Update asciiart.glsl

Data/Sys/Shaders/asciiart.glsl

@@ -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.
+		// 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 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.
+		// 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 ...
+		// 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: 
+		// So the square error of one pixel is: 
-			e = ( t - a⋅f - b⋅(1-f) ) ^ 2
+		// e = ( t - a⋅f - b⋅(1-f) ) ^ 2
 
-			In longer:
+		// 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
+		// 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)
+		// 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
+		// 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":
+		// 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/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
+		// 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;