Question

What is the output of the following code?

def function(n):
if n == 1:
return 1
else:
return function(sqrt(n))

function(3)

Answer 1

The code results in a RecursionError because it contains a recursive function call that does not have a base case which can handle the input of '3', leading to an infinite recursion loop.

Step-by-step explanation:

The output of the given code cannot be determined because there appears to be a recursive function without a proper base case able to handle the input given. The function function(n) is designed to take a number 'n' and check if it is equal to 1. If 'n' is 1, it returns 1. Otherwise, it calls itself with the square root of 'n' as its new argument. However, when the function is called with an argument of 3, as in function(3), the recursive call will occur with the square root of 3, which is not an integer. This will lead to a recursive loop with non-integer inputs that move further away from the base case of 1, and ultimately leads to a RecursionError due to the maximum recursion depth being exceeded in Python.