Question

The ­following Implementation of the Fibonacci function is a correct, but inefficient,

def fibonacci(n):

if n <= 2:

return 1

else:

return fib(n - 1) + fib(n - 2)

In more details, the code shown runs very slowly for even relatively small values of n; it can take minutes or hours to compute even the 40th or 50th Fibonacci number. The code is inefficient because it makes too many recursive calls. It ends up recomputing each Fibonacci number many times. Which of the below recursive modification/s is/are more efficient compared to the aforementioned function.

idef fibonacci (n): if n <= 2: return 1 else: return fibonacci2(n, 3, 1, 1) def fibonacci2 (n, i, prev, curr): if n == i: return prev + curr else: return fibonacci2(n, i + 1, curr, prev + curr) def fibonacci3 (n): i if n not in fibo dict: fibo dict[n] = fibonacci (n-1) + fibonacci (n-2) return fibo dict[n] # global fibonacci dictionary. fibo dict = { } fibo dict[0] = 1 fibo dict[1] = 1

	a) A only
	b) B only
	c) A and B
	d) None of them has a significant improvement

Answer 1

a) A only is more efficient to compute the large Fibonacci number

because, in option B it needs more space i.e for nth fibonacci numbers it needs n space in dictionary.

also in option B there is two recursive function calls and in option A there is one recursive function calls.