Should there be an arrow going from n*f(n-1) to f (around n==0?) ?
The output of the system also depends on n*f(n-1), not just on whether or not n is zero.
The “n==0?” node is intended to be a ternary operator; its output is n*f(n-1) in the case where n is not 0 (and when n is 0, its output is hardcoded to 1).
Should there be an arrow going from n*f(n-1) to f (around n==0?) ? The output of the system also depends on n*f(n-1), not just on whether or not n is zero.
The “n==0?” node is intended to be a ternary operator; its output is n*f(n-1) in the case where n is not 0 (and when n is 0, its output is hardcoded to 1).