Question

This should be a simple question, but I haven't been able to find a way to make it work.

Essentially, I have a silly localhost page that I use in my webdevelopment. When I am surfing between our development server and my local version of the C# code (redirected from the dev url via host file) I have been known to sometimes forget what 'dev.foo.com' points at - local or server.

So I created a page which will run locally as my default web page's default page, so I can easily identify my localhost from the server.

This page does a lot of things randomly (including generating a character's starting stats for D&D), including setting a random background color. I do this by generating 3 random numbers between 0 and 255, and setting them as the RGB value for the body background color in CSS.

Given the 3 ints R, G, and B, how do I determine R2, G2, and B2 such that the second color will have high contrast with the first? I like having the page have random background colors (it keeps me from getting used to the look of the landing page) but I also like to be able to read the text.

Was it helpful?

Solution

You need a difference in brightness for text to be readable, as color vision itself has too low resolution.

So as an algorithm I'd suggest the following:

  • Pick a random background color.

  • Then decide whether it is a light or a dark color. For example you could check whether the average of the three primary colors is greater or equal 128.

  • For a light color use black text, for a dark one white text.

OTHER TIPS

"Contrast" is a loaded word. If you just care about being able to read the text, then one easy way is to work in a luminance-based color space like HSL, and pick foreground and background colors with big differences in luminance.

The conversion between HSL and RGB is well-known--see Wikipedia for the details.

If you're talking about actual color contrast, it's not nearly as cut-and-dried (there are a lot of perceptual factors that, as far as I know, haven't been reduced to a single colors space), but I suspect you don't need that level of sophistication.

Check out this PHP solution: Calculating Color Contrast with PHP by Andreas Gohr. It can be ported to any language of course.

He also has a very nice demonstration of his contrast analyzer where you can find some minimal contrast levels to work with.

You can use method GetBrightness() on Color class. It returns a float value from 0.0 (brightness of black) to 1.0 (white). A simple solution would be:

var color1 = new Color.FromArgb(r1, g1, b1);
var brightness = color1.GetBrightness();

var color2 = brightness > 0.5 ? Color.Black : Color.White;

I did something like this in a Palm OS application. This is what I came up with. It doesn't give you "high contrast" colors but it gives you a background color that's different enough from the text color to be quite readable:

  // Black background is a special case.  It's fairly likely to occur and 
  // the default color shift we do isn't very noticeable with very dark colors.
  if (backColor.r < 0x20 && backColor.g < 0x20 && backColor.b < 0x20)
  {
      textColor.r = backColor.r + 0x20;
      textColor.g = backColor.g + 0x20;
      textColor.b = backColor.b + 0x20;
  }
  else
  {
      textColor.r = backColor.r + ((backColor.r < 128) ? 0x10 : -0x10);
      textColor.g = backColor.g + ((backColor.g < 128) ? 0x10 : -0x10);
      textColor.b = backColor.b + ((backColor.b < 128) ? 0x10 : -0x10);
  }

You might not need to do black as a special case for your purposes - Palm's color handling is a bit funky (16-bit color).

There are some good resources (and algorithms) to address this at http://juicystudio.com/article/luminositycontrastratioalgorithm.php

These answers are more or less suggesting to use one of the two or three color choices based on whether the color is bright or dark.

I use a bit different approach and it worked elegantly in my case. Here is the implementation.

 int color = your_color;
 contrastColor = Color.rgb(255-(color >> 16)&0xFF, 255-(color >> 8)&0xFF, 255- color&0xFF);

It's simple and wonderful.

If you flip all the bits, you will get the "opposite" color which would be pretty good contrast.

I believe it's the ~ operator in C#:

R2 = ~R1;
G2 = ~G1;
B2 = ~B1;

For best contrast use this code

function lumdiff($R1,$G1,$B1,$R2,$G2,$B2){

    $L1 = 0.2126 * pow($R1/255, 2.2) +
          0.7152 * pow($G1/255, 2.2) +
          0.0722 * pow($B1/255, 2.2);

    $L2 = 0.2126 * pow($R2/255, 2.2) +
          0.7152 * pow($G2/255, 2.2) +
          0.0722 * pow($B2/255, 2.2);

    if($L1 > $L2){
        return ($L1+0.05) / ($L2+0.05);
    }else{
        return ($L2+0.05) / ($L1+0.05);
    }
}

function get_the_contrast($c1, $c2) {
    return (lumdiff(hexdec(substr($c1,0,2)),
        hexdec(substr($c1,2,2)),hexdec(substr($c1,4,2)),
        hexdec(substr($c2,0,2)),hexdec(substr($c2,2,2)),
        hexdec(substr($c2,4,2))));
}

The method above ( AVG(red,green,blue) > 128 ) is not realy good.

private Color GetContrastingColor(Color color)
{
    int r = color.R > 0 ? 256 - color.R : 255;
    int g = color.G > 0 ? 256 - color.G : 255;
    int b = color.B > 0 ? 256 - color.B : 255;
    return System.Drawing.Color.FromArgb(r, g, b);
}

Thanks to @starblue !

Here is C# code that I use

 public static string GetContrastBlackOrWhiteColorAsHtmlColorCode(Color c)
        {
            System.Drawing.Color color = System.Drawing.ColorTranslator.FromHtml("transparent");

            try
            {
                if (c.R >= 128 && c.G >= 128 && c.B >= 128)
                {
                    return System.Drawing.ColorTranslator.ToHtml(Color.Black);
                }
                else
                {
                    return System.Drawing.ColorTranslator.ToHtml(Color.White);
                }
            }
            catch (Exception)
            {
            }

            return System.Drawing.ColorTranslator.ToHtml(color);
        }
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