Colors like magenta (red and blue wavelengths together) are nonspectral, having no single wavelength they correspond to. They only exist as a mix, and we see a pinkish color that you can't write down as a wavelength.
Tetrachromats could plausibly see more nonspectral colors by activating their extra color receptor and one of the other ones. There would be no exact equivalent in trichromatic vision.
It depends on what their brains decide to do with it- maybe they would just land on an existing color in between. But there's no reason it has to.
So these colors are something that tetrachromats could maybe see while awake, but usually with lots of other colors mixed in. Asleep, maybe the brain starts to play with them and generates pure versions of them. Trichromats wouldn't be very likely to have these dreams because their brains don't have experience of tetrachromat nonspectral colors at all.
Ah! Now I have theory: We trichromats have a hue circle, and magenta is the point where the lower and higher wavelenghts are joined. However this simple model breaks down for tetrachromats. Perhaps they have two hue circles orthogonal to each other. This means, hue becomes two-dimensional. Really crazy to imagine such a thing.
I imagine in school the teacher tells the poor girl: Mix the blue and yellow colors to get green, and the girl doesn't understand because she has a lot more complex color model than the teacher.
We see magenta as a separate color because we can distinguish it from green- it's not green, because Red and Blue don't activate our green cones. But we can't distinguish between spectral yellow and R+G because there's no yellow cone: the activations of all our cones are more or less identical in both cases.
If you have a yellow cone, you'd conceivably be able to distinguish between both kinds of yellow: redgreen and yellow could look as different as green and redblue do. I'm not sure how that would look as a "wheel", perhaps a figure-eight with the crossing at yellow.
I think this is a very good explanation. You're emphasizing that a trichromat will experience the same activation of their cones to the wavelength of yellow light, and light containing the wavelengths of red and green mixed. So the spectra of these two colors will be different, but appear identical.
Whereas a hypothetical tetrachromat with an additional cone type that activates to the yellow wavelength, will have obviously different activations of their cones for the "wavelength of yellow" and "wavelengths of red and green mixed". And hence experience them as different colors.
I'm not familiar enough with the different definitions of color spaces to define how to model this. You could easily define it as a four-dimensional RGBY color space, but to my understanding the other color spaces are defined because they have better properties for combining different colors, or playing better with interpolation between colors.
It would be a challenge to come up with a good four-dimensional color space that is a useful artistic and visual tool, given that there are very few people around to evaluate it! And also because all computer monitors are trichromatic, so it'd be a job in itself to set up a test system.
I wish someone in the interview had asked some of these people how they experience photos on computers!
You can arrange the three colors in a triangle, which defines a plane, is treated like a circle in color theory, and looks like a weird stretched out "U" when put in a perceptually uniform space (stretched towards green, which we see best, and flat on the magenta side). Perceptual color spaces combine this with luminance, which is sensed by rods instead of cones.
Tetrachromatic colors could be placed in a tetrahedron, defining a space and being approximated as a sphere. A perceptually uniform representation would probably be curved on the three edges that agree with the spectrum and flat on the three that don't. Green is still probably the biggest. Then you'd add luminance as a fourth dimension because it's still rods.
Interesting idea, however I think this is too simple. With three cones we have a three-dimensional space of color. We can express this space with RGB or HSV and so on, just by a three dimensional basis vector we can describe one color point in the color space.
With tetrachromats we get a four-dimensional space of color. So we need something like RGBX or a two-dimensional hue value like for example H1,H2,S,V. That's what I meant with a second hue circle orthogonal to the one we already know.
What we see as yellow is for the tetrachromat a whole second hue circle varying in the value of second hue.
An example: Yellow is a point on the first hue circle. For the tetrachromat these colors are different, but boring yellow for us:
- pure yellow
- mixture of red and green
- mixture of pure yellow, red and green
In other words, what we see as yellow is a whole dimension of different colors for tetrachromats.
They also would probably be able to distinguish yellow+blue from pure green.
Also, if you mix R+B, that's magenta, and if you start mixing in blue it will turn white or gray. But if you can see yellow, RYB would look different from R+a little g+B, which is what RYB would look for trichromats, I think.
I imagine in school the teacher tells the poor girl: Mix the blue and yellow colors to get green, and the girl doesn't understand because she has a lot more complex color model than the teacher.
This is honestly one of the saddest parts of human society. Some people, including those in positions of authority, either don't understand or outright reject the notion that different people perceive the world differently.
What you've described is why I (not a graduate of mathematics) never understood why the 'four colour theorem' is difficult to prove. I accept that it is, because it took so long and initially was only done computationally, but I don't understand it.
In secondary school I argued along the lines you describe, that you can mix three colours in a circle, and then a fourth around the circumference mixes only with at most two of the others - and in terms of the FCT is then a 'barrier' allowing re-use of the first three.
That's quite off-topic, but you suddenly jolted my memory of it. More on-topic, being colour-blind (never told which type and I can't work it out either) I find conversations about colours frustrating, and very strange to wonder if we see anything alike at all, or just learn to call the same physical thing alike despite perceiving it perhaps very differently.
The hard part in proving the four color theorem is proving that there aren't any possible maps that can't be colored with four colors. Perhaps there's one out there that needs five and can't be colored with only four.
Proving some maps can't be colored with three colors is easy, you only need to find one and enumerate all the three-colorings and find none of them work.
It's not really a theorem about colors though, they're just labels.
It's straightforward to prove that you can't have more than four plane regions that all mutually touch; K5 is a nonplanar graph. But you can construct maps (planar graphs) that have no K4 subgraphs that nevertheless have chromatic number 4. C5 plus an additional node connected to all 5 of its nodes, for example. The challenge is to show that the same thing can't happen with chromatic number 5. I think it does happen with chromatic numbers 6 and 7 on the surface of a torus (but you can also just map K7 onto a toroidal map.)
I recognize that this comment includes jargon, but I assume you'll Google it. I'd include diagrams but HN doesn't support them.
Thank you for giving me the terminology - I think I do see why it's hard mathematically, I just struggle to understand that graphically, it seems so obvious on paper - looking by cases at how many existing regions a necessarily new colour touches.
But, I understand and trust the mapping of the problem to a graph, so perhaps if I see why the problem is hard on a graph that should be enough for me.
But my reasoning was along the lines (but now in my newfound terminology) of:
K1 - trivially 1 colour needed. K2 - 2.
If we add another region on the map, we either have K3 or K1,2. In the latter case the third region/vertex need only be different to the one it touches, but K3 clearly needs 3 colours.
Adding the fourth region is similar, and again K4 clearly needs 4.
The fifth region touches at most 3 others, since K5 is non-planar. Thus, it can take the colour of the untouched region.
That's not very formal, but why isn't proof by induction easy from there? i.e. if we can't build a map that ever requires a fifth colour, then it can't be that there exists a map which requires more than four?
But why does it matter if I can tell you how to colour it,* isn't it sufficient to show that we can't construct one that can't be coloured in at most four, as I tried to outline above? Surely it must then be that any exercise is either 4-colourable, or non-planar/otherwise out of scope of the problem?
* It's easy starting from the end, just put one colour in the middle, then alternate second/third colours around the edge until you get back to the start and have to add a fourth. From construction, the worst case is from K1,2 then adding another for K1,3 at which point we need four colours already, but when we add two more on what will be the C5# around they each only connect to one each side and the center. In adding each one you can simply use the colour one hop over on the circumference.
# Since the center of K1,3 must be the center of the final C5-plus-thing, as it already had 3 spokes.
It's correct, as you say, that if we can't build a map that ever requires a fifth colour, then it can't be that there exists a map which requires more than four.
It's just that the C₅-plus-thing requires four colors, even though it doesn't contain any K₄ induced subgraphs. So we could imagine that there might be some larger planar graph that requires 5 colors without containing K₅. As it turns out, there isn't, but that's the thing that's hard to prove.
Hm, many thanks for the pointers. It's still not completely clear to me, but I have a much better idea and know the story of thing to search/read up on. Cheers.
That has nothing to do with my argument. Why would tetrachromats be inclined to dream of colors they do not see in the real world any more than trichromats?
My point is they would see those colors in the real world, but only in combination with other colors, so it wouldn't be that noticable. They wouldn't see pure versions of them much, if at all. The intensity in a dream might be much stronger, enough that it would seem like a qualitatively new color rather than "hmm, that blue has something different going on compared to other blues" they you might get in real life.
You could put them in front of light source that generated pure wavelengths designed to selectively activate their extra cone and ask what they see when you start mixing pure blue or green into it. Maybe they'd recognize those colors from their dreams, maybe not. I don't know. But it's a mechanism that would explain how tetrachromats might dream colors that trichromats don't.
But this assumes the gamut of trichromacy is more anchored in reality (woken life) than the gamut of tetrachromacy. In principle there is nothing different between a trichromat and a tetrachromat sense of color. They see one dimension of color more, just like humans see one dimension more than dogs.
The fourth cone in tetrachromats overlaps quite a lot with the other three (which is why most tetrachromats don't see more colors than trichromats). For people where the fourth cone is meaningfully distinguishable from the others, it doesn't seem weird to me that you wouldn't see many objects that are strongly activating just that extra cone and not the ones next to it (wavelength wise) because they're closer together.
We don't see bright, "pure" magenta very much, and when we do it's usually because we've engineered it. Nobody is engineering pigments for tetrachromats to enjoy.
If I'm right, people who are nearly (but not entirely) colorblind (anomalous trichromats) due to a shifted cone would likewise be able to dream in colors they are capable of seeing only dimly while awake, whereas dichromats wouldn't.
Tetrachromats could plausibly see more nonspectral colors by activating their extra color receptor and one of the other ones. There would be no exact equivalent in trichromatic vision.
It depends on what their brains decide to do with it- maybe they would just land on an existing color in between. But there's no reason it has to.
So these colors are something that tetrachromats could maybe see while awake, but usually with lots of other colors mixed in. Asleep, maybe the brain starts to play with them and generates pure versions of them. Trichromats wouldn't be very likely to have these dreams because their brains don't have experience of tetrachromat nonspectral colors at all.