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@mairod
Last active April 26, 2024 07:12
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Optimised Hue shift function in GLSL
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) );
}
@janpaul123
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janpaul123 commented Nov 26, 2020

@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. 😕

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

@tarikbarri
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tarikbarri commented Jan 22, 2021

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;
}

@vmedea
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vmedea commented Jan 16, 2023

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)

@l375cd
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l375cd commented Aug 30, 2023

vmedea, thank you for the code.

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

@ForeverZer0
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@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);
}

@l375cd
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l375cd commented Aug 30, 2023

@ForeverZer0, thank you so much!

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

@arthurfisher
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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);
}

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