1 votes 1 votes Given two Regular expressions are equal or not ? 1) (1+01*0)* 2) 1*(01*0)* 1* Give proper explanation also. Theory of Computation regular-expression theory-of-computation finite-automata regular-language + – Ashish Roy 1 asked Sep 27, 2018 Ashish Roy 1 2.1k views answer comment Share Follow See all 2 Comments See all 2 2 Comments reply Navneet Kalra commented Sep 27, 2018 reply Follow Share i ma getting a string 0101010 from first statement which i hope second second could'nt generate making them not equal...please correct me if i did a mistake 1 votes 1 votes hitendra singh commented Sep 28, 2018 reply Follow Share 1*(01*0)*1* does this regular expression be understood as ::- while generating the string first 1* will pumped as many times as we want then 01*0 will be pumped then last 1* will be pumped . correct me if I had misinterpreted ? 0 votes 0 votes Please log in or register to add a comment.
Best answer 1 votes 1 votes The two regular expressions are not equivalent. The first regular expressions denotes the set of strings having any combination of 1's and 01*0's. Whereas the second regular expression denotes the set of all strings where any number of 1's are followed by any number of 01*0's followed by any number of 1's. So strings of the form 10101010 are generated by the first regular expression but it isn't generated by the second regular expression. Somoshree Datta 5 answered Sep 27, 2018 • selected Sep 27, 2018 by Ashish Roy 1 Somoshree Datta 5 comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes According to me these two regular expressions are not equivalent. As, on simplifying expression 1, we get 1*(01*0)* which is not equivalent to expression 2 dharm6619 answered Sep 30, 2018 dharm6619 comment Share Follow See all 0 reply Please log in or register to add a comment.