为什么这个函数具有对数时间复杂度（计算数字的第n个根）？

Question

Last week, I was taking a specific MOOC about Computer Science, and the prof. used an inefficient way to calculate the square root of a number (he later showed other ways as well).

Here's a function implemented in C++:

double sqrt(double num)
{
    double eps = 0.001;
    double step = 0.001;

    double result = 0.0;

    while (num - (result * result) > eps)
    {
        result += step;
    }

    return result;
}

I know that the while loop will be executed (square root of num / step ) times.

And I've decided to use matplotlib to draw a plot of this function growth for a range of numbers from 1 to 199 (inclusive), and here's the result:

Then, I compared it to (log(x) / step ) plot, and again here's the result:

And so, I have the following questions:

• Why it's logarithmic? and why this apply for any nth root of a number (using the above method) not only square root?
• What with that gap between sqrt growth and log(x)?

I'm aware that there's more efficient ways to achieve the same results of square root of a number, but I need someone to shed some light on this one.

Answer 1

You are correct when you say that the loop is executed sqrt(num) times, and that makes its complexity √num. However, contrary to the later assumption, a square root isn't logarithmic: it's simply num^(1/2) , which makes it polynomial in the grand scheme of things.

A clear sign and visual help is that it's not a straight line on a logarithmic plot:

The line on the left is a square root, and the line on the right is a base 10 logarithm.

The gap, obviously, is because it's not logarithmic.