12 ms·

I'm curious—is there a point to giving two separate names to two colors which are indistinguishable from one another to the human eye? There are zero people in

I'm curious—is there a point to giving two separate names to two colors which are indistinguishable from one another to the human eye? There are zero people in the world who can tell #FF4500 from #FF4501, so aren't they effectively the same color?

baddox 7y ago
By induction, doesn’t that mean we would only have one color name?
- bb2018 7y ago
  If there are zero people who cannot tell two colors apart, they should have the same name. If there is at least one person who cannot tell two colors apart, they should have different names.
  - roel_v 7y ago
    What does 'telling a color apart' even mean? I did a mini project some time ago, where I basically did some custom color quantization (it was to reduce an image into a configurable amount of discrete colors, to make a pixel quilt with a manageable amount of distinct fabric colors). So I used CieLAB colors to calculate the 'visual distance' between the available colors and then recursively merged colors until I got to the desired amount. But even a simple algorithm like that, it turned out, could be implemented in several way; each of which gave different, yet equally 'valid' results. My point being - it's very hard to give a simple, unambiguous meaning to seemingly simple concepts like 'what is a color', 'when are colors the same' and 'are these two colors more similar than those other two'. There is a lot of abstraction between the physics of light and representations of color e.g. on a screen, or in fabric, or when printing.
    - Firerouge 7y ago
      I think in the case of telling two colors apart, it's can be as simple as presenting them side by side, and seeing if someone can distinguish a difference.
      - Galanwe 7y ago
        Not sure that is so simple unfortunately. First, because if I recursively apply your argument, then no color would have a name. Starting from the very 1st lowest color frequency and the 2nd right next to it, no-one could differentiate them. Apply the same logic between the 2nd and the 3rd, the 3rd and the 4th, etc. So you cannot apply this argument to "any" color pair. You would have to define a set of starting colors, and always do the comparison against them. Now you enter the problem of which starting colors to starts, and the fact that using any pre-existing color names would most likely not be evenly distributed on the color spectrum. Second, it wouldn't work because color recognition is not a transitive process. - If you show me two very close red variations side by side, I may not be able to tell the difference. - If you space the variation, I may be able to tell that there are 2 different reds, but wouldn't be able to tell which one is darker/lighter accurately. - If instead of showing me the two close reds side by side, you show them to me paired with an other color, I may be able to tell the difference. For instance, show me a board with half red1 + yellow, and an other board with red2 + yellow. I may spot than 1 of the red + yellow is matching that good compared to the other one.
        eru 7y ago
        You can get out of that argument by saying that colour names apply in different amounts to different hex-colours. So #FF0000 is red. But is #FA0101 is also red, just slightly less so.
        msclrhd 7y ago
        That's interesting. So what you have is some sort of similarity metric that is colour based. You could define it as being 1 for an exact match, and 0 for not matching, so cyan, etc. would be 0 red, while orange may be slightly red (e.g. 0.1), and pink may be slightly more red (e.g. 0.25). This would be a 3D function/surface. This would allow you to model classes of colours, like purples. Shades of a colour could then be any similar colour in the range 0 to 0.5 (or 0.25).
        roel_v 7y ago
    - js2 7y ago
      > What does 'telling a color apart' even mean? Being able to distinguish colors on a test such as https://xritephoto.com/cool-tools https://xritephoto.com/cool-tools
      - ddoeth 7y ago
        Although I got it right in the end, I found the second row to be insanely hard. But somehow, on all rows, the colors just seemed to snap into place for me, is that a normal feeling?
- ryangittins 7y ago
  I don't follow. Two given colors being indistinguishable doesn't imply that any two colors are indistinguishable.
  - tzs 7y ago
    Your suggestion was that if two colors are indistinguishable, they should have that name. But given any two colors A and B you can construct a sequence C1 = A, C2, C3, ..., Cn = B where adjacent colors Ci and Ci+1 are indistinguishable from each other, and so need the same name, for 1 <= i < n. Hence, A and B have to have the same name.
    - eru 7y ago
      It's just the old paradox of the heap. https://en.wikipedia.org/wiki/Sorites_paradox https://en.wikipedia.org/wiki/Sorites_paradox
    - IgorPartola 7y ago
      So make Ci and Ci + 1 distinguishable. Problem solved.
    - ryangittins 7y ago
      The problem with your argument is that indistinguishable doesn't mean the equal. I'm not saying there is no difference between adjacent colors. I'm saying that the difference is indistinguishable. A lot of indistinguishable differences add to a distinguishable difference. Think about it with numbers: if your sequence A, B, C... are each zero units apart, then yes, A = B = C = ... In actuality, though, they are not zero units apart. They are a small, but positive distance apart. Those tiny differences add up so the distance from A to C is twice as far as the distance from A to B.
      - baddox 7y ago
        Any two adjacent colors must either have the same name or different names.
- 333c 7y ago
  "can't tell apart" isn't a transitive relation. That is, just because I can't tell A from B and I can't tell B from C doesn't mean I can't tell A from C.
  - drdeca 7y ago
    “Has the same name as”, however, is a transitive relation. So, if no two indistinguishable colors have distinct names, and for every pair of colors have a path between them of colors such that adjacent colors in the path are indistinguishable, then all colors would have the same name.
    - zeroimpl 7y ago
      The flaw here is the assumption that colors have a single, well-defined name.
      - drdeca 7y ago
        Even if you associate each color with a set of names, if the graph where vertices are colors and edges are “these colors are indistinguishable” is connected, and any two indistinguishable colors have the same set of names, then all colors would have the same set of names. The solution of “assign a degree of applicability if each name to each color” and allow a color name to have different levels of applicability for a pair of indistinguishable colors, sorta solves the problem? But, in a sense, isn’t “to what degree do each of these names apply to this color” just a kind of identifier like a name is? (Though it has the advantage that we can talk about the identifiers being very close to each other ) We can’t have identifiers for colors always be the same iff the colors are indistinguishable, because “is the same” is transitive while “is visually indistinguishable” is not. Therefore, in order to be able to describe a large variety of colors (with say, rgb), we choose to use an identifying scheme (such as rgb) which has different identifiers for colors which are visually indistinguishable.
- bryanrasmussen 7y ago
  If that were so that would mean induction was a bad thing.
- kristiandupont 7y ago
  Hah, I love it! There's a lot to dislike about HN and the culture here, but questions like these that sort of make me go "Well obviously not, because ... erm" are quite common and always interesting.
- deleted 7y ago
  [deleted]
asdfasgasdgasdg 7y ago
Sure. At a minimum, it seems some folks find it entertaining to do so. That is a point. But naming them for the conventional purpose of indexing into cognitive clusters doesn't work, obviously.
fizixer 7y ago
I can't take it as more than a joke. Or maybe lockdown is really getting to their head.
mirimir 7y ago
Sure. And conversely, there are infinitely more colors than that particular digital representation can express.
bawolff 7y ago
The entire website is obviously a silly joke. Even if the colours were distinguishable you wouldn't go about naming them in this fashion but in some more systematic fashion
- greggman3 7y ago
  I thought the entire thing was an excuse to have yet another node dependency. const namedColors = require('namedColors'); const chalk = require('chalk'); console.log(chalk.hex(nameColors['Hacker News Orange'])( lol, you just downloaded a 200meg dep for named colors ));
  - tiborsaas 7y ago
    Good, let's have it so that the the senior developers can come in and save the day by hardcoding the "hacker news orange" :)
nneonneo 7y ago
I propose that names be considered more like probability distributions. For example the name “red” has a probability distribution that probably peaks in the neighbourhood of #FF0000 but includes a large range of colors; the distribution of the name “crimson” likely has a much narrower distribution. Under this interpretation, and given enough data, one could feasibly extract “the” name for any given colour using a maximum likelihood estimation (what name is most likely to generate the observed colour). And to collect data on this, one could for example ask many random users to choose the range of colours they believe correspond to a given name (“select all red colours”) and merge the results into the distribution. (There are many, many other ways to carry out such a survey; this is but one example). Obviously, you won’t get to 16M names this way, but you could definitely learn quite a lot about where the “boundaries” are between colours from this kind of exercise!
- mikk14 7y ago
  You might expect to hear this, but what you're hinting at was done by xkcd: https://blog.xkcd.com/2010/05/03/color-survey-results/ https://blog.xkcd.com/2010/05/03/color-survey-results/ Particularly, what you're describing is this: http://imgs.xkcd.com/blag/satfaces_map_1024.png http://imgs.xkcd.com/blag/satfaces_map_1024.png :-)
  - owenversteeg 7y ago
    Every time I see that xkcd I'm never surprised that gold is the smallest section. It's consistently a pain in the ass to get something that looks semi-decent gold ish. A shine effect helps, but then that's more work to get right. #b18800, from the xkcd, looks pretty bad; #d4af37 is a good bit better, #cfb53b is semi-passable IMO.
pbhjpbhj 7y ago
Depending on how it's produced it's not even _a_ colour, it's three colours, or two colours, then the same two with a bit of blue. Perhaps my eyes lack a normal number of blue receptors, and a substantial amount of blue still wouldn't alter the perceived colour.

baddox 7y ago

By induction, doesn’t that mean we would only have one color name?

bb2018 7y ago
If there are zero people who cannot tell two colors apart, they should have the same name. If there is at least one person who cannot tell two colors apart, they should have different names.
- roel_v 7y ago
  What does 'telling a color apart' even mean? I did a mini project some time ago, where I basically did some custom color quantization (it was to reduce an image into a configurable amount of discrete colors, to make a pixel quilt with a manageable amount of distinct fabric colors). So I used CieLAB colors to calculate the 'visual distance' between the available colors and then recursively merged colors until I got to the desired amount. But even a simple algorithm like that, it turned out, could be implemented in several way; each of which gave different, yet equally 'valid' results. My point being - it's very hard to give a simple, unambiguous meaning to seemingly simple concepts like 'what is a color', 'when are colors the same' and 'are these two colors more similar than those other two'. There is a lot of abstraction between the physics of light and representations of color e.g. on a screen, or in fabric, or when printing.
  - Firerouge 7y ago
    I think in the case of telling two colors apart, it's can be as simple as presenting them side by side, and seeing if someone can distinguish a difference.
    - Galanwe 7y ago
      Not sure that is so simple unfortunately. First, because if I recursively apply your argument, then no color would have a name. Starting from the very 1st lowest color frequency and the 2nd right next to it, no-one could differentiate them. Apply the same logic between the 2nd and the 3rd, the 3rd and the 4th, etc. So you cannot apply this argument to "any" color pair. You would have to define a set of starting colors, and always do the comparison against them. Now you enter the problem of which starting colors to starts, and the fact that using any pre-existing color names would most likely not be evenly distributed on the color spectrum. Second, it wouldn't work because color recognition is not a transitive process. - If you show me two very close red variations side by side, I may not be able to tell the difference. - If you space the variation, I may be able to tell that there are 2 different reds, but wouldn't be able to tell which one is darker/lighter accurately. - If instead of showing me the two close reds side by side, you show them to me paired with an other color, I may be able to tell the difference. For instance, show me a board with half red1 + yellow, and an other board with red2 + yellow. I may spot than 1 of the red + yellow is matching that good compared to the other one.
      - eru 7y ago
        You can get out of that argument by saying that colour names apply in different amounts to different hex-colours. So #FF0000 is red. But is #FA0101 is also red, just slightly less so.
        msclrhd 7y ago
        That's interesting. So what you have is some sort of similarity metric that is colour based. You could define it as being 1 for an exact match, and 0 for not matching, so cyan, etc. would be 0 red, while orange may be slightly red (e.g. 0.1), and pink may be slightly more red (e.g. 0.25). This would be a 3D function/surface. This would allow you to model classes of colours, like purples. Shades of a colour could then be any similar colour in the range 0 to 0.5 (or 0.25).
        roel_v 7y ago
  - js2 7y ago
    > What does 'telling a color apart' even mean? Being able to distinguish colors on a test such as https://xritephoto.com/cool-tools https://xritephoto.com/cool-tools
    - ddoeth 7y ago
      Although I got it right in the end, I found the second row to be insanely hard. But somehow, on all rows, the colors just seemed to snap into place for me, is that a normal feeling?
ryangittins 7y ago
I don't follow. Two given colors being indistinguishable doesn't imply that any two colors are indistinguishable.
- tzs 7y ago
  Your suggestion was that if two colors are indistinguishable, they should have that name. But given any two colors A and B you can construct a sequence C1 = A, C2, C3, ..., Cn = B where adjacent colors Ci and Ci+1 are indistinguishable from each other, and so need the same name, for 1 <= i < n. Hence, A and B have to have the same name.
  - eru 7y ago
    It's just the old paradox of the heap. https://en.wikipedia.org/wiki/Sorites_paradox https://en.wikipedia.org/wiki/Sorites_paradox
  - IgorPartola 7y ago
    So make Ci and Ci + 1 distinguishable. Problem solved.
  - ryangittins 7y ago
    The problem with your argument is that indistinguishable doesn't mean the equal. I'm not saying there is no difference between adjacent colors. I'm saying that the difference is indistinguishable. A lot of indistinguishable differences add to a distinguishable difference. Think about it with numbers: if your sequence A, B, C... are each zero units apart, then yes, A = B = C = ... In actuality, though, they are not zero units apart. They are a small, but positive distance apart. Those tiny differences add up so the distance from A to C is twice as far as the distance from A to B.
    - baddox 7y ago
      Any two adjacent colors must either have the same name or different names.
333c 7y ago
"can't tell apart" isn't a transitive relation. That is, just because I can't tell A from B and I can't tell B from C doesn't mean I can't tell A from C.
- drdeca 7y ago
  “Has the same name as”, however, is a transitive relation. So, if no two indistinguishable colors have distinct names, and for every pair of colors have a path between them of colors such that adjacent colors in the path are indistinguishable, then all colors would have the same name.
  - zeroimpl 7y ago
    The flaw here is the assumption that colors have a single, well-defined name.
    - drdeca 7y ago
      Even if you associate each color with a set of names, if the graph where vertices are colors and edges are “these colors are indistinguishable” is connected, and any two indistinguishable colors have the same set of names, then all colors would have the same set of names. The solution of “assign a degree of applicability if each name to each color” and allow a color name to have different levels of applicability for a pair of indistinguishable colors, sorta solves the problem? But, in a sense, isn’t “to what degree do each of these names apply to this color” just a kind of identifier like a name is? (Though it has the advantage that we can talk about the identifiers being very close to each other ) We can’t have identifiers for colors always be the same iff the colors are indistinguishable, because “is the same” is transitive while “is visually indistinguishable” is not. Therefore, in order to be able to describe a large variety of colors (with say, rgb), we choose to use an identifying scheme (such as rgb) which has different identifiers for colors which are visually indistinguishable.
bryanrasmussen 7y ago
If that were so that would mean induction was a bad thing.
kristiandupont 7y ago
Hah, I love it! There's a lot to dislike about HN and the culture here, but questions like these that sort of make me go "Well obviously not, because ... erm" are quite common and always interesting.
deleted 7y ago
[deleted]

asdfasgasdgasdg 7y ago

Sure. At a minimum, it seems some folks find it entertaining to do so. That is a point. But naming them for the conventional purpose of indexing into cognitive clusters doesn't work, obviously.

fizixer 7y ago

I can't take it as more than a joke. Or maybe lockdown is really getting to their head.

mirimir 7y ago

Sure. And conversely, there are infinitely more colors than that particular digital representation can express.

bawolff 7y ago

The entire website is obviously a silly joke. Even if the colours were distinguishable you wouldn't go about naming them in this fashion but in some more systematic fashion

greggman3 7y ago
I thought the entire thing was an excuse to have yet another node dependency. const namedColors = require('namedColors'); const chalk = require('chalk'); console.log(chalk.hex(nameColors['Hacker News Orange'])( lol, you just downloaded a 200meg dep for named colors ));
- tiborsaas 7y ago
  Good, let's have it so that the the senior developers can come in and save the day by hardcoding the "hacker news orange" :)

nneonneo 7y ago

I propose that names be considered more like probability distributions. For example the name “red” has a probability distribution that probably peaks in the neighbourhood of #FF0000 but includes a large range of colors; the distribution of the name “crimson” likely has a much narrower distribution. Under this interpretation, and given enough data, one could feasibly extract “the” name for any given colour using a maximum likelihood estimation (what name is most likely to generate the observed colour). And to collect data on this, one could for example ask many random users to choose the range of colours they believe correspond to a given name (“select all red colours”) and merge the results into the distribution. (There are many, many other ways to carry out such a survey; this is but one example). Obviously, you won’t get to 16M names this way, but you could definitely learn quite a lot about where the “boundaries” are between colours from this kind of exercise!

mikk14 7y ago
You might expect to hear this, but what you're hinting at was done by xkcd: https://blog.xkcd.com/2010/05/03/color-survey-results/ https://blog.xkcd.com/2010/05/03/color-survey-results/ Particularly, what you're describing is this: http://imgs.xkcd.com/blag/satfaces_map_1024.png http://imgs.xkcd.com/blag/satfaces_map_1024.png :-)
- owenversteeg 7y ago
  Every time I see that xkcd I'm never surprised that gold is the smallest section. It's consistently a pain in the ass to get something that looks semi-decent gold ish. A shine effect helps, but then that's more work to get right. #b18800, from the xkcd, looks pretty bad; #d4af37 is a good bit better, #cfb53b is semi-passable IMO.

pbhjpbhj 7y ago

Depending on how it's produced it's not even _a_ colour, it's three colours, or two colours, then the same two with a bit of blue. Perhaps my eyes lack a normal number of blue receptors, and a substantial amount of blue still wouldn't alter the perceived colour.