101 lines
3.3 KiB
GLSL
101 lines
3.3 KiB
GLSL
// textures
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uniform sampler2D samp8;
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uniform sampler2D samp9;
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const int char_width = 8;
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const int char_height = 13;
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const int char_count = 95;
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const int char_pixels = char_width*char_height;
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const vec2 char_dim = vec2(char_width, char_height);
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const vec2 font_scale = vec2(1.0/float(char_width)/float(char_count), 1.0/float(char_height));
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out vec4 ocol0;
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in vec2 uv0;
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uniform vec4 resolution;
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void main()
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{
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vec2 char_pos = floor(uv0*resolution.xy/char_dim);
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vec2 pixel_offset = floor(uv0*resolution.xy) - char_pos*char_dim;
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// just a big number
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float mindiff = float(char_width*char_height) * 100.0;
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float minc = 0.0;
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vec4 mina = vec4(0.0, 0.0, 0.0, 0.0);
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vec4 minb = vec4(0.0, 0.0, 0.0, 0.0);
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for(int i=0; i<char_count; i++)
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{
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vec4 ff = vec4(0.0, 0.0, 0.0, 0.0);
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vec4 f = vec4(0.0, 0.0, 0.0, 0.0);
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vec4 ft = vec4(0.0, 0.0, 0.0, 0.0);
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vec4 t = vec4(0.0, 0.0, 0.0, 0.0);
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vec4 tt = vec4(0.0, 0.0, 0.0, 0.0);
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for(int x=0; x<char_width; x++)
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{
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for(int y=0; y<char_height; y++)
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{
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vec2 tex_pos = char_pos*char_dim + vec2(x,y) + 0.5;
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vec4 tex = texture(samp9, tex_pos * resolution.zw);
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vec2 font_pos = vec2(x+i*char_width, y) + 0.5;
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vec4 font = texture(samp8, font_pos * font_scale);
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// generates sum of texture and font and their squares
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ff += font*font;
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f += font;
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ft += font*tex;
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t += tex;
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tt += tex*tex;
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}
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}
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/*
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The next lines are a bit harder, hf :-)
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The idea is to find the perfect char with the perfect background color and the perfect font color.
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As this is an equation with three unknowns, we can't just try all chars and color combinations.
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As criterion how "perfect" the selection is, we compare the "mean squared error" of the resulted colors of all chars.
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So, now the big issue: how to calculate the MSE without knowing the two colors ...
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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
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So the square error of one pixel is:
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e = ( t - a⋅f - b⋅(1-f) ) ^ 2
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In longer:
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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
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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)
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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
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To find the minimum, we have to derive this by "a" and "b":
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d/da sum(e) = 2⋅a⋅ff + 2⋅b⋅f - 2⋅b⋅ff - 2⋅ft
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d/db sum(e) = 2⋅a⋅f - 2⋅a⋅ff - 4⋅b⋅f + 2⋅b⋅ff + 2⋅b⋅sum(1) + 2⋅ft - 2⋅t
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So, both equations must be zero at minimum and there is only one solution.
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*/
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vec4 a = (f*ft - ff*t + f*t - ft*float(char_pixels)) / (f*f - ff*float(char_pixels));
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vec4 b = (f*ft - ff*t) / (f*f - ff*float(char_pixels));
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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;
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float diff_f = dot(diff, vec4(1.0, 1.0, 1.0, 1.0));
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if(diff_f < mindiff) {
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mindiff = diff_f;
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minc = float(i);
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mina = a;
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minb = b;
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}
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}
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vec2 font_pos_res = vec2(minc * float(char_width), 0.0) + pixel_offset + 0.5;
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vec4 col = texture(samp8, font_pos_res * font_scale);
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ocol0 = mina * col + minb * (vec4(1.0,1.0,1.0,1.0) - col);
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}
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