Question

Here is my function. It is a simple one, I’m just not confident on what the answer is.

  int calcul( int n) {
    if(n=1)
      return 1;
    else
      return calcul(n/2) + 1;
  }

Now, to get the complexity, I do:

T(n) = T(n/2) + O(1)

T(n/2) = T(n/4) + O(1)

…

T(1) = O(1)

Now, adding the equations, I’m getting

T(n) = O(1) + O(1)…

so what is the final answer ?

Answer 1

You’re executing the function once for each time you can divide n by 2, which is log n times.

So you get O(log n).

Edit:

The logarithm (of base 2) of a number n is the power 2 has to be raised to get n.

That is, 2^(log n) = n, where ^ indicated exponentiation.

Now, a simple way to calculate an approximation of log n is divide n by 2 while n > 1.

If you’ve divided k times, you get n < 2^k.

Since k - 1 divisions still yielded n > 1, you also have n >= 2^(k-1).

Taking logarithms on each member of 2^(k - 1) <= n < 2^k, you get k - 1 <= log n < k.