0 votes 0 votes wxw ;w,x belongs to {a,b}* this regular or dcfl or cfl??? somewhere it is written as regular but dontknow why?? eyeamgj asked Aug 18, 2018 eyeamgj 486 views answer comment Share Follow See all 4 Comments See all 4 4 Comments reply s9k96 commented Aug 18, 2018 reply Follow Share I think since w,x both belongs to {a,b}*, x can take upto the entire string leaving w with epsilon. so any string can be like (epsilon)(a,b)*(epsilon). hence it is Regular. 0 votes 0 votes Shubhgupta commented Aug 18, 2018 reply Follow Share yes it is regular. x can take up whole string then language will become start and end with same symbol. 1 votes 1 votes eyeamgj commented Aug 18, 2018 reply Follow Share no it is not the case of starting and ending with same symbol 0 votes 0 votes Shubhgupta commented Aug 18, 2018 reply Follow Share ya i haven't checked correctly i thought it was +. so for this it will be (a+b)*. 0 votes 0 votes Please log in or register to add a comment.
1 votes 1 votes Yes, the language is regular because wxw = x only when w, x belong to {a,b}^* Hence the language is (a+b)^* which is regular Vikas Verma answered Aug 18, 2018 Vikas Verma comment Share Follow See all 0 reply Please log in or register to add a comment.
1 votes 1 votes R.E for above language=$a(a+b)^{*}a +b(a+b)^{*}b$ .so it is regular language. BASANT KUMAR answered Aug 19, 2018 BASANT KUMAR comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes Yes, the language is regular because you can eliminate the dependency between two w's by substituting epsilon in place of w. Once the dependency is over the language becomes regular. Now, since language is regular it is also CFL & DCFL. vikas999 answered Aug 18, 2018 vikas999 comment Share Follow See all 3 Comments See all 3 3 Comments reply Mizuki commented Aug 20, 2018 reply Follow Share Isn't that just one case? w being replaced with ε. Also can you tell more about dependency? 0 votes 0 votes Anand. commented Aug 20, 2018 reply Follow Share yes it is the only reason (by maing w as epsilon ) , we can make $L=\left \{ wxw | w,x \,\,\epsilon (a+b)^{*} \right \}$ as regular because making $w$ as $\epsilon $ our $L$ will be nothing but $(a+b)^{*}$ if we try to make $w$ as any other string other than $\epsilon$ ,then we need to keep track of that string which will need a stack. 0 votes 0 votes Mizuki commented Aug 20, 2018 reply Follow Share Thanks @Anand. 1 votes 1 votes Please log in or register to add a comment.