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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 b

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.

roywiggins 7y ago
Here's a badly-worked out theory. 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.
- marvin 7y ago
  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!
  - jtolmar 7y ago
    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.
    - marvin 7y ago
      Great explanation, thanks :)
- _nalply 7y ago
  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.
  - roywiggins 7y ago
    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.
OJFord 7y ago
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.
- roywiggins 7y ago
  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.
  - OJFord 7y ago
    I know it's not about actual colours, I did say it was off-topic. :) Re proving any, I replied to a sibling comment. What I was imagining was the inverse - proving that you can't construct one that requires a fifth.
- kragen 7y ago
  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.
LocalH 7y ago
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.

roywiggins 7y ago

Here's a badly-worked out theory. 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.

marvin 7y ago
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!
- jtolmar 7y ago
  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.
  - marvin 7y ago
    Great explanation, thanks :)
_nalply 7y ago
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.
- roywiggins 7y ago
  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.

OJFord 7y ago

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.

roywiggins 7y ago
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.
- OJFord 7y ago
  I know it's not about actual colours, I did say it was off-topic. :) Re proving any, I replied to a sibling comment. What I was imagining was the inverse - proving that you can't construct one that requires a fifth.
kragen 7y ago
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.

LocalH 7y ago

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.