0 votes 0 votes Is a*b* + b*a* = ( a + b)* ______ Theory of Computation theory-of-computation regular-expression finite-automata + – Ajaaz asked Jul 31, 2018 Ajaaz 1.8k views answer comment Share Follow See all 4 Comments See all 4 4 Comments reply gauravkc commented Jul 31, 2018 reply Follow Share No. a*b* is all strings that have a's followed by b's. b*a* is all strings that have b's followed by a's. However, in (a+b)*, you can have any sequence of a's and b's. Eg abbababa 3 votes 3 votes goxul commented Jul 31, 2018 reply Follow Share In general for such questions, you can try finding strings in one which are not present in the other. In this example, the RHS generates "abab" which is not generated by the LHS. Hence, they are not equivalent. 0 votes 0 votes Spider1896 commented Aug 1, 2018 reply Follow Share No (a+b)*=(a*+b*)*=(a*b*)*=(a*+b) *=(a+b*)*=a*(ba) *=b*(ab)* Hence these are the possibilities and certainly the given expression is wrong 0 votes 0 votes gauravkc commented Aug 2, 2018 reply Follow Share How is (a+b)* = a*(ba)* ? How a*(ba)* generates bbb? 0 votes 0 votes Please log in or register to add a comment.
1 votes 1 votes a*b*= a,b,ab,aab aabb,aabbbb..... b*a*= a,b,ab,ba,bba,bbaaa........ So we can genrate only string like a followed b or b followed a. But we can not genrate syring like aba,bab,aaba,..... So its not equal to (a+b)* abhishekmehta4u answered Aug 2, 2018 abhishekmehta4u comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes No, both LHS !=RHS becoz in a*b* we have a followed by b i.e epsilon,a,b,ab,aa, bb,aabb,abb,aab,abbb in b*a* we have b followed a i.e epsilon,b,a,bb,aa,bbaa,baa,bba,bbbaaa etc (a+b)*= (a*b*)*, (a*+b)*,(a+b*)* any sequence of a and b are acceptable aayushi neeshu answered Aug 4, 2018 aayushi neeshu comment Share Follow See all 0 reply Please log in or register to add a comment.