52 52 votes Given below are the transition diagrams for two finite state machines $M_1$ and $M_2$ recognizing languages $L_1$ and $L_2$ respectively. Display the transition diagram for a machine that recognizes $L_1.L_2$, obtained from transition diagrams for $M_1$ and $M_2$ by adding only $\varepsilon$ transitions and no new states. Modify the transition diagram obtained in part (a) obtain a transition diagram for a machine that recognizes $(L_1.L_2)^*$ by adding only $\varepsilon$ transitions and no new states. (Final states are enclosed in double circles). Theory of Computation gate1996 theory-of-computation finite-automata normal descriptive + – Kathleen 14.7k views answer comment Share Follow Print See all 10 Comments 10 10 Comments reply Show 7 previous comments JAINchiNMay commented Nov 18, 2022 reply Follow flag @yuyutsuin part A in your solution state A wouldn’t be a final state. 0 0 replyShare ROT commented Sep 16, 2024 reply Follow flag It doesn't really make any difference tho 0 0 replyShare Shaik Masthan commented Oct 20, 2024 reply Follow flag @JAINchiNMay, state A should be one of the final state in part A. Because $L_2$ contains epsilon. $L_1. \{\epsilon\} = L_1$However, epsilon transition from state A to state C and State C as one of the final state cover that. 2 2 replyShare Please log in or register to add a comment.
Best answer 54 54 votes We can combine the final state of $M_1$ with the start state of $M_2$ as follows recognizing $L_1L_2$. But before we combine $M_1$ and $M_2$ remove the final state of $M_1$ as the new machine can accept $L_1$ also while it should accept only $L_1.L2$. ~Pic by Praveeen Saini Arjun answered Oct 22, 2014 • edited May 9, 2021 by gatecse Arjun comment Share Follow See all 28 Comments 28 28 Comments reply Show 25 previous comments Shukla_ commented Sep 16, 2023 reply Follow flag Language accepted by M1 is L1=(00+01+10+11)* and M2 is L2= a*b*. So L1.L2=(00+01+10+11)* a*b* In the both the DFA’s of Ques(a) and (b) state A can be final state as there is a null transition from A to C. So making A as final or non final doesn’t change the language accepted by FA. 3 3 replyShare ꧁༒☬ĿọŗԀ 🆂🅷🅸🆅🅰☬༒꧂ commented Sep 22, 2024 reply Follow flag yes that's definitely true here also we are accepting empty string .. 0 0 replyShare Prashant_Dubey commented Oct 11, 2024 reply Follow flag I think for part b , null transition will be from D to A and C to A because in (L1.L2) , the last transition will be either to state C (strings like 0011a) or state D (strings like 0011ab) but we have to again go to L1 to make our DFA accept strings that belongs to (L1.L2)$^*$ . But adding null transition from A to D doesn't make sense because that things can be fulfilled by A to C null transition for strings like { $b,b^2$,.....} 3 3 replyShare Please log in or register to add a comment.
17 17 votes Hi Honey badger answered Sep 19, 2024 • edited Sep 19, 2024 by Honey badger Honey badger comment Share Follow See all 5 Comments 5 5 Comments reply Show 2 previous comments Shaik Masthan commented Oct 20, 2024 reply Follow flag No issues with the answer. That is perfectly fine. Making DFA for those make the answer different from existing comments/answer. 5 5 replyShare Aman Shukla commented Apr 30 reply Follow flag it takes 15 days to understand , why there is epsilon from D to A ..... 2 2 replyShare DΛΞMON commented Jul 19 reply Follow flag Caption 1 1 replyShare Please log in or register to add a comment.