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彩色车轮的功能

This is something I've pseudo-solved many times and never quite found a solution. That's stuck with me. The problem is to come up with a way to generate N colors, that are as distinguishable as possible where N is a parameter.

2016年12月07日10分31秒

My first thought on this is "how generate N vectors in a space that maximize distance from each other." You can see that the RGB (or any other scale you use that forms a basis in color space) are just vectors. Take a look at Random Point Picking. Hope this is a good start for you! Once you have a set of vectors that are maximized a part, you can save them in a hash table or something for later, and just perform random rotations on them to get all the colors you desire that are maximally apart from each other!

Edit: Thinking about this problem more, it would be better to map the colors in a linear manor, possibly (0,0,0) --> (255,255,255) lexicographically, and then distribute them evenly. I really don't know how well this will work, but it should since, lets say:

n = 10 we know we have 16777216 colors (256^3). We can use buckles algorithm 515 to find the lexicographically indexed color.\frac {\binom {256^3} {3}} {n} * i. You'll probably have to edit the algorithm to avoid overflow and probably add some minor speed improvements.

2016年12月07日10分31秒

It would be best to find colors maximally distant in a "perceptually uniform" colorspace, e.g. CIELAB (using Euclidean distance between L*, a*, b* coordinates as your distance metric) and then converting to the colorspace of your choice. Perceptual uniformity is achieved by tweaking the colorspace to approximate the non-linearities in the human visual system.

2016年12月07日10分31秒

Some related resources:

ColorBrewer - Sets of colours designed to be maximally distinguishable for use on maps.

Escaping RGBland: Selecting Colors for Statistical Graphics - A technical report describing a set of algorithms for generating good (i.e. maximally distinguishable) colour sets in the hcl colour space.

2016年12月07日10分31秒

Here is some code to allocate RGB colors evenly around a HSL color wheel of specified luminosity.

class cColorPicker
{
public:
    void Pick( vector<DWORD>&v_picked_cols, int count, int bright = 50 );
private:
    DWORD HSL2RGB( int h, int s, int v );
    unsigned char ToRGB1(float rm1, float rm2, float rh);
};
/**

  Evenly allocate RGB colors around HSL color wheel

  @param[out] v_picked_cols  a vector of colors in RGB format
  @param[in]  count   number of colors required
  @param[in]  bright  0 is all black, 100 is all white, defaults to 50

  based on Fig 3 of http://epub.wu-wien.ac.at/dyn/virlib/wp/eng/mediate/epub-wu-01_c87.pdf?ID=epub-wu-01_c87

*/

void cColorPicker::Pick( vector<DWORD>&v_picked_cols, int count, int bright )
{
    v_picked_cols.clear();
    for( int k_hue = 0; k_hue < 360; k_hue += 360/count )
        v_picked_cols.push_back( HSL2RGB( k_hue, 100, bright ) );
}
/**

  Convert HSL to RGB

  based on http://www.codeguru.com/code/legacy/gdi/colorapp_src.zip

*/

DWORD cColorPicker::HSL2RGB( int h, int s, int l )
{
    DWORD ret = 0;
    unsigned char r,g,b;

    float saturation = s / 100.0f;
    float luminance = l / 100.f;
    float hue = (float)h;

    if (saturation == 0.0) 
    {
      r = g = b = unsigned char(luminance * 255.0);
    }
    else
    {
      float rm1, rm2;

      if (luminance <= 0.5f) rm2 = luminance + luminance * saturation;  
      else                     rm2 = luminance + saturation - luminance * saturation;
      rm1 = 2.0f * luminance - rm2;   
      r   = ToRGB1(rm1, rm2, hue + 120.0f);   
      g = ToRGB1(rm1, rm2, hue);
      b  = ToRGB1(rm1, rm2, hue - 120.0f);
    }

    ret = ((DWORD)(((BYTE)(r)|((WORD)((BYTE)(g))<<8))|(((DWORD)(BYTE)(b))<<16)));

    return ret;
}


unsigned char cColorPicker::ToRGB1(float rm1, float rm2, float rh)
{
  if      (rh > 360.0f) rh -= 360.0f;
  else if (rh <   0.0f) rh += 360.0f;

  if      (rh <  60.0f) rm1 = rm1 + (rm2 - rm1) * rh / 60.0f;   
  else if (rh < 180.0f) rm1 = rm2;
  else if (rh < 240.0f) rm1 = rm1 + (rm2 - rm1) * (240.0f - rh) / 60.0f;      

  return static_cast<unsigned char>(rm1 * 255);
}

int _tmain(int argc, _TCHAR* argv[])
{
    vector<DWORD> myCols;
    cColorPicker colpick;
    colpick.Pick( myCols, 20 );
    for( int k = 0; k < (int)myCols.size(); k++ )
        printf("%d: %d %d %d\n", k+1,
        ( myCols[k] & 0xFF0000 ) >>16,
        ( myCols[k] & 0xFF00 ) >>8,
        ( myCols[k] & 0xFF ) );

    return 0;
}

2016年12月07日10分31秒

Isn't it also a factor which order you set up the colors?

Like if you use Dillie-Os idea you need to mix the colors as much as possible. 0 64 128 256 is from one to the next. but 0 256 64 128 in a wheel would be more "apart"

Does this make sense?

2016年12月07日10分31秒

I've read somewhere the human eye can't distinguish between less than 4 values apart. so This is something to keep in mind. The following algorithm does not compensate for this.

I'm not sure this is exactly what you want, but this is one way to randomly generate non-repeating color values:

(beware, inconsistent pseudo-code ahead)

//colors entered as 0-255 [R, G, B]
colors = []; //holds final colors to be used
rand = new Random();

//assumes n is less than 16,777,216
randomGen(int n){
while (len(colors) < n){
//generate a random number between 0,255 for each color
newRed = rand.next(256);
newGreen = rand.next(256);
newBlue = rand.next(256);
temp = [newRed, newGreen, newBlue];
//only adds new colors to the array
if temp not in colors {
colors.append(temp);
}
}
}

One way you could optimize this for better visibility would be to compare the distance between each new color and all the colors in the array:

for item in color{
itemSq = (item[0]^2 + item[1]^2 + item[2]^2])^(.5);
tempSq = (temp[0]^2 + temp[1]^2 + temp[2]^2])^(.5);
dist = itemSq - tempSq;
dist = abs(dist);
}
//NUMBER can be your chosen distance apart.
if dist < NUMBER and temp not in colors {
colors.append(temp);
}

But this approach would significantly slow down your algorithm.

Another way would be to scrap the randomness and systematically go through every 4 values and add a color to an array in the above example.

2016年12月07日10分31秒

I know this an old post but I found it while looking for a PHP solution to the topic and finally came with a simple solution:

function random_color($i = null, $n = 10, $sat = .5, $br = .7) {
    $i = is_null($i) ? mt_rand(0,$n) : $i;
    $rgb = hsv2rgb(array($i*(360/$n), $sat, $br));
    for ($i=0 ; $i<=2 ; $i++) 
        $rgb[$i] = dechex(ceil($rgb[$i]));
    return implode('', $rgb);
}

function hsv2rgb($c) { 
    list($h,$s,$v)=$c; 
    if ($s==0) 
        return array($v,$v,$v); 
    else { 
        $h=($h%=360)/60; 
        $i=floor($h); 
        $f=$h-$i; 
        $q[0]=$q[1]=$v*(1-$s); 
        $q[2]=$v*(1-$s*(1-$f)); 
        $q[3]=$q[4]=$v; 
        $q[5]=$v*(1-$s*$f); 
        return(array($q[($i+4)%6]*255,$q[($i+2)%6]*255,$q[$i%6]*255)); //[1] 
    } 
}

So just call the random_color() function where $i identifies the color, $n the number of possible colors, $sat the saturation and $br the brightness.

2016年12月07日10分31秒

Last I checked JFreeChart has this precise algorithm and as it is open source you can check out what it does. I do know that the colors I get do not seem to be randomly spaced along some circle or sphere, but rather chosen more specifically.

2016年12月07日10分31秒

To achieve "most distinguishable" we need to use a perceptual color space like Lab (or any other perceptually linear color space) and not RGB. Also, we can quantize this space to reduce the size of the space.

Generate the full 3D space with all possible quantized entries and run the K-means algorithm with k=N. The resulting centers/ "means" should be approximately most distinguishabl from each other.

2016年12月07日10分31秒