$L=\{a^ib^i|i\geq0\}$, cfg for $L^2$

can you write cfg for $L^2$ where $L=\{a^ib^i|i\geq0\}$?

the professor's answer sheet says it's $S\to AA\\ A\to aAb|\lambda$

but I think it is wrong because two $L$ have to be identical with each other.

can you help me figure out if there is cfg describing $L^2$?

  • $\begingroup$ It is your assumption that the “two L” have to be the same which is wrong. Take another look at the definition of $L^2$. $\endgroup$ – Yuval Filmus Oct 17 at 9:05
  • $\begingroup$ @YuvalFilmus after my further study with additional lecture material, I noticed that the concatenation of two language L1 ◦L2 means {xy | x ∈ L1, y ∈ L2}. thank you for your help. $\endgroup$ – Myeongwon Choi Oct 17 at 9:14

The square language $L^2$ is defined as follows: $$ L^2 = LL = \{ xy : x,y \in L \}. $$ There is no requirement that $x = y$. Indeed, in your case $L^2 \supsetneq \{ w^2 : w \in L \}$.


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