Comments: Colors, combinatorics... and common sense

As soon as you started talking about colours you were going to use in *firefox* I thought 'when is going to mention colour blindness?'... not till the end ;-)

On airport runways the traffic lights are horizontal, and only have red and green, no amber. My friend was asked to drive his car across a couple of runways. You can imagine him looking at (to him) two identical bulbs, then looking into the sky to see if it was safe to cross :-) then flooring it.

Obviously in the road, red is at the top. Thats easy to figure out :-)

Colour blindness is interesting. A lot of people don't realise they have it. My Uncle didn't realise untill he was 15 or 16, a very narrow set of colours look the same to him, it's like pale green and pale orange are just grey.

Good post. Fun.

monk.e.boy

Posted by monk.e.boy at March 24, 2006 1:22 AM

"Overall, I don't regret going down this path, as I often do when I make fundamentally wrong assumptions. "

Sometimes making "wrong" assumptions can prove that those assumptionsa arn't actually wrong :)

(From Alex: True, but a bad design is a bad design.)

Posted by mozzila at March 24, 2006 1:41 AM

> Obviously in the road, red is at the top. Thats easy to figure out :-)

Except in places where they mount the lights horizontally, like here in Houston.

Posted by db48x at March 24, 2006 5:30 PM
I started to wonder how I could demonstrate by UI that two sets overlapped and had common members.

Could you do some sort of Venn diagram, maybe using canvas or SVG? Or perhaps, instead of colors, you could use different fill or line styles like diagonal, horizontal, or vertical lines, e.g. ///// \\\\\ ----- |||||. Then where ---- and ||||| overlapped would be +++++. Or I am completely missing what you are trying to do?

(From Alex: For my particular case, where the shading would happen alongside an Inspector-like node tree, it wouldn't have been as useful. I have considered different fill styles, but I don't know how well that would work in XUL trees.)

(For fans of PBS, I first heard about this on the show "Square One", which I still think is one of the best shows there could have been for a young elementary school student.)

I *loved* that show as a child! There was that Pacman-like guy, Mathman (?), but he almost always died. And then there was those detectives, what were they called? Math Squad? Math Net?

(From Alex: MathNet, yes.)

Why not make each selection's shading in the column a different color?... Of course, they couldn't be just different. #ffffff is different than #fffffe, but I dare you to look at a webpage with a background of pure white and a fair-sized square of this odd color, and tell me where the square is, using only your eyes (no mouse, no keys). No, they had to be as different as possible so that people could recognize each one.

You go on to calculate the number of ways you can combine different R, G, and B values to give you some subset of the web-safe colors, but it seems to me here that what you want is to find the number of ways you can combine different web-safe colors (each of which has it own R, G, and B values).

But combinatorics wouldn't help me define a preferred order to these colors. At least, not in so simple a calculation.

Well, if you want to know the number of web-safe colors you can make with different R, G, and B values, that would be equal to the number of ways you can take 3 distinct numbers from a set of 6 numbers (00, 33, 66, 99, CC, and FF), or 6C3 = 6!/(3! 3!) = (6*5*4)/(3*2*1) = 20.

If you want to know the number of combinations you can make with 3 of the 216 web-safe colors, that would be 216C3 = 216!/(213! 3!) = (216*215*214)/(3*2*1) = a lot.

If you're still interested, I can try to calculate how many combinations of "different" web-safe colors there are, if you define what makes two web-safe colors sufficiently different from each other. It was kind of fun to break out the old high school probablity formulas again.


Posted by James Napolitano at March 24, 2006 7:48 PM

Oops, I just realized my first calculation was off. Order does matter, so instead of 6C3, you would use 6P3 (permutations instead of combinations), giving you 6!/3! = 120 different colors, not 20.

Posted by James Napolitano at March 24, 2006 8:17 PM

"how would I choose colors that were the most distinct from one another, so that a little 10-pixel-square marker could clearly be distinguished from another."

By using the best tool analysis for this:

Color contrast analyzer by Juicy Studio

See item #7 at this precise URL:

http://www.mozilla.org/access/authors#validators

Gérard

Posted by Gérard Talbot at May 13, 2006 8:12 AM