O（N）与O（NlogN）

Question

I came accros this example while working on Big-O notation

x=n
while(x>0)
{
    y=x
    while(y>0)
    {   
        y=y-1
    }
    x = x/2
}

Could, you explain me why it seems to have a O(N) complexity ?

This is new for me but I would expect it to be N LogN.

What am I missing ?

Thanks,

EDIT: this piece of code comes from here https://www.quora.com/How-can-we-check-for-the-complexity-log-n-and-n-log-n-for-an-algorithm?encoded_access_token=1b7e0a4d10f84d50b5fea810f5a89cea

Answer 1

Well, let's look how often the inner loop's body is executed:

x =   n:      n
x = n /  2: n /  2
x = n /  4: n /  4
x = n /  8: n /  8
x = n / 16: n / 16
x = n / 32: n / 32
x = n / 64: n / 64
until x < 1

Or put it together:

n + n / 2 + n / 4 + n / 8 + n / 16 + n / 32 + n / 64 ...

Which is easily seen is the same as:

n + n - n / 64

Now we want an upper bound, so we may ignore the last term. And for big-oh, the constant is also insignificant. So:

O(n)

Answer 2

If you find how many times the inner loop runs, you find the complexity of the code. The inner loop runs n + n/2 + n/4 + ... n/k (where n/k>0) times. The maximum value of n/2 + n/4 + ... + n/k part is n-1. Thus, the code can not run more than 2n-1 times making the upper bound 2n-1 which is O(n)