1 votes 1 votes closed with the note: Answered already.. Why option (b) is not equivalent to given regular expression? Theory of Computation regular-expression + – Akhilesh Yadav 1 asked Dec 26, 2016 closed Dec 26, 2016 by Habibkhan Akhilesh Yadav 1 776 views comment Share Follow See all 5 Comments See all 5 5 Comments reply sudsho commented Dec 26, 2016 reply Follow Share b is right; 1 votes 1 votes utk0203 commented Dec 31, 2016 reply Follow Share https://gateoverflow.in/97112/equivalent-regex 0 votes 0 votes bhargav9873 commented Feb 13, 2017 reply Follow Share Even i feel tht option 'b' should be the right answer 0 votes 0 votes balaeinstein commented Apr 14, 2017 reply Follow Share RE=(a+b)*.(a+b+ ε).a =(a+b)*aa + (a+b)*ba + (a+b)*a =(a+b)*a (The remaining 2 terms are redundant). This regular expression denotes the strings over {a,b} such that the strings ending with a.The smallest possible string is a. In option b) the RE is RE=(( ε+a+b*)^+).a This can be rewritten as RE= (ε+a+b*). (ε+a+b*)*.a =(ε+a+b*).(a+b)*.a =(a+b)*.a + a.(a+b)*.a + b*.(a+b)*.a =(a+b)*.a (The remaining 2 terms are redundant). Therefore the option b is same as the RE given in the question. 0 votes 0 votes Rupendra Choudhary commented May 23, 2017 reply Follow Share The asked Reg expression is basically 'Strings whose end with a' and yes option B too generate the same expression then certainly the answer is wrong . 0 votes 0 votes Please log in or register to add a comment.