Graphics Reference
In-Depth Information
arguments we made earlier don't tell the whole story. The key idea that Kajiya
(and Cook et al.) put forth is that while the time to trace any given ray is about
the same, the amount of
information
that you get from some rays is greater than
for others, and you'd like to favor those. If you measure the difference from the
true mean of a pixel in a path-traced image that used
r
rays (total, not just primary
rays) and an image produced using
r
rays in the exponential fan-out pattern, you'll
find that the path-traced image is usually closer to the true mean, that is, the bias is
smaller. Kajiya actually leverages the variable importance of rays in several ways;
here we'll just focus on the strategy of sending one (terminal) direct illumination
ray and one (recursive) indirect illumination ray at each surface.
We indicated earlier that computing the series solution
L
=(
1
+
T
+
T
2
+
)
L
e
...
(31.76)
=
L
e
+
∞
T
k
L
e
(31.77)
k
=
1
to the rendering equation
v
o
)+
S
2
v
o
)=
L
e
(
P
,
L
(
P
,
L
(
P
,
−
v
i
)
f
s
(
P
,
v
i
,
v
o
)
|
v
i
·
n
|
d
v
i
,
(31.78)
where
L
e
is the emitted radiance,
L
is the radiance, and
T
is the transport operator,
was closely related to computing
x
=
b
+
∞
M
k
b
,
(31.79)
k
=
1
as the solution to the equation
x
=
b
+
Mx
,
(31.80)
where
b
and
x
are
n
-vectors and
M
is an
n
n
matrix. The analogy is that
T
is
a linear operator on the space of all possible radiance functions, while multipli-
cation by
M
is a linear operation on the vector space
R
n
; in the case where we
approximated the space of possible radiance functions by those that are piecewise
constant on a fixed mesh, and where the transport operation involved only Lam-
bertian reflection (the radiosity approximation), the analogy was exact: Instead of
solving an integral equation, we had to solve a finite-dimensional matrix equation.
We'll consider the case where
M
is a 2
×
2 matrix, chosen so that its eigen-
values are both less than one in magnitude, which makes the series converge,
and so that its entries are all non-negative, because they're meant to correspond,
roughly, to values of the BSDF in the rendering equation, and these values are
never negative.
In the next several pages we'll estimate the value of
x
1
, the first entry of the
solution. We can, of course, solve Equation 31.80 directly, given any particular
2
×
2matrix
M
, but you should imagine that
M
could have a thousand or a trillion
rows rather than just two; it's in that case that the methods described here work
best.
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