computing RGBA to match an RGB color

Question

If I have an RGB color, with 100% opacity.

I want that same color (or close to it) with a transparent alpha channel. I will paint the transparent color over a white background.

How do I compute the RGBA color?

I guess what I am asking is the opposite of this question.

Answer 1

You mean you want the RGBA color with maximum transparency which, when drawn on top of a white background, gives the original RGB color?

Let R₀, G₀ and B₀ be the components of the original color, each ranging from 0.0 to 1.0, and let R, G, B and A be the components of the new RGBA color (with A = 1 denoting 100% opacity). We know that the colors must satisfy:

R₀ = A·R + (1 − A)
G₀ = A·G + (1 − A)
B₀ = A·B + (1 − A)

which, if we knew A, we could easily solve for R, G and B:

R = (R₀ − 1 + A) / A = 1 − (1 − R₀) / A
G = (G₀ − 1 + A) / A = 1 − (1 − G₀) / A
B = (B₀ − 1 + A) / A = 1 − (1 − B₀) / A

Since we require that R ≥ 0, G ≥ 0 and B ≥ 0, it follows that 1 − R₀ ≥ A, 1 − G₀ ≥ A and 1 − B₀ ≥ A, and therefore the smallest possible value for A is:

A = max( 1 − R₀, 1 − G₀, 1 − B₀ ) = 1 − min( R₀, G₀, B₀ )

Thus, the color we want is:

A = 1 − min( R₀, G₀, B₀ )
R = 1 − (1 − R₀) / A
G = 1 − (1 − G₀) / A
B = 1 − (1 − B₀) / A

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Ps. For a black background, the same formulas would be even simpler:

A = max( R₀, G₀, B₀ )
R = R₀ / A
G = G₀ / A
B = B₀ / A

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Pps. Just to clarify, all the formulas above are for non-premultiplied RGBA colors. For premultiplied alpha, just multiply R, G and B as calculated above by A, giving:

R = A · ( 1 − (1 − R₀) / A ) = R₀ − (1 − A)
G = A · ( 1 − (1 − G₀) / A ) = G₀ − (1 − A)
B = A · ( 1 − (1 − B₀) / A ) = B₀ − (1 − A)

(or, for a black background, simply R = R₀, G = G₀ and B = B₀.)