How to tell if a color is imaginary/impossible?

Question

Short version

How can I tell if a color (e.g. XYZ) is impossible? (Wikipedia: Imposible color)

For example, this color is impossible:

• XYZ: (15.96, 84.04, 0)
• xyY: (0.1595, 0.8404, 0.8404)
• Lab: (93, -196, 161) (D65 whitepoint)

It's impossible because it lies outside of the chromacity diagram:

How can I know that?

Incorrect code

The goal is for someone to fill in the function:

Boolean IsImaginaryColor(Single X, Single Y, Single Z)
{
   //...TODO: Get someone to answer the question.
}

Right now we know that if any of the components of a corresponding LMS color are negative, then the color is imaginary.

That is a necessary, but not sufficient, condition for a color to be real. You can have all three components of LMS be positive, but it still be an imaginary color.

Boolean IsImaginaryColor(Single X, Single Y, Single Z)
{
   //If any component of LMS color is negative, 
   //then the color is definitely imaginary.
   LMSColor lms = XYZtoLMS(X, Y, Z);
   if ((lms.L < 0) or (lms.M < 0) or (lms.S < 0))
      return true;

   //The color may still be imaginary, 
   //but i don't know how to solve that problem
   //So as a first approximation i'll say it's real
   return false; 
}

LMSColor XYZtoLMS(Single X, Single Y, Single Z)
{
   //perform Matrix multiplication:
   //
   // LMS = M * XYZ
   //
   // Where M is the M_CAT02 transformation matrix from CIECAM02
   //
   //    0.7328, 0.4296, -0.1624
   //   -0.7036, 1.6975,  0.0061
   //    0.0030, 0.0136,  0.9834

   LMSColor result;

   result.L =  0.7328*X + 0.4296*Y + -0.1624*Z
   result.M = -0.7036*X + 1.6975*Y +  0.0061*Z
   result.S =  0.0030*X + 0.0136*Y +  0.9834*Z
}

In the xy color plane, this gives a good first-approximation (and nice visual indication) of impossible colors:

But the calculation still gives colors outside the chromacity diagram *(technically they're outside the "spectral locus"). So obviously only checking for negative components in LMS is incomplete.

Long Version

I am rendering a color picker. For example:

• to pick an Lab color
• you pick an ab color
• for a given L plane

This is similar to what you can already do in Photoshop:

So in this case I've picked the color:

• Lab: (72, -58, 119)

That color (assuming the D65 whitepoint) corresponds to the XYZ color:

• Lab: (72, -58, 119)
• XYZ: (25.22, 43.66, 0.36)

You can tell if a real color is outside the sRGB color gamut if one of its components is either:

• less than 0
• greater than 255

This XYZ color lies outside of the sRGB color space because one of it's components is negative:

• XYZ: (25.22, 43.66, 0.36)
• Lab: (72, -58, 119) (D65)
• RGB: (106.1, 199.6, -234.7) (sRGB)

Photoshop already knows if a color is outside the sRGB color gamut, and will display a gumut warning:

But I'd like to go one step further

I can already know if a color is outside the sRGB color gamut.

But now i want to know if a color is imaginary, so i can continue to show the gamut, but hide completely impossible colors. A conceptual mockup might be:

Warning: I have no idea which of those colors actually are impossible. This is only the idea of the concept.

So what I need to know is if a color is impossible.

Background Theory - What is an example of an impossible color?

The Wikipedia page on Impossible colors notes that while the primaries for the sRGB color space all lie inside the spectral locus - and so are all real colors:

The ProPhotoRGB color space does use some primaries that are impossible:

The ProPhoto RGB color space uses imaginary green and blue primaries to obtain a larger gamut (space inside the triangle) than would be possible with three real primaries. However, some real colors are still irreproducible.

So now I have a concrete example of an impossible color: the green primary of the ProPhoto RGB color space:

| Color | CIE x  | CIE y  |
|-------|--------|--------|
| red   | 0.7347 | 0.2653 |
| green | 0.1596 | 0.8404 | <--- this one
| blue  | 0.0366 | 0.0001 |
| white | 0.3457 | 0.3585 |

This impossible color, given different color spaces, is:

• xyY: (0.1596, 0.8404, 0.8404)
• XYZ: (15.96, 84.04, 0)
• LMS: (47.80, 131.43, 1.19)
• Lab: (93.4679, -195.9973, 161.1515)
• LCHab: (93.4679, 253.7415, 140.5725)

How can I tell that this color is impossible?

Given an XYZ color, how can I tell that it is impossible? E.g.:

• XYZ: 15.96, 84.04, 0

Bonus Chatter

It's important to note the difference between

• colors existing outside some gamut
• and imaginary colors

A quick single-image primer would be:

• Gamut: a color may not be displayable on your monitor, or printer, or phone, but it is still a real color - you could get a combination of Electromagnetic Waves of various wavelengths and intensities to generate the color
• Imaginary: No combination of EM waves, of any intensities, of any wavelengths, can generate that response in the Long, Medium, and Short human cones

I already know how to tell if a color exists outside a particular color gamut.

I want to know if a color also exists outside the spectral locus.

In other words: i want to know if it is imaginary.

Bruce Lindbloom has a nice graphic that raises the issues of colors outside the Lab color space when you arbitrary choose to arbitrarily limit the a and b component values to +- 128:

Bonus Reading

• https://physics.stackexchange.com/q/94375/
• Determine that a Luv Color is non-imaginary
• https://physics.stackexchange.com/questions/420614

Answer 1

This is a duplicate of the answer I gave here: Determine that a Luv Color is non-imaginary which relate to https://stackoverflow.com/a/48396021/931625

I think the safe way is to compute the XYZ volume boundaries and check if you are within or outside.

Answer 2

Some thoughts how this could be tackled numerically:

The edge of the domain of allowed colors is defined by the function

f: \lambda -> (x, y)

where \lambda is a wavelength and (x, y) is the corresponding spectral color locus.

From a practical point of view, you could approximate the curve defined by \lambda in, say, [0, 1000nm], i.e., the outer edge of the xy-color diagram with a polygon and then check if a given color point is inside that polygon (proper color) or not (imaginary color).

If you want it more accurate (but also computationally more involved), you can define a ray starting at the white point in the direction of the xy-point in question. Then you numerically solve for the point on the edge of the region of allowed colors, i.e., for the wavelength \lambda, such that f(\lambda) is on the ray. You can use for example a Newton solver to this end.