# Getting Composition of a Mixture of Colors

I wanted to validate a thought on blending of colours digitally.

Imagine I have two colours, $C_1$ $(R_1,G_1,B_1)$ and $C_2$ $(R_2,G_2,B_2)$. Now, I create colour $C3$ as follows: $C_3 = \alpha C_1 + (1-\alpha) C_2$

Where $\alpha$ is a value between $0$ and $1$. So, yes, basically, I am trying to put one object over another with a transparency associated with it.

What I want to understand is that, if I have $C_3$, $\alpha$ and $C_1$ in the above equation, will I get an accurate value of $C_2$?

Math suggests that I should, but I just want to confirm that I am not missing anything. Also, can I keep a really high but $< 1$ value for $\alpha$ and still accurately get $C_2$?

You actually have 3 independent equations.
If one of them is solvable, all three are.

Now, the equation is given by:

$$z = \alpha x + \left( 1 - \alpha \right) y$$

Assume $\alpha, x, z$ are known, then:

$$y = \frac{z - \alpha x}{1 - \alpha}$$

For any value of $\alpha \neq 1$ the solution is valid.
Moreover, if $alpha = 1$ then $z = x$ and $y$ is irrelevant.

Looking at the other case, restoring $x$, yields:

$$x = \frac{z - \left( 1- \alpha \right) y}{\alpha}$$

For any value of $\alpha \neq 0$ the solution is valid.
Moreover, if $alpha = 0$ then $z = y$ and $x$ is irrelevant.

As you wrote, in order to have a Linear Interpolation the parameter $\alpha$ must be in the range $\left[ 0, 1 \right]$.

• Yes. I ended up writing some code to try it out and it works perfectly fine. – TheBlueNotebook Dec 5 '15 at 4:58
• Hi, Great to hear. It would be great to mark the answer so the question will be marked as solved. Enjoy... – Royi Dec 5 '15 at 18:43