0 votes 0 votes $L = \{a^ma^n: m\geq 1, n\geq 1\}$ $L = \{a^ma^n: m\geq 0, n\geq 0\}$ Theory of Computation theory-of-computation identify-class-language + – iarnav asked Sep 11, 2017 • edited Jun 16, 2022 by Arjun iarnav 469 views answer comment Share Follow See all 6 Comments See all 6 6 Comments reply joshi_nitish commented Sep 11, 2017 reply Follow Share 1) L = aaa* = aa+ 2) L= a* 3 votes 3 votes iarnav commented Sep 11, 2017 reply Follow Share @ joshi_nitish So, both are regular L? Also, please tell me, is it necessary that if a L is Regular then it should/must generate epsilon i.e empty string? 0 votes 0 votes Rishabh Gupta 2 commented Sep 11, 2017 reply Follow Share It is not necessary that a regular language will generate empty string. e.g take the language which takes strings on (0+1)* which have only one 0. You can think of many examples where the language does not contains empty string. 1 votes 1 votes iarnav commented Sep 12, 2017 reply Follow Share @Rishabh Gupta 2 (0+1)* which have only one 0. one 0?! How, if we put *=0 in (0+1)* then we can generate epsilon. I'm sorry, I'm not following you! 0 votes 0 votes Rishabh Gupta 2 commented Sep 12, 2017 reply Follow Share Language which contains exactly one 0. Is this a regular language? Yes. You can easily draw fda for it. And it's regular expression will be 1*01* So this is a regular language which does not contain empty string. This is an example which shows that if a language is regular than it is not necessary that it contains empty string. Got it? 0 votes 0 votes iarnav commented Sep 12, 2017 reply Follow Share @Rishabh Gupta 2 Yes Sir, understood. Thanks 0 votes 0 votes Please log in or register to add a comment.
1 votes 1 votes L1 = aa+ L2= a* Divyendi answered Sep 30, 2018 Divyendi comment Share Follow See all 0 reply Please log in or register to add a comment.