I would argue it isn't tractable. f(n) = f(n-1) + f(n-1) has runtime O(2^n); f(n) = f(n-1) + f(n-2) has runtmie O(~1.62^n); f(n) = f(n/2) + f(n/2) has O(n log n); for i = 1 to f(n) has runtime f(n).. good luck…
I would argue it isn't tractable. f(n) = f(n-1) + f(n-1) has runtime O(2^n); f(n) = f(n-1) + f(n-2) has runtmie O(~1.62^n); f(n) = f(n/2) + f(n/2) has O(n log n); for i = 1 to f(n) has runtime f(n).. good luck…