1 votes 1 votes Find a regular expression for the complement of the language in $L (r) =$ {$a^{2n}b^{2m+1}: n ≥ 0, m ≥ 0$}. Theory of Computation peter-linz peter-linz-edition4 regular-expression theory-of-computation regular-language + – Naveen Kumar 3 asked Mar 31, 2019 Naveen Kumar 3 489 views answer comment Share Follow See 1 comment See all 1 1 comment reply Harsh Saini_1 commented Feb 23, 2023 i reshown by Harsh Saini_1 Feb 24, 2023 reply Follow Share @Deepak Poonia Sir how do I approach this question? What I can think of right now is in $L(r)$ complement : We’ll have strings of the sort $a^{2n+1} b^{2m}$ where $n$ ≥ 0 & $m$ ≥ 0 We’ll also have all the strings that basically end with $a$. Regex : $(a+b)^* a$ After this i’m unable to think how to proceed further 0 votes 0 votes Please log in or register to add a comment.
2 votes 2 votes It would be easy [ but takes time ] if u can draw the DFA for this Language and Replace NON FINAL TO FINAL and FINAL TO NON FINAL states so that we can get the REGULAR EXPRESSION for the complement of the given language [ Jiren ] answered Feb 24, 2023 [ Jiren ] comment Share Follow See all 6 Comments See all 6 6 Comments reply Show 3 previous comments Sujith48 commented Jun 11, 2023 reply Follow Share In Case 1, we should also be considering the occurrences of null string, so the regular expression has to be ((a+b)* ba (a+b)*)* isn’t? 0 votes 0 votes [ Jiren ] commented Jun 11, 2023 reply Follow Share @Sujith48 Null String will be covered in Case 4 1 votes 1 votes Sujith48 commented Jun 11, 2023 i edited by Sujith48 Jun 12, 2023 reply Follow Share Oh yes, I missed that part , thanks @[ Jiren ] 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes Another way of thinking :) Genius answered Mar 4, 2023 Genius comment Share Follow See all 0 reply Please log in or register to add a comment.