Quick Procedural Geometry Notes

This past week was crazy busy and I didn’t have any time to sit down and code. Nonetheless, I had the time to work out the math and jot down a quick piece of pseudocode on how to procedurally generate a cone.

Math and Pseudocode for generating a cone procedurally.

Math and Pseudocode for generating a cone procedurally.

Here, I’ve chosen to place the cone’s center at (0,0,0) in Object Space. This will allow me to always rotate the cone on its apex, something that will be extremely helpful for the effect we’re trying to achieve.

I haven’t had the time to test this idea in code yet, however, the plan is to build it into Vortex as part of the vtx::procedural package.

I’m not giving away any more details this week! You’ll have to stay tuned for more!

Implementing sqrt

I was reading some post about interview questions of 2011 and came across one that stated “find the square root of a number”.

Assuming we can’t use the sqrtf function of the standard C math library, let’s see how we can calculate the square root of a number x.

Given n, we know that its square root is a number x that holds:

\sqrt{n} = x

Let’s work on this equation a little. Raise both sides to the second power:

n = x^2

Move to the left of the equality:

0 = x^2 - n

If we found the roots of this last equation somehow, we would have found the square root of n. We can do this by using the Newton-Raphson iteration.

The Newton-Raphson iteration states that we can find the root of an equation using the following formula iteratively:

x_{n+1} = x_n - \frac{f(x)}{f'(x)}

Where f'(x) is the derivative of function f(x). We will approximate the derivative using the definition of derivative at a point (we could also note that the derivative could be trivially calculated; this method is more general).

f'(x) = \frac{f(x+h) - f(x)}{h}

The error of the Newton-Raphson iteration is given by:

|x_{n+1} - x_n|

Starting with a hardcoded seed value, we will perform this iteration in a loop until the error is less than a given value. I have chosen to iterate until the error is less than 1×10^(-10): 0.00000000001.

Let us see what a tentative “pythonesque” pseudocode for this loop could be:

def sqrt(n):
    f = function(x*x - n)
    x = 1 # seed
    xant = 0
        f1 = (f.eval(x+h) - f.eval(x)) / h
        xant = x
        x = x - f.eval(x)/f1
    while abs(x - xant) > err;
    return x

Assuming we have a symbolic function type, that loop does not seem too difficult. In order to code this in C, since the equation is always the same, I will hardcode it as a plain function.

typedef double real; // change to float for single precision

real f(real x, real n)
    return x*x - n;

real sqrt(real n)
    real err = 0.00000000001f;
    real h = 0.01f;

    real x = 1.0f; // seed
    real xant = 0.0f;

        xant = x;
        real df = (f(x+h, n) - f(x, n))/h;
        x = x - f(x, n)/df;
    while (abs(x - xant) > err);

    return x;

Here are the results of running our custom square root function, compared to the standard version provided with the C programming language:

[ale@syaoran sqrt]$ ./sqrt 1.0
Custom sqrt of: 1 = 1
libm sqrt of: 1 = 1

[ale@syaoran sqrt]$ ./sqrt 2.0
Custom sqrt of: 2 = 1.41421
libm sqrt of: 2 = 1.41421

[ale@syaoran sqrt]$ ./sqrt 4.0
Custom sqrt of: 4 = 2
libm sqrt of: 4 = 2

[ale@syaoran sqrt]$ ./sqrt 16.0
Custom sqrt of: 16 = 4
libm sqrt of: 16 = 4

[ale@syaoran sqrt]$ ./sqrt 32.0
Custom sqrt of: 32 = 5.65685
libm sqrt of: 32 = 5.65685

[ale@syaoran sqrt]$ ./sqrt 100.0
Custom sqrt of: 100 = 10
libm sqrt of: 100 = 10

[ale@syaoran sqrt]$ ./sqrt 1000000.0
Custom sqrt of: 1e+06 = 1000
libm sqrt of: 1e+06 = 1000