TNCG15: Advanced Global Illumination and
Rendering, Lecture 6 (Monte-Carlo raytracing)
Mark Eric Dieckmann,
MIT group, ITN
Campus Norrköping
SE-60174 Norrköping
September 26, 2016
Outline of the lecture
◮
We have previously discussed how raytracing works and we have
addressed the special case of Whitted raytracing.
◮
Whitted raytracing works well for specular or transparent objects,
but it can not be used for general BRDFs.
◮
We have seen that Monte-Carlo integration is a fast method to
integrate multi-dimensional integrals.
Hence it can be used for computing the integral in the rendering
equation.
◮
Now we show how we do that in practice.
Basic idea
Our Whitted ray-tracer computed the reflected and refracted rays using
physical laws: we obtain photorealistic renderings of mirrors and glas.
This is not possible for general BRDF’s: we use instead the Monte-Carlo
scheme to integrate the rendering equation.
The starting point of Whitted ray-tracing is the same as that of MC ray
tracing.
The first importance ray is launched at the eye position e and goes
through a point on the pixel.
We compute the intersection of the ray with an object or with a wall.
Basic idea
The point x is the first point of intersection between the camera ray and
a surface point.
The camera is located in the direction ωout . We can use this direction to
set up the local coordinate system around x:
~ N
~
~v1 = ωout − (ωout · N)
Our aim is to compute the radiance
L(x → ωout ) using L(x ← ωin ).
ωin = (ϕi , θi ) has random values
for the azimuth ϕi (relative to ~v1 )
and inclination θi .
N
ω out
ω in
θi
ρi
V1
X
The rendering equation
The rendering equation is
L(x → ωout ) = Le (x → ωout ) +
Z
fr (x, ωin , ωout ) L(x ← ωin ) cos θin dωin
We can compute the emitted radiance L(x → ωout ) once we know the
self-emission Le (x → ωout ) of the point x (light source term) and once
we have solved the integral.
Monte-Carlo
integration approximates a two-dimensional
integral
RR
PN
I =
f (x, y ) dx dy by an estimator hI i = N −1 i =1 f (xi , yi )/p(xi , yi ).
We need to choose a suitable PDF and determine how we transform the
integral into a form that matches the equation for MC integration.
Random numbers and probability distributions
We must choose a suitable probability distribution p(x, y ) for the random
numbers. Two questions spring to mind:
(1) Does it matter, which PDF we take for the evaluation of
PN
hI i = N −1 i =1 f (xi , yi )/p(xi , yi )?
The estimator hI i for the integral I is linked to the expected value of a
function of random numbers and the variance σ 2 estimates its accuracy.
(2) How do we compute random numbers with a distribution p(xi , yi )?
What kind of random numbers can we get from a computer?
The ideal probability distribution
The expected value of the estimator hI i of an integral I in 1D is
E [hI i] = E [N
−1
N
X
f (xi )/p(xi )] = I
i =1
The variance determines how close the hI i are to I .
2
2
σ = E [(hI i − I ) ] = E [
N
−1
N
X
f (xi )
p(xi )
i =1
!
−I
!2
The variance becomes in the limit of large N
1
σ =
N
2
Z f (x)
−I
p(x)
2
p(x)dx
The variance σ 2 = 0 if p(x) = f (x)/I . Can we use this PDF?
]
The ideal probability distribution
The variance σ 2 = 0 for the PDF p(x) = f (x)/I .
R∞
We would need to fulfill 1 = −∞ p(x) dx and we can not integrate f (x).
f(x)
0
0
a
p(x)
Choose an integrable PDF p(x)
that is similar to f (x).
c
d
b x
Draw random positions xi from
p(x).
We get more xi in the domain
[c, d] than in [a, c] or [d, b].
a
c
d
b x
We thus place most samples xi into those x-intervals where f (x) is
largest. These intervals contribute most to I .
Getting random numbers with the CDF
How can we compute random numbers xi with a PDF p(x) using
uniformly distributed random numbers (p(x) = constant)?
p(x)
Given is a continuous
R xPDF p(x). Its cumulative distribution function
(CDF) is cfd(x) = −∞ p(x̃) d x̃.
2
1.5
1
0.5
0
0
0.2
0.4
0.6
0.8
1
cfd(x)
x
1
0.8
0.6
0.4
0.2
0
0
0.2
0.4
The value of cfd(x) grows fastest
where p(x) is large.
0.6
x
The values of
cfd(x) range from 0
R∞
to 1 due to −∞ p(x) dx = 1 and
p(x) ≥ 0.
0.8
1
cfd(x) increases strictly with x in
intervals with p(x) > 0.
If a function increases strictly in an interval, it can be inverted.
Sometimes we can find an analytic expression for the inverse.
Getting random numbers with the CDF
The cumulative distribution function is cfd(x) =
cfd(x)
−∞
p(x̃ ) d x̃.
The red horizontal lines have a
constant spacing along the vertical
y-axis.
1
0.8
0.6
0.4
0.2
0
0
Rx
0.2
0.4
0.6
0.8
1
The vertical blue lines go through
the intersection of the red lines
with cfd(x).
x
If we place lines with equal spacing along y = cfd(x), the spacing of the
vertical lines along x is not equal.
The blue lines are concentrated in intervals with a large slope of the CDF.
The slope of the CDF is proportional to p(x).
Getting random numbers
A fast way to get random numbers with a probability distribution
function p(x) is:
Step 1: Integrate the
R x PDF, which gives the cumulative distribution
function cfd(x) = −∞ p(x) dx.
Step 2: Invert the CDF: cfd(x) maps values on the x-axis onto
y = cfd(x). Its inverse maps the y-axis onto x = cfd −1 (y ).
Step 3: Calculate random numbers yi using a random number generator
with a uniform PDF p(y ) = 1 on 0 ≤ y ≤ 1.
Step 4: Map the uniform random numbers onto random numbers with
the PDF p(x) using xi = cfd −1 (yi ).
If you use in your project a PDF p(x) that gives an invertible CDF then
you can easily compute random numbers that are distributed like p(x)
using uniform random numbers.
Rejection sampling
If the elegant way does not work, use the brute-force one.
Take the 2D array of size: [ a , b ] × [ 0 , M ], where p(x) ≤ M.
Calculate pairs x, y in the 2D field. If for a pair xi , yi we get yi > p(xi ),
the sample is rejected.
The xi are then distributed according to p(x).
The probability distribution for the renderer
We want to compute the integral of the rendering equation with an MC
scheme.
Z
fr (x, ωin , ωout ) L(x ← ωin ) cos θin dωin
The distribution L(x ← ωin ) is not initially known and we can not use it
as the PDF.
The BRDF fr (x, ωin , ωout ) rarely gives an invertible CDF (works for
Phong’s reflection models and some others).
You should use in your project the remaining term p(θ, ϕ) = π −1 cos θ
for the Lambertian and Oren-Nayar reflectors.
Radiance flow (almost) along the surface normal contributes most to the
irradiance at the point x. The PDF above sends most rays that way.
The probability distribution for the renderer
R
The CDF of p(θ, ϕ) = π −1 cos θ is F = π −1 cos θ dω, where ω is the
solid angle and dω = sin θ dθ dϕ.
Rϕ
′ R θ
′
′
′
We integrate it as F (θ , ϕ) = π −1 0 dϕ 0 cos θ sin θ dθ .
′
= (ϕ/π) (− cos2 θ /2)|θ0 = (ϕ/2π) (1 − cos2 θ).
We can separate the function into two terms Fϕ = (ϕ/2π) and
Fθ = 1 − cos2 θ, which we invert separately.
We draw a pair of random numbers u1 , u2 ǫ [0, 1).
Using them as the argument of the inverse CDF gives us the random
numbers that follow p(θ, ϕ):
√
The azimuth ϕi = 2πu1 and the elevation θi = cos−1 ( u2 ).
Computing the reflected importance
The rendering integral with no source term for radiance is
Z
L(x → ωout ) =
fr (x, ωin , ωout ) L(x ← ωin ) cos θin dωin .
We draw random numbers with the PDF p(ϕ, θ) = cos θin /π.
Z
L(x → ωout ) = π
fr (x, ωin , ωout ) L(x ← ωin ) p(ϕ, θ) dωin
≈
N
π X
fr (x, ωin , ωout ) L(x ← ωin )
N
i =1
We use only one random ray with direction ωin = (ϕi , ωi ):
L(x → ωout ) = πfr (x, ωin , ωout )L(x ← ωin )
Example of Lambertian reflector: fr (x, ωin , ωout ) = ρ/π.
Computing the reflected importance
We have approximated the rendering integral as
L(x → ωout ) = πfr (x, ωin , ωout )L(x ← ωin ).
Since L(x → ωout )W (x ← ωout ) = L(x ← ωin )W (x → ωin ):
W (x → ωin )
= πfr (x, ωin , ωout ).
W (x ← ωout )
We have defined ωin and ωout with respect to the radiance flow.
Importance flows the other way:
W2 = πfr (x, ωin , ωout )W1
We give the importance W2 to ray r2 and we use this expression to
compute the importances of all consecutive rays.
Ray termination
Our Monte-Carlo scheme allows us to track rays as they move through
the scene.
We need to specify a ray termination strategy for our renderer:
We will not terminate rays when they interact with reflecting or
transparent surfaces.
We will terminate rays on area light sources and on Lambertian /
Oren-Nayar reflectors (walls and objects).
We terminate them with an unbiased scheme.
Ray termination
First idea: select an absorption probability α for a ray that hits the
surface. The value of α can be connected to surface properties.
Draw a random number 0 ≤ t ≤ 1. If t > 1 − α we terminate the ray.
Let’s see what happens.
We approximate I =
R
f (x) dx by hI i = N −1
PN
i =1
f (xi )/p(xi ).
The expected value of the random estimator is E (hI i) = I .
Let us set NP
= 4. We draw 4 random values x1 , x2 , x3 , x4 to compute
4
hI i = (1/4) i =1 f (xi )/p(xi ).
We compute a large number M of estimators (each giving a random
PM
value hI ij ) and average them to get E (hI i) ≈ M −1 j=1 hI ij .
Ray termination
We cut off rays with a probability of α = 0.25.
This value forPα removes on average one ray from every estimator
hI i∗ = (1/4) 3i =1 f (xi )/p(xi ).
The expected value obtained from the averaging of M estimators hI ij is
now E (hI i∗ ) = (3/4)E (hI i) = 3I /4.
The estimator hI i∗ underestimates the integral value and it is thus biased.
We can compensate this bias by multiplying f (xi ) by 4/3.
Ray termination
R
′
We take the integral I = (4/3) f (x) dx, which gives the 4-sample
estimator
′
hI i = (1/4)
4 X
(4/3)f (xi )
i =1
p(xi )
We remove systematically one ray and we get
′′
hI i = (1/4)
3 X
(4/3)f (xi )
i =1
p(xi )
′
34
3
E (hI i) =
E (hI i) = E (hI i)
4
43
Russian roulette follows up on this idea in a systematic way.
′′
E (hI i) =
Ray termination: Russian roulette
Let us take the integral I =
R1
0
f (x) dx.
We rescale x and the function by a
factor 0 < P < 1.
The x-interval shrinks to [0, P].
The function value is multiplied by
1/P.
The integral value is preserved:
RP
R1
0 f (x/P)/P dx = 0 f (x) dx.
We approximate the rescaled integral by the estimator:
I =
Z
0
P
N
f (x/P)
1 X f (xi /P)
dx ≈
P
PN i =1 p(xi )
Russian roulette in the project
The rendering integral for the radiance at the surface point x has been
approximated by a Monte-Carlo scheme that uses one ray.
L(x → ωout ) = πfr (x, ωin , ωout )L(x ← ωin ).
We want to terminate the ray path - we launch a ray W (x → ωin ) with a
probability P and we do not launch one with a probability (1 − P).
To remove the bias we use the equation
L(x → ωout ) =
π
fr (x, ωin , ωout )L(x ← ωin ).
P
We modify the equation for the random azimuth angle to ϕi = (2π/P) u
where the random value 0 ≤ u ≤ 1. If ϕi > 2π we terminate the ray
path, if 0 ≤ ϕi ≤ 2π we spawn a ray.
The tree
We develop a ray tree like we did for Whitted ray tracing.
If the ray interacts with an intransparent and non-specular surface a new
ray is spawned unless Russian roulette terminates it.
If the ray hits a specular reflector (mirror) one new ray is spawned (no
Russian roulette).
If the ray hits a transparent surface one reflected and one refracted ray
are spawned (no Russian roulette).
If the ray hits an area light source it is always terminated.
Eventually Russian roulette has terminated all rays and we have
completed the ray tree.
Introducing radiance into the tree
We have developed the tree for the importance rays. We follow one ray
path until the termination ray rN . The direction of the ray ri is ωi
We use area light sources in MC
raytracing. (We use triangles).
Rays can hit an area light source.
We still need direct light for the
ray-surface intersections.
Area light
rN
Shadow
rays
r N−2
O2
r N−1
O1
The ray rN ends here on the point xN+1 on an area light source and
Le (xN+1 → −ωN ) > 0.
We use Lambertian emitters in our project: Le (xN+1 → −ωN ) = L0 with
L0 being the same for all −ωN and xN+1 .
Reflected radiance
Our rendering equation has been reduced for our case into the form
L(x → ωout ) = LD (x → ωout ) +
At xN : L(xN → −ωN−1 ) =
LD (xN → −ωN−1 ) +
π
f (xN , ωN , −ωN−1 )L(xN ← ωN )
P r
The value Pπ fr (xN , ωN , −ωN−1 )
corresponds to the ratio of the
importances of rN and rN−1 .
π
fr (x, ωin , ωout )L(x ← ωin )
P
x N+1
L0
rN
Shadow
rays
r N−2
xN
O2
r N−1
O1
The same procedure is true for xN−1 , xN−2 , .., x0 .
All that remains to be done is to compute LD using shadow rays.
Direct light
We want to compute the direct light component (light from the light
source that is reflected once):
Z
LD (xN → −ωN−1) = fr (xN , ωin , −ωN−1)L(xN ← ωin ) cos θin dωin
S2
The integration can be
approximated with an MC scheme.
S1
Light
rN
We launch shadow rays and
perform a visibility test.
Use a number M ≫ 1 of shadow
rays. 10-20 might be a good
compromise.
S3
xN
r N−1
O2
Our light source is a Lambertian emitter and Le (q → ωout ) = L0 for all
points q and light directions ωout of the light source.
Direct light
We want to approximate the integral
Z
LD (xN → −ωN−1) = fr (xN , ωin , −ωN−1)L(xN ← ωin ) cos θin dωin
by a Monte-Carlo estimator that uses M shadow rays.
The PDF p(ϕ, θ) of our light source is nontrivial in hemispherical
coordinates. We thus use the integral over the triangle’s surface A.
Z
LD (xN → −ωN−1 ) = L0 fr (xN , ωin , −ωN−1) V (xN , q) G (xN , q) dA
A
V (xN , q) is the visibility function for the end point q of a shadow ray.
G (xN , q) = cos α cos β/d 2 is the geometric term. The length of the
shadow ray is d and α, β are the inclination angles relative to the normals
at its start- and endpoint.
Computing shadow rays
Our lightsource is a triangle with the corner points v0 , v1 and v2 .
The triangle surface can be computed by
A = 0.5 · ||(v1 − v0 ) × (v2 − v0 )||.
You can compute A by hand from the triangle list that defines your scene.
We pick random points q on its surface with a uniform PDF p(q) = 1/A.
Suggestion: Draw two uniform random numbers u, v . If u + v < 1 you
construct a point qi on the triangle with barycentric coordinates
qi = (1 − u − v )v0 + uv1 + v v2 . Otherwise you draw new u, v . You draw
M valid points on your light source.
The endpoint of each shadow ray is qi and the starting point xN of all M
shadow rays is located on the surface where you launched it.
Direct light
We have obtained M shadow rays that go from xN to M points qi on the
light source surface. The shadow ray direction is ωS,i and di is its length.
We can approximate
LD (xN → −ωN−1) = L0
Z
A
fr (xN , ωS,i , −ωN−1 ) V (xN , qi ) G (xN , qi ) dA,
by the estimator
< LD (xN → −ωN−1) >=
M
AL0 X
fr (xN , ωS,i , −ωN−1) V (xN , qi ) G (xN , qi )
M
i =1
We add the computed radiance value hLD i to the radiance that has been
reflected by the point xN .
Lecture summary
◮
We have discussed practical issues related to Monte-Carlo
raytracing.
◮
We have discussed how we choose the probability distribution
function (PDF) that sets the random azimuth and inclination angles
for the rays.
◮
We have shown how the cumulative distribution function gives us
suitable random numbers.
◮
We have discussed how we build up the ray tree and how we can
compute the direct light contributions.
◮
We have completed our discussion of basic Monte-Carlo raytracing.