Graphics Reference
In-Depth Information
In code, we might see this:
x
= uniform(0, 1);
y
= uniform(0, 1);
// U = (x, y)
x
=2
*
π
1
2
3
4
5
*
x;
y
=y;
// V= (x', y')
x
,
y
)
(
return
There's no mention of the pdf in the code, but it matters nonetheless when we
want to analyze the code.
The ability to easily draw samples from a distribution (i.e., to produce many inde-
pendent random variables all having the same specified pdf) will be critical in
rendering. In estimating the reflectance integral, we sometimes want to pick a ran-
dom direction
in the positive hemisphere with probability density proportional
to the bidirectional reflectance distribution function (BRDF), with the first argu-
ment held constant at the incoming direction, that is, the function
v
)
.
(Notice that this function is
not
in general a probability density on the hemisphere:
Its integral is not 1.) Unfortunately, sampling in proportion to a general function
on the hemisphere is not often easy.
We now examine two examples where such direct sampling is possible.
Example 1: The Lambertian BRDF. We need to choose a direction on the
hemisphere uniformly at random (i.e., the probability that the direction lies in any
solid angle
Ω
v
→
f
r
(
v
i
,
v
S
+
is proportional to the measure of
Ω
.) In fact, since the total
area of the hemisphere is 2
⊂
, we need the probability that the direction lies in a
solid angle
Ω
to be exactly
m
(Ω)
2
π
π
.
Fortunately, as we saw in Section 26.6.4, the map
(
x
1
y
2
,
y
,
z
1
g
:(
x
,
y
,
z
)
→
−
−
y
2
)
(30.44)
from the unit cylinder around the
y
-axis to the unit sphere is area-preserving
(although it's not length- or angle-preserving). By restricting our attention to the
half-cylinder
(
x
,
y
,
z
):
x
2
+
z
2
=
1, 0
H
=
{
≤
y
≤
1
}
,
(30.45)
we get an area-preserving map from
H
to
S
+
. And the map
f
:[
0, 1
]
×
[
0, 1
]
→
H
:(
u
,
v
)
→
(cos(
2
π
u
)
,
v
,
sin(
2
π
u
))
(30.46)
multiplies areas by exactly 2
π
, as you can check by computing its Jacobian. So
the composition
Y
=
g
f
of the two gives us a map from the unit square to
S
+
that multiplies areas by 2
◦
π
. The density for
Y
is therefore the constant function
on
S
+
.
In the program shown in Listing 30.2, the returned point is a random point on
the hemisphere with density
1
2
π
1
.
π
2
Listing 30.2: Producing a uniformly distributed random sample on a hemisphere.
1
2
3
4
5
Point3 randhemi()
u
= uniform(0, 1)
v
= uniform(0, 1)
r
= sqrt(1 - y
*
y)
return Point3(
r
cos(
2
π
u
)
,
y
,
r
sin(
2
π
u
)
);
