colormath: Desaturation

hmmmm . . . 0.3, 0.59, 0.11 ? i don't understand...

Just multiply every color component with 0.3 for red, 0.59 for green, 0.11 for blue.

this are the values for color conversion to grayscale. search with google for color conversion and yo will find many other algorithms

here are some links:

http://www.easyrgb.com/math.html
http://astronomy.swin.edu.au/~pbourke/colour/

sure? how can this be?? if you multiply plain white (1.0,1.0,1.0) with these values you get (0.3,0.59,0.11) which is green!?

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Original post by genesys sure? how can this be?? if you multiply plain white (1.0,1.0,1.0) with these values you get (0.3,0.59,0.11) which is green!?

No, you do a dot product of the color and (0.3, 0.59, 0.11) to get a grayscale intensity value. If you were converting to grayscale you would set each color channel to this intensity. What the AP was suggesting is to get this grayscale intensity and then linear interpolate between it and the original color to get your desaturation.

Quote:

Original post by genesys sure? how can this be?? if you multiply plain white (1.0,1.0,1.0) with these values you get (0.3,0.59,0.11) which is green!?

1 * .3 = .3
1 * .59 = .59
1 * .11 = .11
----------------+
intensity = .3 + .59 + .11 = 1

r = intensity
g = intensity
b = intensity

if you're just trying to change the image into grayscale, I believe that taking the average of the r,g,b channels and replacing them with it, should work.

m = .33333f*(r+g+b);r=g=b=m;

I don't want to desaturate the colors completely! just a lil bit!

In that case blend a little between the completely desaturated colour and the original colour. Do something like 0.9 * original_color + 0.1 * desaturated_color.

oh ok! i understand!

taht's working :) thanks!

just one mor question: perhaps there is a less accurate but (for small saturation changes) working formula with less operations?

For more colour manipulation ideas using interpolation/extrapolation, see the following: http://www.sgi.com/misc/grafica/interp/index.html

Quote:

just one mor question: perhaps there is a less accurate but (for small saturation changes) working formula with less operations?

The linear interpolations posted above can be done with a single pixel shader 1.1 instruction, you won't find much less than 1![smile]

HLSL:
const float3 coef = {0.3, 0.59, 0.11};
out.rgb = lerp(in.rgb, coef.rgb, amount);

PS1.1:
def c0, amount, 0, 0, 0
def c1, 0.3, 0.59, 0.11, 1
lrp r0.rgb, c0.r, r0.rgb, c1.rgb ; r0=input and output

Quote:

Original post by Anonymous Poster Anonymous Poster: Not one instruction, you forgot to do the dot product: (Without knowing exact HLSL syntax) HLSL: const float3 coef = {0.3, 0.59, 0.11}; out.rgb = lerp(in.rgb, dot3(coef.rgb, in.rgb), amount); PS1.1: def c0, amount, 0, 0, 0 def c1, 0.3, 0.59, 0.11, 1 ; r0=input and output dot3 r1.rgb, c1.rgb, r0.rgb lrp r0.rgb, c0.r, r0.rgb, r1.rgb

AP: Oops, my bad, yep you're quite right.


OP: So if you don't already have the luminance of the source image elsewhere in your shader, you'll have to dot the input colour against the coefficients to get it.

That said, luminance being scalar does open up some opportunities for co-issuing, so in most non-trivial pixel shaders, I'd still expect the whole desaturation operation to take only 1-1.5 effective instruction slots (or to be able to tag some extra functionality onto the operation to save elsewhere).