Why does blending order matter? Is there a way to avoid this?

[drawtheliner]

The brown I get by adding Red, Blue and Yellow is different depending on how I add them.

If I add them all at once, I get the color on the left.
If I blend Yellow and Red (1/2), then blend that to blue (1/3), I get the color on the right.

All at once:

		var z1 = Mixbox.rgb_to_latent(Color(1,1,0,1))
		var z2 = Mixbox.rgb_to_latent(Color(1,0,0,1))
		var z3 = Mixbox.rgb_to_latent(Color(0,0,1,1))
		var z_mix = []
		z_mix.resize(Mixbox.LATENT_SIZE)
		for i in z_mix.size():
			z_mix[i] = (1.0/3.0*z1[i] +
						1.0/3.0*z2[i] +
						1.0/3.0*z3[i])
		color_mix = Mixbox.latent_to_rgb(z_mix)				

In two blends:

		color_mix2 = Mixbox.lerp(Color(1,1,0,1),Color(1,0,0,1),0.5)
		color_mix2 = Mixbox.lerp(color_mix2,Color(0,0,1,1),1.0/3.0)

Here's another attempt with two blends, this time doing R+B, then that +Y:
(note the brown on the right is different from both, the all-at-once mix on the left, and the previous 2-step blend)

[knokelmaat]

I believe that this is because while RGB -> LATENT -> RGB usually gives the same RGB value as the one you started with, LATENT -> RGB -> LATENT does not.

When doing all three in latent space you get only one traversal of this conversion:
RGB -> LATENT -> RGB

When you mix 2 followed by the third, you get two traversals of this conversion:
RGB -> LATENT -> RGB -> LATENT-> RGB

It is this extra conversion that seems to add an error to the calculations.

I don't know if this behavior is due to an error in the latent_to_rgb or rgb_to_latent code or if it is just inherent to the calculation method used...

[drawtheliner]

Thank you! Removing one of the conversions eliminated the difference.