Reputation: 1051
I'm trying to display the Julia set with Newton iteration, but I get a result shown below. What could be the problem? Here's my EDIT: FIXED, WORKING code:
#version 130
in vec3 vs_out_col;
in vec3 vs_out_pos;
out vec4 fs_out_col;
vec2 cplx_mul(vec2 z1, vec2 z2)
{
return vec2(z1.x * z2.x - z1.y * z2.y, z1.y * z2.x + z1.x * z2.y);
}
vec2 cplx_div(vec2 z1, vec2 z2)
{
float denom = z2.x * z2.x + z2.y * z2.y;
return vec2(
(z1.x * z2.x + z1.y * z2.y) / denom,
(z1.y * z2.x - z1.x * z2.y) / denom
);
}
vec2 f(vec2 z)
{
vec2 res = cplx_mul(cplx_mul(z, z), z);
return vec2(res.x - 1.0f, res.y);
}
vec2 f_der(vec2 z)
{
return 3 * cplx_mul(z, z);
}
void main()
{
vec2 z = vs_out_pos.xy;
for(int i = 0; i < 30; ++i){
z = z - cplx_div(f(z), f_der(z));
}
vec2 root1 = vec2( 1.0f , 0.0f);
vec2 root2 = vec2(-1.0f/2.0f, 1.0f/2.0f * sqrt(3.0f));
vec2 root3 = vec2(-1.0f/2.0f, -1.0f/2.0f * sqrt(3.0f));
if(abs(length(z - root1)) < 0.5f){
fs_out_col = vec4 (1, 0, 0, 1);
}
else if(abs(length(z - root2)) < 0.01f){
fs_out_col = vec4 (0, 1, 0, 1);
}
else if(abs(length(z - root3)) < 0.01f){
fs_out_col = vec4 (0, 0, 1, 1);
}
else{
fs_out_col = vec4 (0, 0, 0, 1);
}
}
And here's the result:
And the Fixed result:
Upvotes: 2
Views: 1063
Reputation: 25992
You should compute the derivative correctly, of f(z)=z^3-1 the derivative is f'(z)=3*z^2, the factor 3 is missing.
And does the subtraction of the constant 1.0f really work that simply?
Upvotes: 2