Knight Probability - Base Cases

$1.$ If $(r, c)$ is off the board: return $0$. The knight has fallen off.

$2.$ If $k = 0$ and $(r, c)$ is on the board: return $1$. No more moves needed, and you're still on the board. These two cases handle all boundaries. Off-board positions contribute nothing. On-board positions with no moves left contribute certainty. This is where recursion stops. Without it, you get infinite recursion and a stack overflow. Every recursive solution needs at least one base case.