简单 Python 函数的时间复杂度

Question

I have been given a simply Python programme as such:

def boo(n):
    if n == 1:
        return 1
    else:
        return boo(n/2) + boo(n/2)

To boo, I have to find out the time complexity of the function. My idea is that since each n makes a call to two more boo 's, the time complexity is O(2^n) . However, it has been revealed that this is wrong. The answer has been revealed to me that the actual time complexity is O(n) without rationale given. My hunch is that boo(n/2) + boo(n/2) can be treated as 2*boo(n/2) , but I am highly suspicious of this as I have seen similar scenarios before, and I vaguely remember that boo(n/2) + boo(n/2) cannot be equated to 2*boo(n/2) in terms of processing steps.

What would be a more appropriate way to understand this? I am still struggling with the ideas in order of growth (space and time complexity), so if anyone has any general advice or material that can possibly help me, I greatly welcome them. Thank you very much!

Answer 1

This code is O(∞) , as a non-power-of-two input will never complete (it will go from > 1 to < 1 without ever being == 1 ).

For powers of 2 (or if properly rewritten to do boo(n // 2) and not recurse infinitely), it's O(n) , as each step halves the input, but then does that halved work twice, leading to overall linear performance (to be precise, boo is called 2n - 1 times for n being a power of 2). Changing the final line to return boo(n // 2) * 2 would get it to O(log n) performance.