0 votes 0 votes which of the following statement is True? a) r* and r+ are always different. b) r* and r+ may be equal. Please explain briefly. Theory of Computation theory-of-computation + – Siddharth Bhardawaj asked May 1, 2017 • retagged May 20, 2021 by Shiva Sagar Rao Siddharth Bhardawaj 1.5k views answer comment Share Follow See 1 comment See all 1 1 comment reply papesh commented May 1, 2017 reply Follow Share Take r= epsilon 1 votes 1 votes Please log in or register to add a comment.
0 votes 0 votes According to me option A is true Alok Kumar 1 answered May 1, 2017 Alok Kumar 1 comment Share Follow See all 6 Comments See all 6 6 Comments reply Show 3 previous comments balaeinstein commented May 1, 2017 reply Follow Share ε*= ε It is not equal to Φ . Please read the identities of regular expressions 0 votes 0 votes Alok Kumar 1 commented May 1, 2017 reply Follow Share yes, u r right. if r=ϵ,then r*={ϵ} and r+={ϵ}. so ,Options b is True . 0 votes 0 votes Abhinandan Sharma commented May 1, 2017 reply Follow Share yup i was wrong its actually Φ*=ϵ thanks for correcting 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes a) r* and r+ are always different. b) r* and r+ may be equal. if r= epsilon epsilon* = epsilon epsilon+ = epsilon from above b is correct but if u place r= phi phi*= epsilon phi+ = phi which is different so it is corrected as fore some condition r+ and r* are smae and for some they are different akankshadewangan24 answered May 1, 2017 akankshadewangan24 comment Share Follow See 1 comment See all 1 1 comment reply gabbar commented May 6, 2017 reply Follow Share both are correct? 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes r* means *>= 0, i.e. r* = {Ɛ,r,rr,rrr,rrrr.....} but in the case of r+, r can not be equal to 0 i.e. in simple words + denotes that language contains atleast one r. i.e. r+ = {r,rr,rrr,rrrr.....} Hence r* and r+ are always different. Sachi Saxena answered May 5, 2017 Sachi Saxena comment Share Follow See all 0 reply Please log in or register to add a comment.