Optimised Hue shift function in GLSL

Webgl shader Hue Shift function

vec3 hueShift( vec3 color, float hueAdjust ){

    const vec3  kRGBToYPrime = vec3 (0.299, 0.587, 0.114);
    const vec3  kRGBToI      = vec3 (0.596, -0.275, -0.321);
    const vec3  kRGBToQ      = vec3 (0.212, -0.523, 0.311);

    const vec3  kYIQToR     = vec3 (1.0, 0.956, 0.621);
    const vec3  kYIQToG     = vec3 (1.0, -0.272, -0.647);
    const vec3  kYIQToB     = vec3 (1.0, -1.107, 1.704);

    float   YPrime  = dot (color, kRGBToYPrime);
    float   I       = dot (color, kRGBToI);
    float   Q       = dot (color, kRGBToQ);
    float   hue     = atan (Q, I);
    float   chroma  = sqrt (I * I + Q * Q);

    hue += hueAdjust;

    Q = chroma * sin (hue);
    I = chroma * cos (hue);

    vec3    yIQ   = vec3 (YPrime, I, Q);

    return vec3( dot (yIQ, kYIQToR), dot (yIQ, kYIQToG), dot (yIQ, kYIQToB) );

}

For future visitors, the hue value should be in radians, not in degrees (0.0 to 360.0), so if your hue value is in degrees, you will need to convert before passing to the shader add it the shader code.

Do you have any recommended reading to explain this? I don't understand how any of it works, especially the constants. It works perfectly though.

I know this is an old comment, but someone might still find it useful.

Basically the shader uses the YIQ color mode. The constants are conversion matrices between RGB and YIQ, but wikipedia can explain that better than I do:
https://en.wikipedia.org/wiki/YIQ

Hue is the angle on the IQ-plane. In other words you fix the Y value, and get a two dimensional IQ-plane, then calculate the angle between those two values. The shift is simply performed by adding the shift value to this angle.
Once this is done you then simply need to convert it back to RGB, for that you need the new I and Q values. You can split the hue by using sin and cos, you just need to scale them to match the previous length on the IQ plane. That "length" is called "chroma" in the shader and is better known as "saturation". You obtain it by using the pythagorean theorem.

This is an incredibly slow implementation.

Here is the same thing, but it works several times faster. Written on HLSL. Porting to GLSL is easy.

float3 ApplyHue(float3 col, float hueAdjust)
{
    const float3 k = float3(0.57735, 0.57735, 0.57735);
    half cosAngle = cos(hueAdjust);
    return col * cosAngle + cross(k, col) * sin(hueAdjust) + k * dot(k, col) * (1.0 - cosAngle);
}

@viruseg Indeed, it is a more elegant solution, and simple to convert to GLSL, thanks.

vec3 hueShift(vec3 color, float hue) {
    const vec3 k = vec3(0.57735, 0.57735, 0.57735);
    float cosAngle = cos(hue);
    return vec3(color * cosAngle + cross(k, color) * sin(hue) + k * dot(k, color) * (1.0 - cosAngle));
}

There was a typo, this one works for me:

vec3 hueShift(vec3 color, float hue)
{
const vec3 k = vec3(0.57735, 0.57735, 0.57735);
float cosAngle = cos(hue);
return vec3(color * cosAngle + cross(k, color) * sin(hue) + k * dot(k, color) * (1.0 - cosAngle));
}

Does anyone know where the 0.57735 come from?

@RichardBray it's sqrt(3)/3 or 1/sqrt(3). It's one of those values that you see often in trigonometry, see e.g. http://mathforum.org/dr.math/faq/formulas/faq.trig.html (Not super often though; I had to do some digging to find this one too 😅). Not sure how to arrive at this formula from scratch though. 😕

Ah: https://en.wikipedia.org/wiki/Rodrigues%27_rotation_formula

@janpaul123 Thanks that makes more sense. But yeah I still can't figure out how that formula was reached.

Thank you so much for this! Contrary to what you seem to think though, your original function produces much more natural looking/beautiful results than the cheaper hueshift function provided by @viruseg. Of course it depends on context whether or not that's needed, but I for one deeply appreciate it!
EDIT but, I believe this is a faster implementation ->

vec2 rotate2(vec2 v, float fi) {
return v*mat2(cos(fi), -sin(fi), sin(fi), cos(fi));
}

// YIQ color rotation/hue shift
vec3 hueShiftYIQ(vec3 rgb, float hs) {
float rotAngle = hs*-6.28318530718;
const mat3 rgb2yiq = mat3(0.299, 0.596, 0.211,
0.587, -0.274, -0.523,
0.114, -0.322, 0.312);
const mat3 yiq2rgb = mat3(1, 1, 1,
0.956, -0.272, -1.106,
0.621, -0.647, 1.703);
vec3 yiq = rgb2yiq * rgb;
yiq.yz *= rot(rotAngle);
return yiq2rgb * yiq;
}

Thank you so much for this! Contrary to what you seem to think though, your original function produces much more natural looking/beautiful results than the cheaper hueshift function provided by @viruseg.

I had the same conclusion. The idea to use Rodrigues rotation formula is clever, but the problem is that the RGB to YIQ translation isn't a pure rotation. It also has skew and scaling components. So the output won't be the same.

The most compact version I could come up with, with some symbolic manipulation (along the same lines as Rodrigues' but replacing the cross product), that should be equivalent, is:

vec3 hue_shift(vec3 color, float dhue) {
	float s = sin(dhue);
	float c = cos(dhue);
	return (color * c) + (color * s) * mat3(
		vec3(0.167444, 0.329213, -0.496657),
		vec3(-0.327948, 0.035669, 0.292279),
		vec3(1.250268, -1.047561, -0.202707)
	) + dot(vec3(0.299, 0.587, 0.114), color) * (1.0 - c);
}

(Godot shader, but should be compatible to GLSL)

vmedea, thank you for the code.

How to adapt your code fro RGBA (alpha) image, vec4?

@l375cd It doesn't do anything with alpha channel, so just use GLSL swizzling to only pass in the RGB channels, and then re-apply the alpha channel to the result. Something like the pseudo-code below.

in vec2 texCoord;
out vec4 outFrag;

uniform sampler2D image;
uniform float hue

void main()
    vec4 color = texture(image, texCoord);
    outFrag = vec4(hueShift(color.rgb, hue), color.a);
}

@ForeverZer0, thank you so much!

Will it work with sRGB color format, or should I convert my color to RGB before using this function?

My version of hue shift I did for some codegolf shader.

vec3 h(vec3 c, float s){  
    return c * mat3(c = .66 * cos(s + vec3(0, 2.09, 4.18)) + .33, c.zxy, c.yzx);
}