Question

wxw is REGULAR or CSL ? for w belongs to (0,1)+ & x belongs to (0,1)* ?

Comments

What if W and X belongs to (1+0)+

 

can we say that the language is regular, considering the argument that we can extend X as much as we want

Answer 1

It is a CSL but not CFL. 

Have a look at this compilations. 

Answer 2

Regular

We can write it as 1(0+1)*1 +0(0+1)*0

Comments

I think you have missed that its WxW not Wx(W^R). 

---

yes

I think I missed wxw but x belongs to (0+1)⁺

and w belongs to (0+1)*

then it will be regular . right?

---

yes, in that case any string of length at least $1$ will be in $L$. This is because we can always consider $w = \epsilon$ and take all letters as part of $x$.

---

yes it is csl can not be regular as 01(0+1)^(*)01 can not be generated

Answer 3

It is regular and its regular expression is 0 (0+1)* 0 + 1 (0+1)* 1 + 0 (0+1)* 1 + 1(0+1)* 0

Comments

001 is not in $L$ but is generated by your regular expression. Your r.e. is actually generating $(0+1)^+(0+1)^+$.

---

U r rt sir my mistake but re i written not generating 0 or 1 then how it is (0+1)+

---

I was wrong - corrected now :)