while L contains 1 b's. So there is no common string L2-L= L2 guptashekhar54 answered Jul 14, 2018 • edited Aug 31, 2018 by MiNiPanda guptashekhar54 comment Share Follow See all 5 Comments See all 5 5 Comments reply Show 2 previous comments surbhijain93 commented Aug 31, 2018 reply Follow Share For (anb)(anb) , will the no. of a's be equal? if yes, then how would a dfa be possible? 0 votes 0 votes MiNiPanda commented Aug 31, 2018 reply Follow Share Not equal..the reg exp is a*ba*b.. The answer needs an edit.. 1 votes 1 votes Apoorva Jain commented Aug 31, 2018 reply Follow Share No of a's need not be equal in L2 1 votes 1 votes Please log in or register to add a comment.
0 votes 0 votes Here, L={b,ab,aab,aaab,...................} Now, L2 ={bb,bab,baab,..........................} As we can see there is no common element Angkit answered May 5, 2018 Angkit comment Share Follow See all 0 reply Please log in or register to add a comment.