1 votes 1 votes Wxw / w belogs to (a, b) ^*, and x belongs to (a, b) ^+ Regular? Csl? Gunvantwavhal asked Sep 4, 2017 Gunvantwavhal 427 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes i think ...this is a regular language.... and RE=a(a+b)+a + b(a+b)+b +(a+b)+ hs_yadav answered Sep 4, 2017 hs_yadav comment Share Follow See all 2 Comments See all 2 2 Comments reply Gunvantwavhal commented Sep 4, 2017 reply Follow Share I think regular expression is (a+ b) ^+ 0 votes 0 votes hs_yadav commented Sep 4, 2017 reply Follow Share yes sure....final RE =(a+b)+ ..... i write it like this...just for better understanding..... 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes No this is NOT a regular expression. this expression is "wxw" meaning it should start with w and end with w. So even if we were to consume symbols of w by 'x' we may have a different symbol at the end. Consider the example : abbbc b abbbc (divided as w x w) if x consumes bbbc of either side of w then it leaves a and c at both ends. But we want to have the ends same. Warlock lord answered Sep 4, 2017 Warlock lord comment Share Follow See all 2 Comments See all 2 2 Comments reply joshi_nitish commented Sep 4, 2017 reply Follow Share @Warlock lord it is regular take w = epsilon and x = (a+b)+ , now, one string in language is (a+b)+ and it will absorb all other strings you can think of. if w ϵ (a+b)+ , then it would not be regular and for that your explanation would be correct. 0 votes 0 votes Warlock lord commented Sep 4, 2017 reply Follow Share oh yes I didn't pay heed to the * My bad 0 votes 0 votes Please log in or register to add a comment.