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Update asciiart.glsl

This commit is contained in:
myownfriend 2014-05-21 00:05:13 -04:00
parent d13e9c5aa8
commit f59c7f25da
1 changed files with 21 additions and 22 deletions
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43
Data/Sys/Shaders/asciiart.glsl
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@ -26,7 +26,7 @@ void main()
vec4 mina = vec4(0.0, 0.0, 0.0, 0.0);
vec4 minb = vec4(0.0, 0.0, 0.0, 0.0);
for(int i=0; i<char_count; i++)
for (int i=0; i<char_count; i++)
{
vec4 ff = vec4(0.0, 0.0, 0.0, 0.0);
vec4 f = vec4(0.0, 0.0, 0.0, 0.0);
@ -34,9 +34,9 @@ void main()
vec4 t = vec4(0.0, 0.0, 0.0, 0.0);
vec4 tt = vec4(0.0, 0.0, 0.0, 0.0);
for(int x=0; x<char_width; x++)
for (int x=0; x<char_width; x++)
{
for(int y=0; y<char_height; y++)
for (int y=0; y<char_height; y++)
{
vec2 tex_pos = char_pos*char_dim + vec2(x,y) + 0.5;
vec4 tex = texture(samp9, tex_pos * resolution.zw);
@ -53,39 +53,38 @@ void main()
}
}
/*
The next lines are a bit harder, hf :-)
// The next lines are a bit harder, hf :-)
The idea is to find the perfect char with the perfect background color and the perfect font color.
As this is an equation with three unknowns, we can't just try all chars and color combinations.
// The idea is to find the perfect char with the perfect background color and the perfect font color.
// As this is an equation with three unknowns, we can't just try all chars and color combinations.
As criterion how "perfect" the selection is, we compare the "mean squared error" of the resulted colors of all chars.
So, now the big issue: how to calculate the MSE without knowing the two colors ...
// As criterion how "perfect" the selection is, we compare the "mean squared error" of the resulted colors of all chars.
// So, now the big issue: how to calculate the MSE without knowing the two colors ...
In the next steps, "a" is the font color, "b" is the background color, "f" is the font value at this pixel, "t" is the texture value
// In the next steps, "a" is the font color, "b" is the background color, "f" is the font value at this pixel, "t" is the texture value
So the square error of one pixel is:
e = ( t - a⋅f - b⋅(1-f) ) ^ 2
// So the square error of one pixel is:
// e = ( t - a⋅f - b⋅(1-f) ) ^ 2
In longer:
e = a^2⋅f^2 - 2⋅a⋅b⋅f^2 + 2⋅a⋅b⋅f - 2⋅a⋅f⋅t + b^2⋅f^2 - 2⋅b^2⋅f + b^2 + 2⋅b⋅f⋅t - 2⋅b⋅t + t^2
// In longer:
// e = a^2⋅f^2 - 2⋅a⋅b⋅f^2 + 2⋅a⋅b⋅f - 2⋅a⋅f⋅t + b^2⋅f^2 - 2⋅b^2⋅f + b^2 + 2⋅b⋅f⋅t - 2⋅b⋅t + t^2
The sum of all errors is: (as shortcut, ff,f,ft,t,tt are now the sums like declared above, sum(1) is the count of pixels)
sum(e) = a^2⋅ff - 2⋅a^2⋅ff + 2⋅a⋅b⋅f - 2⋅a⋅ft + b^2⋅ff - 2⋅b^2⋅f + b^2⋅sum(1) + 2⋅b⋅ft - 2⋅b⋅t + tt
// The sum of all errors is: (as shortcut, ff,f,ft,t,tt are now the sums like declared above, sum(1) is the count of pixels)
// sum(e) = a^2⋅ff - 2⋅a^2⋅ff + 2⋅a⋅b⋅f - 2⋅a⋅ft + b^2⋅ff - 2⋅b^2⋅f + b^2⋅sum(1) + 2⋅b⋅ft - 2⋅b⋅t + tt
To find the minimum, we have to derive this by "a" and "b":
d/da sum(e) = 2⋅a⋅ff + 2⋅b⋅f - 2⋅b⋅ff - 2⋅ft
d/db sum(e) = 2⋅a⋅f - 2⋅a⋅ff - 4⋅b⋅f + 2⋅b⋅ff + 2⋅b⋅sum(1) + 2⋅ft - 2⋅t
// To find the minimum, we have to derive this by "a" and "b":
// d/da sum(e) = 2⋅a⋅ff + 2⋅b⋅f - 2⋅b⋅ff - 2⋅ft
// d/db sum(e) = 2⋅a⋅f - 2⋅a⋅ff - 4⋅b⋅f + 2⋅b⋅ff + 2⋅b⋅sum(1) + 2⋅ft - 2⋅t
// So, both equations must be zero at minimum and there is only one solution.
So, both equations must be zero at minimum and there is only one solution.
*/
vec4 a = (f*ft - ff*t + f*t - ft*float(char_pixels)) / (f*f - ff*float(char_pixels));
vec4 b = (f*ft - ff*t) / (f*f - ff*float(char_pixels));
vec4 diff = a*a*ff + 2.0*a*b*f - 2.0*a*b*ff - 2.0*a*ft + b*b *(-2.0*f + ff + float(char_pixels)) + 2.0*b*ft - 2.0*b*t + tt;
float diff_f = dot(diff, vec4(1.0, 1.0, 1.0, 1.0));
if(diff_f < mindiff) {
if (diff_f < mindiff) {
mindiff = diff_f;
minc = float(i);
mina = a;