2. Power of 2

In [5]:

def power_of_2(n):
    # 1. write the base case
    if n == 0:
        return 1

    # 2. Reccurence relation
    smaller_problem = power_of_2(n - 1)
    bigger_problem = 2 * smaller_problem

    return bigger_problem

def power_of_2(n): # 1. write the base case if n == 0: return 1 # 2. Reccurence relation smaller_problem = power_of_2(n - 1) bigger_problem = 2 * smaller_problem return bigger_problem

In [6]:

power_of_2(100)

power_of_2(100)

Out[6]:

1267650600228229401496703205376

In [ ]: