I also could not found a solution on the web today, so I tried to derive it.
Firstly the notations of a 3D Perlin noise is defined.
Notation
Assume the 3D Perlin noise is computed by the trilinear interpolation as
n = Lerp(
Lerp(
Lerp(dot000, dot100, u),
Lerp(dot010, dot110, u),
v),
Lerp(
Lerp(dot001, dot101, u),
Lerp(dot011, dot111, u),
v),
w)
where u
, v
, w
are the interpolation factors by the quintic polynomial of fraction coordinates (i.e., improved Perlin noise):
x0 = frac(x)
y0 = frac(y)
z0 = frac(z)
x1 = x0 - 1
y1 = y0 - 1
z1 = z0 - 1
u = x0 * x0 * x0 * (x0 * (6 * x0 - 15) + 10)
v = y0 * y0 * y0 * (y0 * (6 * y0 - 15) + 10)
w = z0 * z0 * z0 * (z0 * (6 * z0 - 15) + 10)
and dot___
s are dot products of the gradient vectors (gx___, gy___, gz___)
s at lattice points and the fraction coordinates:
dot000 = gx000 * x0 + gy000 * y0 + gz000 * z0
dot100 = gx100 * x1 + gy100 * y0 + gz100 * z0
dot010 = gx010 * x0 + gy010 * y1 + gz010 * z0
dot110 = gx110 * x1 + gy110 * y1 + gz110 * z0
dot001 = gx001 * x0 + gy001 * y0 + gz001 * z1
dot101 = gx101 * x1 + gy101 * y0 + gz101 * z1
dot011 = gx011 * x0 + gy011 * y1 + gz011 * z1
dot111 = gx111 * x1 + gy111 * y1 + gz111 * z1
Computing the derivatives
First, compute derivatives of u
, v
and w
u' = 30 * x0 * x0 * (x0 - 1) * (x0 - 1)
v' = 30 * y0 * y0 * (y0 - 1) * (y0 - 1)
w' = 30 * z0 * z0 * (z0 - 1) * (z0 - 1)
By expanding n
with Lerp(a, b, t) = a + (b - a) * t
,
n = dot000
+ u(dot100 - dot000)
+ v(dot010 - dot000)
+ w(dot001 - dot000)
+ uv(dot110 - dot010 - dot100 + dot000)
+ uw(dot101 - dot001 - dot100 + dot000)
+ vw(dot011 - dot001 - dot010 + dot000)
+ uvw(dot111 - dot011 - dot101 + dot001 - dot110 + dot010 + dot100 - dot000)
Then take partial derivatives of n
,
nx = gx000
+ u' (dot100 - dot000)
+ u (gx100 - gx000)
+ v (gx010 - gx000)
+ w (gx001 - gx000)
+ u'v (dot110 - dot010 - dot100 + dot000)
+ uv (gx110 - gx010 - gx100 + gx000)
+ u'w (dot101 - dot001 - dot100 + dot000)
+ uw (gx101 - gx001 - gx100 - gx000)
+ vw (gx011 - gx001 - gx010 + gx000)
+ u'vw(dot111 - dot011 - dot101 + dot001 - dot110 + dot010 + dot100 - dot000)
+ uvw (gx111 - gx011 - gx101 + gx001 - gx110 + gx010 + gx100 - gx000)
,
ny = gy000
+ u (gy100 - gy000)
+ v' (dot010 - dot000)
+ v (gy010 - gy000)
+ w (gy001 - gy000)
+ uv' (dot110 - dot010 - dot100 + dot000)
+ uv (gy110 - gy010 - gy100 + gy000)
+ uw (gy101 - gy001 - gy100 + gy000)
+ v'w (dot011 - dot001 - dot010 + dot000)
+ vw (gy011 - gy001 - gy010 + gy000)
+ uv'w(dot111 - dot011 - dot101 + dot001 - dot110 + dot010 + dot100 - dot000)
+ uvw (gy111 - gy011 - gy101 + gy001 - gy110 + gy010 + gy100 - gy000)
,
nz = gz000
+ u (gz100 - gz000)
+ v (gz010 - gz000)
+ w' (dot001 - dot000)
+ w (gz001 - gz000)
+ uv (gz110 - gz010 - gz100 + gz000)
+ uw' (dot101 - dot001 - dot100 + dot000)
+ uw (gz101 - gz001 - gz100 + gz000)
+ vw' (dot011 - dot001 - dot010 + dot000)
+ vw (gz011 - gz001 - gz010 + gz000)
+ uvw'(dot111 - dot011 - dot101 + dot001 - dot110 + dot010 + dot100 - dot000)
+ uvw (gz111 - gz011 - gz101 + gz001 - gz110 + gz010 + gz100 - gz000)
Then (nx, ny, nz)
is the gradient vector (partial derivatives) of the noise function.
Optimization
Some common sub-expression can be factored out, if the compiler cannot handle it. For example:
uv = u * v
vw = v * w
uw = u * w
uvw = uv * w
The coefficients in the expanded n
are reused multiple times. They can be computed by:
k0 = dot100 - dot000
k1 = dot010 - dot000
k2 = dot001 - dot000
k3 = dot110 - dot010 - k0
k4 = dot101 - dot001 - k0
k5 = dot011 - dot001 - k1
k6 = (dot111 - dot011) - (dot101 - dot001) - k3
Also the derivatives has similar coefficients,
gxk0 = gx100 - gx000
gxk1 = gx010 - gx000
...
The computation of n
can uses the expanded form with k0
, ... k6
as well.
Final words
This solution has been verified against central difference method.
Although this solution looks clumsy, my experiment (CPU only, SSE) showed that, computing these derivatives by this solution only incurs about 50% extra time to computing a single 3D Perlin noise sample.
Finite difference will at least need 300% extra time (doing extra 3 samples) or 600% (doing 6 samples for central difference).
Therefore, this solution is better in performance, and should also be more numerically stable.