# sRGB↔L*a*b*↔LCh_{ab} conversions

Posted by Michał ‘mina86’ Nazarewicz on 21st of March 2021

After writing about conversion between sRGB and XYZ colour spaces I’ve been asked about a related process: moving between sRGB and CIELAB (perhaps better known as L*a*b*). As this may be of interest to others, I’ve decided to go ahead and make an article out of it. I’ll also touch on CIELCh_{ab} which is a closely related colour representation.

The L*a*b* colour space was intended to be perceptually uniform. While it’s not truly uniform it’s nonetheless useful and widely used in the industry. For example, it’s the basis of the ΔE*_{00} colour difference metric. LCh_{ab} aim to make L*a*b* easier to interpret by replacing a* and b* axes with more intuitive chroma and hue parameters.

Importantly, the conversion between sRGB and L*a*b* goes through XYZ colour space. As such, the full process has multiple steps with a round trip conversion being: sRGB→XYZ→L*a*b*→XYZ→sRGB. Because of that structure I will describe each of the steps separately.

## sRGB to XYZ

For brevity I’ll assume herein that we’re working with 8-bit colour depth, i.e. that red, green and blue components are integers in the \(\{0..255\}\) range. This does not need to be true in general and in cases when this differs the equations presented herein need to be adjusted accordingly.

When doing any conversions from sRGB, we first need to recall that it uses gamma correction. Each component of the colour is compressed and needs to be expanded to move to linear RGB space. Only then the conversion matrix \(M\) can be used to switch from linear RGB to XYZ. Having an \(\langle r_8, g_8, b_8 \rangle\) colour, the following formualæ may be used to get corresponding XYZ coordinates: $$ \begin{align} \begin{bmatrix}X \\ Y \\ Z\end{bmatrix} &= M \begin{bmatrix}R \\ G \\ B\end{bmatrix} \\[.5em] R &= g_e(r_8) \qquad \\ G &= g_e(g_8) \qquad \\ B &= g_e(b_8) \qquad \\[.5em] g_e(c_8) &= \begin{cases} {c_8 \over 3294.6} & \text{if } c_8 ≤ 10 \\ \left({c_8 + 14.025 \over 269.025}\right)^{2.4} & \text{otherwise} \end{cases} \\[.5em] \langle r_8, g_8, b_8 \rangle &∈ \{0 .. 255\}^3 \end{align} $$

Expressing all that in code, specifically Rust, we get:

/// Converts sRGB colour given as a triple of 8-bit integers /// into XYZ colour space (with white’s Y value equal 1). fn xyz_from_srgb(srgb: [u8; 3]) -> [f32; 3] { // Gamma expansion for 8-bit sRGB components. fn gamma_expansion(value255: u8) -> f32 { return if value255 <= 10 { value255 as f32 / 3294.6 } else { ((value255 as f32 + 14.025) / 269.025).powf(2.4) } } // Convert all components to linear RGB. let r = gamma_expansion(srgb[0]); let g = gamma_expansion(srgb[1]); let b = gamma_expansion(srgb[2]); // Switch to XYZ. [ r*0.4124108464885388 + g*0.3575845678529519 + b*0.18045380393360833, r*0.21264934272065283 + g*0.7151691357059038 + b*0.07218152157344333, r*0.019331758429150258 + g*0.11919485595098397 + b*0.9503900340503373, ] }

As mentioned, when working with different colour depths a different \(g_e\) or `gamma_expansion`

function is necessary. It’s *not* enough to scale all numbers since the \(c_8 ≤ 10\) condition has been chosen specifically to work with 8-bit integers. The base formula — for normalised RGB components in the \([0, 1]\) range — and further discussion of this topic is present in the aforementioned article on gamma correction.

## XYZ to L*a*b*

While the sRGB standard defines a white point — which is incorporated into the conversion matrix — CIELAB can be used with different illuminants. For simplicity, I’m going to assume the same reference — standard illuminant D_{65} — that sRGB uses and which is widely used when working with emissive displays. If necessary that can be easily adjusted by changing \(X_r\), \(Y_r\) and \(Z_r\) values below to those corresponding to one’s favourite white point. With all that said, equations for calculating L*a*b* coordinates for a colour given in XYZ space are as follows: $$ \begin{align} L^* &= 116 f_y - 16 \\ a^* &= 500 (f_x - f_y) \\ b^* &= 200 (f_y - f_z) \\[.5em] f_x &= f(X / X_r) \\ f_y &= f(Y / Y_r) \\ f_z &= f(Z / Z_r) \\[.5em] f(v) &= \begin{cases} (κ v + 16) / 116 & \text{if }v ≤ ε \\ \sqrt[3]{v} & \text{otherwise} \end{cases} \\[.5em] ε &= {216 \over 24389} \\ κ &= {24389 \over 27} \\[.5em] X_r &= 0.9504492182750991 \\ Y_r &= 1 \\ Z_r &= 1.0889166484304715 \end{align} $$

Translating the formulæ into code yields:

/// Converts colour from XYZ space to CIELAB (a.k.a. L*a*b) using /// D65 reference white point. fn lab_from_xyz(xyz: [f32; 3]) -> [f32; 3] { const EPSILON: f32 = 216.0 / 24389.0; const KAPPA: f32 = 24389.0 / 27.0; fn f(v: f32) -> f32 { if v > EPSILON { v.powf(1.0 / 3.0) } else { (KAPPA * v + 16.0) / 116.0 } } let fx = f(xyz[0] / 0.9504492182750991); let fy = f(xyz[1]); let fz = f(xyz[2] / 1.0889166484304715); [ 116.0 * fy - 16.0, 500.0 * (fx - fy), 200.0 * (fy - fz), ] }

## L*a*b* to XYZ

Conversion from L*a*b* to sRGB involves performing all the steps in reverse order. First going from CIELAB to XYZ, then to linear RGB and finally to sRGB. The first step of this process is governed by the following equations: $$ \begin{align} X &= f^{-1}(f_x) X_r \\ Y &= f^{-1}(f_y) Y_r \\ Z &= f^{-1}(f_z) Z_r \\[.5em] f_y &= {(L^* + 16) \over 116} \\ f_x &= {a^* \over 500} + f_y \\ f_z &= f_y - {b^* \over 200} \\[.5em] f^{-1}(v) &= \begin{cases} v^3 & \text{if }v > \sqrt[3]{ε} \\ (116 v − 16) / κ & \text{otherwise} \end{cases} \\[.5em] \sqrt[3]{ε} &= \sqrt[3]{216 \over 24389} = {6 \over 29} \end{align} $$

To optimise calculations it’s also worth to observe that: $$ \begin{align} f^{-1}(f_y) &= \begin{cases} f_y^3 & \text{if }(L^* + 16) / 116 > \sqrt[3]{ε} \\ (116 (L^* + 16) / 116 − 16) / κ & \text{otherwise} \end{cases} \\ &= \begin{cases} f_y^3 & \text{if }L^* > 116\sqrt[3]{ε}-16 \\ L^* / κ & \text{otherwise} \end{cases} \\ 116\sqrt[3]{ε}-16 &= 116{6 \over 29} - 16 = 8 \end{align} $$

Taking that optimisations into consideration, the equations translate into the following Rust function:

/// Converts colour from CIELAB (a.k.a. L*a*b) space /// using D65 reference white point to XYZ. fn xyz_from_lab(lab: [f32; 3]) -> [f32; 3] { const CBRT_EPSILON: f32 = 0.20689655172413796; const KAPPA: f32 = 24389.0 / 27.0; fn f_inv(v: f32) -> f32 { if v > CBRT_EPSILON: { v.powi(3) } else { (v * 116.0 - 16.0) / KAPPA }; } let fy = (lab[0] + 16.0) / 116.0; let fx = (lab[1] / 500.0) + fy; let fz = fy - (lab[2] / 200.0); let x = f_inv(fx) * 0.9504492182750991; let y = if lab[0] > 8 { fy.powi(3) } else { lab[0] / KAPPA } let z = f_inv(fz) * 1.0889166484304715; [x, y, z] }

## XYZ to sRGB

The second step is moving to sRGB colour space. Similarly to conversion in the other direction, this involves switching from XYZ to linear RGB first (by using the inverse conversion matrix) and then applying gamma compression function to get final sRGB coordinates. $$ \begin{align} r_8 &= g_c(R) \\ g_8 &= g_c(G) \\ b_8 &= g_c(B) \\[.5em] \begin{bmatrix}R \\ G \\ B\end{bmatrix} &= M^{-1} \begin{bmatrix}X \\ Y \\ Z\end{bmatrix}^T \\[.5em] g_c(s) &= g_e^{-1}(s) \\ & = \begin{cases} ⌊3294.6 s⌉ & \text{if } s ≤ S_0 \\ ⌊269.025 s^{1/2.4} - 14.025⌉ & \text{otherwise} \end{cases} \\ S_0 &= 0.00313066844250060782371 \end{align} $$

Like before, this assumes 8-bit colour depth output. For general form of the \(g_c\) function which can be adjusted to other depths see discussion in the sRGB↔XYZ conversion article. In the meanwhile, the above formulæ expressed in code take the following form:

/// Converts colour given in XYZ colour space (with white’s Y value /// equal 1) into an sRGB colour specified as a triple of 8-bit integers. fn srgb_from_xyz(xyz: [f32; 3]) -> [u8; 3] { // Gamma compression from linear [0, 1] value to 8-bit integer. fn gamma_compression(linear: f32): u8 { let v = if linear <= 0.00313066844250060782371 { 3294.6 * linear } else { 269.025 * linear.pow(5.0 / 12.0) - 14.025 }; v.round().min(255.0).max(0.0) as u8 } let x = xyz[0]; let y = xyz[1]; let z = xyz[2]; let r = x* 3.240812398895283 - y*1.5373084456298136 - z*0.4985865229069666; let g = x*-0.9692430170086407 + y*1.8759663029085742 + z*0.04155503085668564; let b = x* 0.055638398436112804 - y*0.20400746093241362 + z*1.0571295702861434; [ gamma_compression(r), gamma_compression(g), gamma_compression(b), ] }

And that’s it as far as converting between L*a*b* and sRGB colour spaces is concerned. With the presented code, `lab_from_xyz(xyz_from_srgb([r, g, b]))`

gets an sRGB colour converted into L*a*b* and `srgb_from_xyz(xyz_from_lab([l, a, b]))`

moves in the opposite direction. There’s still a closely related colour representation — CIELCh_{ab} or LCh_{ab} — which we’ll look at next.

## L*a*b* and LCh_{ab}

The a* and b* components of the L*a*b* colour are not easy to interpret. Even knowing that they component represents green-red and blue-yellow axes respectively, it’s tricky to visualise the colour space. To make it easier for humans to work with, CIE defined an LCh_{ab} colour space as a conversion of L*a*b* into a cylindrical coordinate system. It replaces a* and b* components with more intuitive chroma and hue. The relations between all the components is as follows: $$ \begin{align} C^* &= \sqrt{a^{*2} + b^{*2}} \\ h &= \text{atan}_2(b^*, a^*) \\[.5em] a^* &= C^* \cos(h) \\ b^* &= C^* \sin(h) \end{align} $$

Depending on use case it may be better to keep hue expressed in radians or to convert it to degrees. The former is usually easier for computation since mathematical instructions typically operate in radians. On the other hand, the latter is more intuitive as it’s easier to think in degrees than radians.

Below are the formulæ expressed in code with conversion to degrees present:

/// Converts colour given in CIELAB (a.k.a. L*a*b) space to /// CIELCh(ab) with hue expressed in degrees. fn lch_from_lab(lab: [f32; 3]) -> [f32; 3] { [ lab[0], lab[1].hypot(lab[2]), lab[2].atan2(lab[1]) * 360.0 / std::f32::consts::TAU, ] } /// Converts colour given in CIELCh(ab) space with hue /// expressed in degrees to CIELAB (a.k.a. L*a*b*). fn lab_from_lch(lch: [f32; 3]) -> [f32; 3] { let h_rad = lch[2] * std::f32::consts::TAU / 360.0; [ lch[0], lch[1] * h_rad.cos(), lch[1] * h_rad.sin(), ] }

## Accuracy benchmark and possible improvements

To finish up, let’s test just how good the presented code is. It’s not easy to benchmark precision of the calculations in the conversion code. There is no authoritative set of colours with their coordinates given in RGB and L*a*b* colour spaces. Still, there is one thing that’s easy to check.

Colours with a* and b* components equal to zero represent shades of grey. In other words, for any \(c\), \(\langle c, c, c \rangle\) sRGB colour converted to L*a*b* should have its last two components equal zero with L* going from zero (for black) and a hundred (for white). That is, of course, so long as the L*a*b* uses D65 reference white point. This observation allows a simple benchmark to be written as follows:

fn main() { let mut count: usize = 0; let mut error: f64 = 0.0; let mut max: f32 = 0.0; for i in 0..256 { let lab = lab_from_xyz(xyz_from_srgb([i as u8, i as u8, i as u8])); if lab[1] != 0.0 || lab[2] != 0.0 { eprintln!("{:3} {:20} {:20} {:20}", i, lab[0], lab[1], lab[2]); count += 1; error = (lab[1] as f64).mul_add(lab[1] as f64, error); error = (lab[2] as f64).mul_add(lab[2] as f64, error); max = max.max(lab[1].abs()); max = max.max(lab[2].abs()); } } println!("{:3} {}e-9 {}e-6", count, error * 1e9, max * 1e6); }

When executed the code shows that 48 colours weren’t converted correctly with the maximum error of less than 30 millionth and total square error of less than 21 billionth. With floating point numbers some imprecision is always to be expected so this isn’t a bad result by no means. However, on architectures with fused multiply-add instruction accuracy can be improved by improving the matrix multiplications at the end of `xyz_from_srgb`

and at the beginning of `srgb_from_xyz`

function. I’ll leave that as an exercise to the reader.