Given an R,G,B triplet and a factor F, how do I calculate a “watermark” version of the color?

I have an (R, G, B) triplet, where each color is between 0.0 and 1.0 . Given a factor F (0.0 means the original color and 1.0 means white), I want to calculate a new triplet that is the “watermarked” version of the color.

I use the following expression (pseudo-code):

for each c in R, G, B:
    new_c ← c + F × (1 - c)

This produces something that looks okayish, but I understand this introduces deviations to the hue of the color (checking the HSV equivalent before and after the transformation), and I don't know if this is to be expected.

Is there a “standard” (with or without quotes) algorithm to calculate the “watermarked” version of the color? If yes, which is it? If开发者_如何学编程 not, what other algorithms to the same effect can you tell me?

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Actually this looks like it should give the correct hue, minus small variations for arithmetic rounding errors.

This is certainly a reasonable, simple was to achieve a watermark effect. I don't know of any other "standard" ones, there are a few ways you could do it.

Alternatives are:
Blend with white but do it non-linearly on F, e.g. new_c = c + sqrt(F)*(1-c), or you could use other non-linear functions - it might help the watermark look more or less "flat".

You could do it more efficiently by doing the following (where F takes the range 0..INF):

new_c = 1 - (1-c)/pow(2, F)

for real pixel values (0..255) this would convert into:

new_c = 255 - (255-c)>>F

Not only is that reasonably fast in integer arithmetic, but you may be able to do it in a 32b integer in parallel.

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Why not just?

new_c = F*c

I think you should go first over watermarking pixels and figure out if it should be darker or lighter.

For lighter the formula might be new_c=1-F*(c-1)