0 votes 0 votes ADITYA CHAURASIYA 5 asked Sep 24, 2017 ADITYA CHAURASIYA 5 324 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
1 votes 1 votes I think the answer should be B. The state machine in Option A will accept 111 as an input and C will accept both 11 and 111 as its input. rahul saxena answered Oct 14, 2017 rahul saxena comment Share Follow See all 4 Comments See all 4 4 Comments reply sourav. commented Oct 14, 2017 reply Follow Share why $B$? $B$ too will accept $11$ or $111$ see at last of B, there is $(0+1)^{*}$ 0 votes 0 votes saxena0612 commented Oct 14, 2017 reply Follow Share @ sourav But you have to accept four preceding symbols which make string other than given right? 0 votes 0 votes rahul saxena commented Oct 14, 2017 reply Follow Share In option B, its of this type X . Y is a Regular Expression corresponding to the language L(X) . L(Y) where L(X.Y) = L(X) . L(Y) The X in option B is (0 + 1(0+1(...))) and Y is (0 +1)*. The languages it will accept will contain both X and Y. The "+" is used as union and "." is used for intersection. Both things must be included. Here's a link to a good resource. My explanation is not much clear but I hope these hints might have helped you get an idea about my approach. Does it help? You 0 votes 0 votes sourav. commented Oct 14, 2017 reply Follow Share yes you are right actually i thought of $11$,$111$ as substring 1 votes 1 votes Please log in or register to add a comment.