Related
I found an implementation of the HEALpix algorithm this is the dokumentation
And the output looks very nice.
The following images show the latitude / longitude conversion to HEALpix areas.
The x-axe goes from 0 to 2 * pi. The y-axe goes from 0 to pi. The grey color represents the HEALpix pixel encoded in grey.
Nside = 1
Nside = 2
Nside = 4
Nside = 8
The different grey values are the IDs for the texture I have to use. That means, that each HEALpix pixel represents one texture. The missing part is the UV mapping within each of the HEALpix pixels like shown below:
nSide = 1 with UV mapping
Right now I am using the function:
void ang2pix_ring( const long nside, double theta, double phi, long *ipix)
Which gives me the correct texture ID. But I've no idea how to calculate the UV mapping for each HEALpix pixel.
Is there a way to calculate all four corners in lat/lon coordinates of a HEALpix pixel? Or even better a direct calculation to the UV coordinates?
BTW: I am using the RING scheme. But if the NESTED scheme is simpler to calculate I also would change to that.
After a lot of research I came to a solution for this problem:
First of all, I've changed the scheme to NESTED. With the NESTED scheme and a very high nSide value (8192), the returned value from the
void ang2pix_ring( const long nside, double theta, double phi, long *ipix)
function gives back a long value where the UV coordinates can be read out in the following way:
Bit 26 till 30 represents the level 0 (only the 12 HEALPix pixels).
By using higher levels, the Bits from 30 till 26 - (level * 2) represents the HEALPix pixels.
The leftover 26 - (level * 2) - 1 till bit 1 encode the UV texture-coordinates in the following way:
Each second odd bit shrink together represents the U coordinate and the even once represents the V coordinate.
To normalize these UV-coordinates the responding shrinked values need to be divided by the value of pow(2, (26 - level * 2) / 2).
Code says more than 1000 words:
unsigned long ignoreEverySecondBit(unsigned long value, bool odd, unsigned int countBits)
{
unsigned long result = 0;
unsigned long mask = odd == true ? 0b1 : 0b10;
countBits = countBits / 2;
for (int i = 0; i < countBits; ++i)
{
if ((value & mask) != 0)
{
result += std::pow(2, i);
}
mask = mask << 2;
}
return result;
}
//calculate the HEALPix values:
latLonToHealPixNESTED(nSide, theta, phi, &pix);
result.level = level;
result.texture = pix >> (26 - level * 2);
result.u = static_cast<float>(ignoreEverySecondBit(pix, true, 26 - level * 2));
result.v = static_cast<float>(ignoreEverySecondBit(pix, false, 26 - level * 2));
result.u = result.u / pow(2, (26 - level * 2) / 2);
result.v = result.v / pow(2, (26 - level * 2) / 2);
And of cause a few images to show the results. The blue value represents the textureID, the red value represents the U-coordinate and the green value represents the V-coordinate:
Level 0
Level 1
Level 2
Level 3
Level 4
I hope this solution will help others too.
BOUNTY STATUS UPDATE:
I discovered how to map a linear lens, from destination coordinates to source coordinates.
How do you calculate the radial distance from the centre to go from fisheye to rectilinear?
1). I actually struggle to reverse it, and to map source coordinates to destination coordinates. What is the inverse, in code in the style of the converting functions I posted?
2). I also see that my undistortion is imperfect on some lenses - presumably those that are not strictly linear. What is the equivalent to-and-from source-and-destination coordinates for those lenses? Again, more code than just mathematical formulae please...
Question as originally stated:
I have some points that describe positions in a picture taken with a fisheye lens.
I want to convert these points to rectilinear coordinates. I want to undistort the image.
I've found this description of how to generate a fisheye effect, but not how to reverse it.
There's also a blog post that describes how to use tools to do it; these pictures are from that:
(1) : SOURCE Original photo link
Input : Original image with fish-eye distortion to fix.
(2) : DESTINATION Original photo link
Output : Corrected image (technically also with perspective correction, but that's a separate step).
How do you calculate the radial distance from the centre to go from fisheye to rectilinear?
My function stub looks like this:
Point correct_fisheye(const Point& p,const Size& img) {
// to polar
const Point centre = {img.width/2,img.height/2};
const Point rel = {p.x-centre.x,p.y-centre.y};
const double theta = atan2(rel.y,rel.x);
double R = sqrt((rel.x*rel.x)+(rel.y*rel.y));
// fisheye undistortion in here please
//... change R ...
// back to rectangular
const Point ret = Point(centre.x+R*cos(theta),centre.y+R*sin(theta));
fprintf(stderr,"(%d,%d) in (%d,%d) = %f,%f = (%d,%d)\n",p.x,p.y,img.width,img.height,theta,R,ret.x,ret.y);
return ret;
}
Alternatively, I could somehow convert the image from fisheye to rectilinear before finding the points, but I'm completely befuddled by the OpenCV documentation. Is there a straightforward way to do it in OpenCV, and does it perform well enough to do it to a live video feed?
The description you mention states that the projection by a pin-hole camera (one that does not introduce lens distortion) is modeled by
R_u = f*tan(theta)
and the projection by common fisheye lens cameras (that is, distorted) is modeled by
R_d = 2*f*sin(theta/2)
You already know R_d and theta and if you knew the camera's focal length (represented by f) then correcting the image would amount to computing R_u in terms of R_d and theta. In other words,
R_u = f*tan(2*asin(R_d/(2*f)))
is the formula you're looking for. Estimating the focal length f can be solved by calibrating the camera or other means such as letting the user provide feedback on how well the image is corrected or using knowledge from the original scene.
In order to solve the same problem using OpenCV, you would have to obtain the camera's intrinsic parameters and lens distortion coefficients. See, for example, Chapter 11 of Learning OpenCV (don't forget to check the correction). Then you can use a program such as this one (written with the Python bindings for OpenCV) in order to reverse lens distortion:
#!/usr/bin/python
# ./undistort 0_0000.jpg 1367.451167 1367.451167 0 0 -0.246065 0.193617 -0.002004 -0.002056
import sys
import cv
def main(argv):
if len(argv) < 10:
print 'Usage: %s input-file fx fy cx cy k1 k2 p1 p2 output-file' % argv[0]
sys.exit(-1)
src = argv[1]
fx, fy, cx, cy, k1, k2, p1, p2, output = argv[2:]
intrinsics = cv.CreateMat(3, 3, cv.CV_64FC1)
cv.Zero(intrinsics)
intrinsics[0, 0] = float(fx)
intrinsics[1, 1] = float(fy)
intrinsics[2, 2] = 1.0
intrinsics[0, 2] = float(cx)
intrinsics[1, 2] = float(cy)
dist_coeffs = cv.CreateMat(1, 4, cv.CV_64FC1)
cv.Zero(dist_coeffs)
dist_coeffs[0, 0] = float(k1)
dist_coeffs[0, 1] = float(k2)
dist_coeffs[0, 2] = float(p1)
dist_coeffs[0, 3] = float(p2)
src = cv.LoadImage(src)
dst = cv.CreateImage(cv.GetSize(src), src.depth, src.nChannels)
mapx = cv.CreateImage(cv.GetSize(src), cv.IPL_DEPTH_32F, 1)
mapy = cv.CreateImage(cv.GetSize(src), cv.IPL_DEPTH_32F, 1)
cv.InitUndistortMap(intrinsics, dist_coeffs, mapx, mapy)
cv.Remap(src, dst, mapx, mapy, cv.CV_INTER_LINEAR + cv.CV_WARP_FILL_OUTLIERS, cv.ScalarAll(0))
# cv.Undistort2(src, dst, intrinsics, dist_coeffs)
cv.SaveImage(output, dst)
if __name__ == '__main__':
main(sys.argv)
Also note that OpenCV uses a very different lens distortion model to the one in the web page you linked to.
(Original poster, providing an alternative)
The following function maps destination (rectilinear) coordinates to source (fisheye-distorted) coordinates. (I'd appreciate help in reversing it)
I got to this point through trial-and-error: I don't fundamentally grasp why this code is working, explanations and improved accuracy appreciated!
def dist(x,y):
return sqrt(x*x+y*y)
def correct_fisheye(src_size,dest_size,dx,dy,factor):
""" returns a tuple of source coordinates (sx,sy)
(note: values can be out of range)"""
# convert dx,dy to relative coordinates
rx, ry = dx-(dest_size[0]/2), dy-(dest_size[1]/2)
# calc theta
r = dist(rx,ry)/(dist(src_size[0],src_size[1])/factor)
if 0==r:
theta = 1.0
else:
theta = atan(r)/r
# back to absolute coordinates
sx, sy = (src_size[0]/2)+theta*rx, (src_size[1]/2)+theta*ry
# done
return (int(round(sx)),int(round(sy)))
When used with a factor of 3.0, it successfully undistorts the images used as examples (I made no attempt at quality interpolation):
Dead link
(And this is from the blog post, for comparison:)
If you think your formulas are exact, you can comput an exact formula with trig, like so:
Rin = 2 f sin(w/2) -> sin(w/2)= Rin/2f
Rout= f tan(w) -> tan(w)= Rout/f
(Rin/2f)^2 = [sin(w/2)]^2 = (1 - cos(w))/2 -> cos(w) = 1 - 2(Rin/2f)^2
(Rout/f)^2 = [tan(w)]^2 = 1/[cos(w)]^2 - 1
-> (Rout/f)^2 = 1/(1-2[Rin/2f]^2)^2 - 1
However, as #jmbr says, the actual camera distortion will depend on the lens and the zoom. Rather than rely on a fixed formula, you might want to try a polynomial expansion:
Rout = Rin*(1 + A*Rin^2 + B*Rin^4 + ...)
By tweaking first A, then higher-order coefficients, you can compute any reasonable local function (the form of the expansion takes advantage of the symmetry of the problem). In particular, it should be possible to compute initial coefficients to approximate the theoretical function above.
Also, for good results, you will need to use an interpolation filter to generate your corrected image. As long as the distortion is not too great, you can use the kind of filter you would use to rescale the image linearly without much problem.
Edit: as per your request, the equivalent scaling factor for the above formula:
(Rout/f)^2 = 1/(1-2[Rin/2f]^2)^2 - 1
-> Rout/f = [Rin/f] * sqrt(1-[Rin/f]^2/4)/(1-[Rin/f]^2/2)
If you plot the above formula alongside tan(Rin/f), you can see that they are very similar in shape. Basically, distortion from the tangent becomes severe before sin(w) becomes much different from w.
The inverse formula should be something like:
Rin/f = [Rout/f] / sqrt( sqrt(([Rout/f]^2+1) * (sqrt([Rout/f]^2+1) + 1) / 2 )
I blindly implemented the formulas from here, so I cannot guarantee it would do what you need.
Use auto_zoom to get the value for the zoom parameter.
def dist(x,y):
return sqrt(x*x+y*y)
def fisheye_to_rectilinear(src_size,dest_size,sx,sy,crop_factor,zoom):
""" returns a tuple of dest coordinates (dx,dy)
(note: values can be out of range)
crop_factor is ratio of sphere diameter to diagonal of the source image"""
# convert sx,sy to relative coordinates
rx, ry = sx-(src_size[0]/2), sy-(src_size[1]/2)
r = dist(rx,ry)
# focal distance = radius of the sphere
pi = 3.1415926535
f = dist(src_size[0],src_size[1])*factor/pi
# calc theta 1) linear mapping (older Nikon)
theta = r / f
# calc theta 2) nonlinear mapping
# theta = asin ( r / ( 2 * f ) ) * 2
# calc new radius
nr = tan(theta) * zoom
# back to absolute coordinates
dx, dy = (dest_size[0]/2)+rx/r*nr, (dest_size[1]/2)+ry/r*nr
# done
return (int(round(dx)),int(round(dy)))
def fisheye_auto_zoom(src_size,dest_size,crop_factor):
""" calculate zoom such that left edge of source image matches left edge of dest image """
# Try to see what happens with zoom=1
dx, dy = fisheye_to_rectilinear(src_size, dest_size, 0, src_size[1]/2, crop_factor, 1)
# Calculate zoom so the result is what we wanted
obtained_r = dest_size[0]/2 - dx
required_r = dest_size[0]/2
zoom = required_r / obtained_r
return zoom
I took what JMBR did and basically reversed it. He took the radius of the distorted image (Rd, that is, the distance in pixels from the center of the image) and found a formula for Ru, the radius of the undistorted image.
You want to go the other way. For each pixel in the undistorted (processed image), you want to know what the corresponding pixel is in the distorted image.
In other words, given (xu, yu) --> (xd, yd). You then replace each pixel in the undistorted image with its corresponding pixel from the distorted image.
Starting where JMBR did, I do the reverse, finding Rd as a function of Ru. I get:
Rd = f * sqrt(2) * sqrt( 1 - 1/sqrt(r^2 +1))
where f is the focal length in pixels (I'll explain later), and r = Ru/f.
The focal length for my camera was 2.5 mm. The size of each pixel on my CCD was 6 um square. f was therefore 2500/6 = 417 pixels. This can be found by trial and error.
Finding Rd allows you to find the corresponding pixel in the distorted image using polar coordinates.
The angle of each pixel from the center point is the same:
theta = arctan( (yu-yc)/(xu-xc) ) where xc, yc are the center points.
Then,
xd = Rd * cos(theta) + xc
yd = Rd * sin(theta) + yc
Make sure you know which quadrant you are in.
Here is the C# code I used
public class Analyzer
{
private ArrayList mFisheyeCorrect;
private int mFELimit = 1500;
private double mScaleFESize = 0.9;
public Analyzer()
{
//A lookup table so we don't have to calculate Rdistorted over and over
//The values will be multiplied by focal length in pixels to
//get the Rdistorted
mFisheyeCorrect = new ArrayList(mFELimit);
//i corresponds to Rundist/focalLengthInPixels * 1000 (to get integers)
for (int i = 0; i < mFELimit; i++)
{
double result = Math.Sqrt(1 - 1 / Math.Sqrt(1.0 + (double)i * i / 1000000.0)) * 1.4142136;
mFisheyeCorrect.Add(result);
}
}
public Bitmap RemoveFisheye(ref Bitmap aImage, double aFocalLinPixels)
{
Bitmap correctedImage = new Bitmap(aImage.Width, aImage.Height);
//The center points of the image
double xc = aImage.Width / 2.0;
double yc = aImage.Height / 2.0;
Boolean xpos, ypos;
//Move through the pixels in the corrected image;
//set to corresponding pixels in distorted image
for (int i = 0; i < correctedImage.Width; i++)
{
for (int j = 0; j < correctedImage.Height; j++)
{
//which quadrant are we in?
xpos = i > xc;
ypos = j > yc;
//Find the distance from the center
double xdif = i-xc;
double ydif = j-yc;
//The distance squared
double Rusquare = xdif * xdif + ydif * ydif;
//the angle from the center
double theta = Math.Atan2(ydif, xdif);
//find index for lookup table
int index = (int)(Math.Sqrt(Rusquare) / aFocalLinPixels * 1000);
if (index >= mFELimit) index = mFELimit - 1;
//calculated Rdistorted
double Rd = aFocalLinPixels * (double)mFisheyeCorrect[index]
/mScaleFESize;
//calculate x and y distances
double xdelta = Math.Abs(Rd*Math.Cos(theta));
double ydelta = Math.Abs(Rd * Math.Sin(theta));
//convert to pixel coordinates
int xd = (int)(xc + (xpos ? xdelta : -xdelta));
int yd = (int)(yc + (ypos ? ydelta : -ydelta));
xd = Math.Max(0, Math.Min(xd, aImage.Width-1));
yd = Math.Max(0, Math.Min(yd, aImage.Height-1));
//set the corrected pixel value from the distorted image
correctedImage.SetPixel(i, j, aImage.GetPixel(xd, yd));
}
}
return correctedImage;
}
}
I found this pdf file and I have proved that the maths are correct (except for the line vd = *xd**fv+v0 which should say vd = **yd**+fv+v0).
http://perception.inrialpes.fr/CAVA_Dataset/Site/files/Calibration_OpenCV.pdf
It does not use all of the latest co-efficients that OpenCV has available but I am sure that it could be adapted fairly easily.
double k1 = cameraIntrinsic.distortion[0];
double k2 = cameraIntrinsic.distortion[1];
double p1 = cameraIntrinsic.distortion[2];
double p2 = cameraIntrinsic.distortion[3];
double k3 = cameraIntrinsic.distortion[4];
double fu = cameraIntrinsic.focalLength[0];
double fv = cameraIntrinsic.focalLength[1];
double u0 = cameraIntrinsic.principalPoint[0];
double v0 = cameraIntrinsic.principalPoint[1];
double u, v;
u = thisPoint->x; // the undistorted point
v = thisPoint->y;
double x = ( u - u0 )/fu;
double y = ( v - v0 )/fv;
double r2 = (x*x) + (y*y);
double r4 = r2*r2;
double cDist = 1 + (k1*r2) + (k2*r4);
double xr = x*cDist;
double yr = y*cDist;
double a1 = 2*x*y;
double a2 = r2 + (2*(x*x));
double a3 = r2 + (2*(y*y));
double dx = (a1*p1) + (a2*p2);
double dy = (a3*p1) + (a1*p2);
double xd = xr + dx;
double yd = yr + dy;
double ud = (xd*fu) + u0;
double vd = (yd*fv) + v0;
thisPoint->x = ud; // the distorted point
thisPoint->y = vd;
This can be solved as an optimization problem. Simply draw on curves in images that are supposed to be straight lines. Store the contour points for each of those curves. Now we can solve the fish eye matrix as a minimization problem. Minimize the curve in points and that will give us a fisheye matrix. It works.
It can be done manually by adjusting the fish eye matrix using trackbars! Here is a fish eye GUI code using OpenCV for manual calibration.
What do 2 & 3 mean in this and how can I change them?
CvMat* rot = cvCreateMat(2,3,CV_32FC1)
When I change these two values I get an openCV GUI error handler.
size of input arguments do not match()
in function cvConvertScale.\cxconvert.cpp(1601)
I want to understand what that means
Update:
The code is:
#include <cv.h>
#include <highgui.h>
int main()
{
CvMat* rot = cvCreateMat(2,3,CV_32FC1);
IplImage *src, *dst;
src=cvLoadImage("doda.jpg");
// make acopy of gray image(src)
dst = cvCloneImage( src );
dst->origin = src->origin;
// make dstof zeros
cvZero( dst );
// Compute rotation matrix
double x=0.0;
// loop to get rotation from 0 to 360 by 4 press on anykey
for(int i=1;i<=5;i++)
{
CvPoint2D32f center = cvPoint2D32f(src->width/2,src->height/2);
double angle = 0+x;
double scale = 0.6;
cv2DRotationMatrix( center, angle, scale, rot );
// Do the transformation
cvWarpAffine( src, dst, rot);
cvNamedWindow( "Affine_Transform", 1 );
cvShowImage( "Affine_Transform", dst );
if (i<=4)
x=x+90.0;
else
x=0.0;
cvWaitKey();
}
cvReleaseImage( &dst );
cvReleaseMat( &rot );
return 0;
}
2 and 3 are the row and column counts of the matrix you're creating.
From Introduction to programming with OpenCV:
Allocate a matrix:
CvMat* cvCreateMat(int rows, int cols, int type);
type: Type of the matrix elements. Specified in form
CV_<bit_depth>(S|U|F)C<number_of_channels>. E.g.: CV_8UC1 means an
8-bit unsigned single-channel matrix, CV_32SC2 means a 32-bit signed
matrix with two channels.
Example:
CvMat* M = cvCreateMat(4,4,CV_32FC1);
Changing them is as simple as substituting different values. But I guess you should already know that.
2 = number of rows and 3 = number of columns in your matrix, rot.
Can you post the entire code? Or maybe tell us what you want to achieve? Are you trying to rotate an image?
Also, I'd recommend upgrading to OpenCV 2.0 which has a C++ interface. With the new version, you can extensively use the Mat class which handles everything (matrices,images,etc.) and makes things much simpler.
You get an error using any other shape than 2x3 because it is then meaningless for opencv when you use rot for rotation.
Take a look at Jacob's answer.
He describes the rotation matrix components in details.
Motivation
I'd like to find a way to take an arbitrary color and lighten it a few shades, so that I can programatically create a nice gradient from the one color to a lighter version. The gradient will be used as a background in a UI.
Possibility 1
Obviously I can just split out the RGB values and increase them individually by a certain amount. Is this actually what I want?
Possibility 2
My second thought was to convert the RGB to HSV/HSB/HSL (Hue, Saturation, Value/Brightness/Lightness), increase the brightness a bit, decrease the saturation a bit, and then convert it back to RGB. Will this have the desired effect in general?
As Wedge said, you want to multiply to make things brighter, but that only works until one of the colors becomes saturated (i.e. hits 255 or greater). At that point, you can just clamp the values to 255, but you'll be subtly changing the hue as you get lighter. To keep the hue, you want to maintain the ratio of (middle-lowest)/(highest-lowest).
Here are two functions in Python. The first implements the naive approach which just clamps the RGB values to 255 if they go over. The second redistributes the excess values to keep the hue intact.
def clamp_rgb(r, g, b):
return min(255, int(r)), min(255, int(g)), min(255, int(b))
def redistribute_rgb(r, g, b):
threshold = 255.999
m = max(r, g, b)
if m <= threshold:
return int(r), int(g), int(b)
total = r + g + b
if total >= 3 * threshold:
return int(threshold), int(threshold), int(threshold)
x = (3 * threshold - total) / (3 * m - total)
gray = threshold - x * m
return int(gray + x * r), int(gray + x * g), int(gray + x * b)
I created a gradient starting with the RGB value (224,128,0) and multiplying it by 1.0, 1.1, 1.2, etc. up to 2.0. The upper half is the result using clamp_rgb and the bottom half is the result with redistribute_rgb. I think it's easy to see that redistributing the overflows gives a much better result, without having to leave the RGB color space.
For comparison, here's the same gradient in the HLS and HSV color spaces, as implemented by Python's colorsys module. Only the L component was modified, and clamping was performed on the resulting RGB values. The results are similar, but require color space conversions for every pixel.
I would go for the second option. Generally speaking the RGB space is not really good for doing color manipulation (creating transition from one color to an other, lightening / darkening a color, etc). Below are two sites I've found with a quick search to convert from/to RGB to/from HSL:
from the "Fundamentals of Computer Graphics"
some sourcecode in C# - should be easy to adapt to other programming languages.
In C#:
public static Color Lighten(Color inColor, double inAmount)
{
return Color.FromArgb(
inColor.A,
(int) Math.Min(255, inColor.R + 255 * inAmount),
(int) Math.Min(255, inColor.G + 255 * inAmount),
(int) Math.Min(255, inColor.B + 255 * inAmount) );
}
I've used this all over the place.
ControlPaint class in System.Windows.Forms namespace has static methods Light and Dark:
public static Color Dark(Color baseColor, float percOfDarkDark);
These methods use private implementation of HLSColor. I wish this struct was public and in System.Drawing.
Alternatively, you can use GetHue, GetSaturation, GetBrightness on Color struct to get HSB components. Unfortunately, I didn't find the reverse conversion.
Convert it to RGB and linearly interpolate between the original color and the target color (often white). So, if you want 16 shades between two colors, you do:
for(i = 0; i < 16; i++)
{
colors[i].R = start.R + (i * (end.R - start.R)) / 15;
colors[i].G = start.G + (i * (end.G - start.G)) / 15;
colors[i].B = start.B + (i * (end.B - start.B)) / 15;
}
In order to get a lighter or a darker version of a given color you should modify its brightness. You can do this easily even without converting your color to HSL or HSB color. For example to make a color lighter you can use the following code:
float correctionFactor = 0.5f;
float red = (255 - color.R) * correctionFactor + color.R;
float green = (255 - color.G) * correctionFactor + color.G;
float blue = (255 - color.B) * correctionFactor + color.B;
Color lighterColor = Color.FromArgb(color.A, (int)red, (int)green, (int)blue);
If you need more details, read the full story on my blog.
Converting to HS(LVB), increasing the brightness and then converting back to RGB is the only way to reliably lighten the colour without effecting the hue and saturation values (ie to only lighten the colour without changing it in any other way).
A very similar question, with useful answers, was asked previously:
How do I determine darker or lighter color variant of a given color?
Short answer: multiply the RGB values by a constant if you just need "good enough", translate to HSV if you require accuracy.
I used Andrew's answer and Mark's answer to make this (as of 1/2013 no range input for ff).
function calcLightness(l, r, g, b) {
var tmp_r = r;
var tmp_g = g;
var tmp_b = b;
tmp_r = (255 - r) * l + r;
tmp_g = (255 - g) * l + g;
tmp_b = (255 - b) * l + b;
if (tmp_r > 255 || tmp_g > 255 || tmp_b > 255)
return { r: r, g: g, b: b };
else
return { r:parseInt(tmp_r), g:parseInt(tmp_g), b:parseInt(tmp_b) }
}
I've done this both ways -- you get much better results with Possibility 2.
Any simple algorithm you construct for Possibility 1 will probably work well only for a limited range of starting saturations.
You would want to look into Poss 1 if (1) you can restrict the colors and brightnesses used, and (2) you are performing the calculation a lot in a rendering.
Generating the background for a UI won't need very many shading calculations, so I suggest Poss 2.
-Al.
IF you want to produce a gradient fade-out, I would suggest the following optimization: Rather than doing RGB->HSB->RGB for each individual color you should only calculate the target color. Once you know the target RGB, you can simply calculate the intermediate values in RGB space without having to convert back and forth. Whether you calculate a linear transition of use some sort of curve is up to you.
Method 1: Convert RGB to HSL, adjust HSL, convert back to RGB.
Method 2: Lerp the RGB colour values - http://en.wikipedia.org/wiki/Lerp_(computing)
See my answer to this similar question for a C# implementation of method 2.
Pretend that you alpha blended to white:
oneMinus = 1.0 - amount
r = amount + oneMinus * r
g = amount + oneMinus * g
b = amount + oneMinus * b
where amount is from 0 to 1, with 0 returning the original color and 1 returning white.
You might want to blend with whatever the background color is if you are lightening to display something disabled:
oneMinus = 1.0 - amount
r = amount * dest_r + oneMinus * r
g = amount * dest_g + oneMinus * g
b = amount * dest_b + oneMinus * b
where (dest_r, dest_g, dest_b) is the color being blended to and amount is from 0 to 1, with zero returning (r, g, b) and 1 returning (dest.r, dest.g, dest.b)
I didn't find this question until after it became a related question to my original question.
However, using insight from these great answers. I pieced together a nice two-liner function for this:
Programmatically Lighten or Darken a hex color (or rgb, and blend colors)
Its a version of method 1. But with over saturation taken into account. Like Keith said in his answer above; use Lerp to seemly solve the same problem Mark mentioned, but without redistribution. The results of shadeColor2 should be much closer to doing it the right way with HSL, but without the overhead.
A bit late to the party, but if you use javascript or nodejs, you can use tinycolor library, and manipulate the color the way you want:
tinycolor("red").lighten().desaturate().toHexString() // "#f53d3d"
I would have tried number #1 first, but #2 sounds pretty good. Try doing it yourself and see if you're satisfied with the results, it sounds like it'll take you maybe 10 minutes to whip up a test.
Technically, I don't think either is correct, but I believe you want a variant of option #2. The problem being that taken RGB 990000 and "lightening" it would really just add onto the Red channel (Value, Brightness, Lightness) until you got to FF. After that (solid red), it would be taking down the saturation to go all the way to solid white.
The conversions get annoying, especially since you can't go direct to and from RGB and Lab, but I think you really want to separate the chrominance and luminence values, and just modify the luminence to really achieve what you want.
Here's an example of lightening an RGB colour in Python:
def lighten(hex, amount):
""" Lighten an RGB color by an amount (between 0 and 1),
e.g. lighten('#4290e5', .5) = #C1FFFF
"""
hex = hex.replace('#','')
red = min(255, int(hex[0:2], 16) + 255 * amount)
green = min(255, int(hex[2:4], 16) + 255 * amount)
blue = min(255, int(hex[4:6], 16) + 255 * amount)
return "#%X%X%X" % (int(red), int(green), int(blue))
This is based on Mark Ransom's answer.
Where the clampRGB function tries to maintain the hue, it however miscalculates the scaling to keep the same luminance. This is because the calculation directly uses sRGB values which are not linear.
Here's a Java version that does the same as clampRGB (although with values ranging from 0 to 1) that maintains luminance as well:
private static Color convertToDesiredLuminance(Color input, double desiredLuminance) {
if(desiredLuminance > 1.0) {
return Color.WHITE;
}
if(desiredLuminance < 0.0) {
return Color.BLACK;
}
double ratio = desiredLuminance / luminance(input);
double r = Double.isInfinite(ratio) ? desiredLuminance : toLinear(input.getRed()) * ratio;
double g = Double.isInfinite(ratio) ? desiredLuminance : toLinear(input.getGreen()) * ratio;
double b = Double.isInfinite(ratio) ? desiredLuminance : toLinear(input.getBlue()) * ratio;
if(r > 1.0 || g > 1.0 || b > 1.0) { // anything outside range?
double br = Math.min(r, 1.0); // base values
double bg = Math.min(g, 1.0);
double bb = Math.min(b, 1.0);
double rr = 1.0 - br; // ratios between RGB components to maintain
double rg = 1.0 - bg;
double rb = 1.0 - bb;
double x = (desiredLuminance - luminance(br, bg, bb)) / luminance(rr, rg, rb);
r = 0.0001 * Math.round(10000.0 * (br + rr * x));
g = 0.0001 * Math.round(10000.0 * (bg + rg * x));
b = 0.0001 * Math.round(10000.0 * (bb + rb * x));
}
return Color.color(toGamma(r), toGamma(g), toGamma(b));
}
And supporting functions:
private static double toLinear(double v) { // inverse is #toGamma
return v <= 0.04045 ? v / 12.92 : Math.pow((v + 0.055) / 1.055, 2.4);
}
private static double toGamma(double v) { // inverse is #toLinear
return v <= 0.0031308 ? v * 12.92 : 1.055 * Math.pow(v, 1.0 / 2.4) - 0.055;
}
private static double luminance(Color c) {
return luminance(toLinear(c.getRed()), toLinear(c.getGreen()), toLinear(c.getBlue()));
}
private static double luminance(double r, double g, double b) {
return r * 0.2126 + g * 0.7152 + b * 0.0722;
}
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This is something I've pseudo-solved many times and have never quite found a solution for.
The problem is to come up with a way to generate N colors, that are as distinguishable as possible where N is a parameter.
My first thought on this is "how to 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. Once you have a set of vectors that are maximized apart, 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!
Thinking about this problem more, it would be better to map the colors in a linear manner, 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, let us say:
n = 10
we know we have 16777216 colors (256^3).
We can use Buckles Algorithm 515 to find the lexicographically indexed color.. You'll probably have to edit the algorithm to avoid overflow and probably add some minor speed improvements.
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.
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.
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;
}
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?
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.
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.
To achieve "most distinguishable" we need to use a perceptual color space like Lab (or any other perceptually linear color space) other than 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 distinguishable from each other.