Shift hue of an RGB Color

Question

I'm trying to write a function to shift the hue of an RGB color. Specifically I'm using it in an iOS app, but the math is universal.

The graph below shows how the R, G, and B values change with respect to the hue.

Looking at that it seems like it should be a relatively simple to write a function to shift the hue without doing any nasty conversions to a different color format which would introduce more error (which could be an issue if continue applying small shifts to a color), and I suspect would be more computationally expensive.

Here is what I have so far which sort of works. It works perfectly if you're shifting from pure yellow or cyan or magenta but otherwise it gets a little squiffy in some places.

Color4f ShiftHue(Color4f c, float d) {
    if (d==0) {
        return c;
    }
    while (d<0) {
        d+=1;
    }

    d *= 3;

    float original[] = {c.red, c.green, c.blue};
    float returned[] = {c.red, c.green, c.blue};

    // big shifts
    for (int i=0; i<3; i++) {
        returned[i] = original[(i+((int) d))%3];
    }
    d -= (float) ((int) d);
    original[0] = returned[0];
    original[1] = returned[1];
    original[2] = returned[2];

    float lower = MIN(MIN(c.red, c.green), c.blue);
    float upper = MAX(MAX(c.red, c.green), c.blue);

    float spread = upper - lower;
    float shift  = spread * d * 2;

    // little shift
    for (int i = 0; i < 3; ++i) {
        // if middle value
        if (original[(i+2)%3]==upper && original[(i+1)%3]==lower) {
            returned[i] -= shift;
            if (returned[i]<lower) {
                returned[(i+1)%3] += lower - returned[i];
                returned[i]=lower;
            } else
                if (returned[i]>upper) {
                    returned[(i+2)%3] -= returned[i] - upper;
                    returned[i]=upper;
                }
            break;
        }
    }

    return Color4fMake(returned[0], returned[1], returned[2], c.alpha);
}

I know you can do this with UIColors and shift the hue with something like this:

CGFloat hue;
CGFloat sat;
CGFloat bri;
[[UIColor colorWithRed:parent.color.red green:parent.color.green blue:parent.color.blue alpha:1] getHue:&hue saturation:&sat brightness:&bri alpha:nil];
hue -= .03;
if (hue<0) {
    hue+=1;
}
UIColor *tempColor = [UIColor colorWithHue:hue saturation:sat brightness:bri alpha:1];
const float* components= CGColorGetComponents(tempColor.CGColor);
color = Color4fMake(components[0], components[1], components[2], 1);

but I'm not crazy about that as It only works in iOS 5, and between allocating a number of color objects and converting from RGB to HSB and then back it seems pretty overkill.

I might end up using a lookup table or pre-calculate the colors in my application, but I'm really curious if there's a way to make my code work. Thanks!

Answer 1

Basically there are two options:

1. Convert RGB -> HSV, change hue, convert HSV -> RGB
2. Change the hue directly with a linear transformation

I'm not really sure about how to implement 2, but basically you'll have to create a transformation matrix and filter the image through this matrix. However, this will re-color the image instead of changing only the hue. If this is ok for you, then this could be an option but if not a conversion cannot be avoided.

Edit

A little research shows this, which confirms my thoughts. To summarize: The conversion from RGB to HSV should be preferred, if an exact result is desired. Modifying the original RGB image by a linear transform also leads to a result but this rather tints the image. The difference is explained as follows: The conversion from RGB to HSV is non-linear, whereas the transform is linear.

Answer 2

It seems converting to HSV makes the most sense. Sass provides some amazing color helpers. It's in ruby, but it might provide useful.

http://sass-lang.com/docs/yardoc/Sass/Script/Functions.html

Answer 3

Excelent code, but, i wonder that it can be faster if you simply don´t use self.matrix[2][0], self.matrix[2][1], self.matrix[2][1]

Therefore, set_hue_rotation can be written simply as:

def set_hue_rotation(self, degrees):
    cosA = cos(radians(degrees))
    sinA = sin(radians(degrees))
    self.matrix[0][0] = cosA + (1.0 - cosA) / 3.0
    self.matrix[0][1] = 1./3. * (1.0 - cosA) - sqrt(1./3.) * sinA
    self.matrix[0][2] = 1./3. * (1.0 - cosA) + sqrt(1./3.) * sinA
    self.matrix[1][0] = self.matrix[0][2] <---Not sure, if this is the right code, but i think you got the idea
    self.matrix[1][1] = self.matrix[0][0]
    self.matrix[1][2] = self.matrix[0][1]

Answer 4

The RGB color space describes a cube. It is possible to rotate this cube around the diagonal axis from (0,0,0) to (255,255,255) to effect a change of hue. Note that some of the results will lie outside of the 0 to 255 range and will need to be clipped.

I finally got a chance to code this algorithm. It's in Python but it should be easy to translate to the language of your choice. The formula for 3D rotation came from http://en.wikipedia.org/wiki/Rotation_matrix#Rotation_matrix_from_axis_and_angle

Edit: If you saw the code I posted previously, please ignore it. I was so anxious to find a formula for the rotation that I converted a matrix-based solution into a formula, not realizing that the matrix was the best form all along. I've still simplified the calculation of the matrix using the constant sqrt(1/3) for axis unit vector values, but this is much closer in spirit to the reference and simpler in the per-pixel calculation apply as well.

from math import sqrt,cos,sin,radians

def clamp(v):
    if v < 0:
        return 0
    if v > 255:
        return 255
    return int(v + 0.5)

class RGBRotate(object):
    def __init__(self):
        self.matrix = [[1,0,0],[0,1,0],[0,0,1]]

    def set_hue_rotation(self, degrees):
        cosA = cos(radians(degrees))
        sinA = sin(radians(degrees))
        self.matrix[0][0] = cosA + (1.0 - cosA) / 3.0
        self.matrix[0][1] = 1./3. * (1.0 - cosA) - sqrt(1./3.) * sinA
        self.matrix[0][2] = 1./3. * (1.0 - cosA) + sqrt(1./3.) * sinA
        self.matrix[1][0] = 1./3. * (1.0 - cosA) + sqrt(1./3.) * sinA
        self.matrix[1][1] = cosA + 1./3.*(1.0 - cosA)
        self.matrix[1][2] = 1./3. * (1.0 - cosA) - sqrt(1./3.) * sinA
        self.matrix[2][0] = 1./3. * (1.0 - cosA) - sqrt(1./3.) * sinA
        self.matrix[2][1] = 1./3. * (1.0 - cosA) + sqrt(1./3.) * sinA
        self.matrix[2][2] = cosA + 1./3. * (1.0 - cosA)

    def apply(self, r, g, b):
        rx = r * self.matrix[0][0] + g * self.matrix[0][1] + b * self.matrix[0][2]
        gx = r * self.matrix[1][0] + g * self.matrix[1][1] + b * self.matrix[1][2]
        bx = r * self.matrix[2][0] + g * self.matrix[2][1] + b * self.matrix[2][2]
        return clamp(rx), clamp(gx), clamp(bx)

Here are some results from the above:

You can find a different implementation of the same idea at http://www.graficaobscura.com/matrix/index.html

Answer 5

Scott....not exactly. The algo seems to work the same as in HSL/HSV, but faster. Also, if you simply multiply the 1st 3 elements of the array with the factor for grey, you add/decrease luma.

Example...Greyscale from Rec709 have those values [GrayRedFactor_Rec709: R$ 0.212671 GrayGreenFactor_Rec709: R$ 0.715160 GrayBlueFactor_Rec709: R$ 0.072169]

When you multply self.matrix[x][x] with the GreyFactor correspondent you decrease luma without touching saturation Ex:

def set_hue_rotation(self, degrees):
    cosA = cos(radians(degrees))
    sinA = sin(radians(degrees))
    self.matrix[0][0] = (cosA + (1.0 - cosA) / 3.0) * 0.212671
    self.matrix[0][1] = (1./3. * (1.0 - cosA) - sqrt(1./3.) * sinA) * 0.715160
    self.matrix[0][2] = (1./3. * (1.0 - cosA) + sqrt(1./3.) * sinA) * 0.072169
    self.matrix[1][0] = self.matrix[0][2] <---Not sure, if this is the right code, but i think you got the idea
    self.matrix[1][1] = self.matrix[0][0]
    self.matrix[1][2] = self.matrix[0][1]

And the opposite is also true.If you divide instead multiply, the luminosity is increased dramatically.

From what i´m testing this algorithms can be a wonderfull replacement for HSL, as long you don´t need saturation, of course.

Try doing this...rotate the hue to only 1 degree (Just to force the algo to work properly while keeping the same perception sensitivity of the image), and multiply by those factors.