Graphics Reference
In-Depth Information
For the moment, let's pretend that the problem we care about really
is
finite
dimensional: that
L
is a vector of
n
elements, for some large
n
, and so is
L
e
, while
I
−
×
T
is an
n
n
matrix. The solution to the equation is then
T
)
−
1
(
L
e
)
.
L
=(
I
−
(31.49)
In general, computing the inverse of an
n
n
matrix is quite expensive and prone
to numerical error, particularly for large
n
. But there's a useful trick. We observe
that
×
T
)(
I
+
T
+
T
2
+
+
T
k
)=
I
T
k
+
1
.
(
I
−
...
−
(31.50)
Inline Exercise 31.5:
Verify Equation 31.50 by multiplying everything out.
Remember that matrix multiplication isn't generally commutative. Why is it
OK to swap the order of multiplications in this case?
Suppose that as
k
gets large,
T
k
+
1
gets very small (i.e., all entries of
T
k
+
1
approach zero). Then the sum of all powers of
T
ends up being the inverse of
(
I
−
T
)
, that is, in this special case we can in fact write
T
)
−
1
=
I
+
T
+
T
2
+
(
I
−
...
.
(31.51)
Multiplying both sides by
L
e
, we get
T
)
−
1
L
e
;
L
=(
I
−
(31.52)
=
IL
e
+
TL
e
+
T
2
L
e
+
...
.
(31.53)
In words, this says that the light in the scene consists of that emitted from the
luminaires (
IL
e
), plus the light emitted from luminaires and scattered once (
TL
e
),
plus that emitted from luminaires and scattered twice (
T
2
L
e
), etc.
Our fanciful reasoning, in which we assumed everything was finite-
dimensional, has led us to a very plausible conclusion. In fact, the reasoning
is
valid even for transformations on infinite-dimensional spaces. The only restric-
tion is that
T
2
must be interpreted as “apply the operator
T
twice” rather than
“square the matrix
T
.”
We
did
have to assume that
T
k
0as
k
gets large, however. When
T
is a
matrix, this simply means that all entries of
T
k
go toward zero as
k
gets large. For
a linear operator on an infinite dimensional space, the corresponding statement
is that
T
k
H
goes to zero as
k
gets large, where
H
is an arbitrary element of the
domain of
T
. (In our case, this means that for any initial emission values, if we
trace the light through enough bounces, it gets dimmer and dimmer.)
We'll assume, from now on, that the scattering operator
T
has the property
that
T
k
→
→
→∞
0as
k
so that the series solution of the rendering equation will
produce valid results.
Of course, the series solution has infinitely many terms to sum up, each of
them expensive to compute, so it's not, as written, a practical method for rendering
a scene. On the other hand, as we've already seen with radiosity, there are practical
approximations to be made based on this series solution.
