Is $a^mb^n$ where $m=n^2$ a CFL?
I have a doubt regrading this problem. Say if we pop $n$ number of $a's$ from the stack for each $b$ then it is a CFL (to be exact DCFL) right?
On the other hand I have this doubt, do we know the number of $b's$ we are provided with, if not how can we pop $n$ number of $a's$ for each $b$? So this case doesn't make it a CFL right?
Which of these 2 cases are applicable? Are we always aware of the number of $b's$ we have? Then shouldn't it be a regular language?