Virtual Gate Test Series: Theory Of Computation - CFL Self Concatenation

Question

Let $L$ be a given context-free language over the alphabet $\{a, b\}$ then $L_{2} = L·L$ is  $\text{CFL.}$

Is $\text{CFL.}$ in general closed under $\text{self-concatenation?}$

If $L={ a^nb^n }$ then $L.L= { a^nb^na^nb^n }$ $\text{(or)}$  $L.L= { a^nb^na^mb^m } ?$

Answer 1

If L={ aⁿbⁿ } then L.L= { aⁿbⁿaⁿbⁿ } is wrong and  L.L= { aⁿbⁿa^mb^m } is correct!

Comments

@Vijay thanks. But does this also imply that CFL will be closed under self-concatenation always?

---

Yes! moreover CFL is closed under concatenataion.
PS: remember CFL is closed under concatenation but not DCFL

---

Sir i think DCFL is closed under concatenation also, DCFL is a strict subset of CFL it follows closure properties of CFL including closure umder complement also.