Graphics Reference
In-Depth Information
Inline Exercise 25.10: Carry out the same computation as in the preceding
exercise, but this time start with s ( 0 )= 2; confirm that all other values shift
by
2.
Similarly, we can add a linear variation to the signal (a steady increase or
decrease in value) by changing just s ( 0 ) . Thus, the second-derivative information
captures the aspects of the signal that are unaffected by translation or shearing of
the signal.
The analog, for meshes, is provided by the mesh Laplacians. Suppose that
we have a mesh, and at each vertex we have a real number, which we'll denote
s ( v ) in analogy with the one-dimensional discrete-signal case. There's no longer
a notion of the previous or next signal value, but there is still a notion of adjacent
values. Letting N ( v ) denote the 1-ring consisting of vertices adjacent to v , and
|
N ( v )
|
denote the size of this set, we define the Laplacian of s at v to be
L ( s )( v )= C
w
( s ( v )
s ( w )) ,
(25.11)
N ( v )
where the constant C is unimportant for now. We can bring the constant s ( v ) out-
side the sum, to get
1
s ( w )= C ( s ( v )
L ( s )( v )= C
|
N ( v )
|
s ( v )
s ( w )) ,
(25.12)
|
N ( v )
|
w N ( v )
w N ( v )
where we've absorbed the number
of neighbors of v into the constant C to
make C . In this form, we see the Laplacian expresses the difference between the
signal value at the vertex v and the average of the signal values at the neighbors
of v .
To be clear: The Laplacian is a function from “signals on the mesh” to “signals
on the mesh.” Thus, if s is a signal, so is L ( s ) , and L ( s )( v ) denotes the value of
that signal at a particular vertex.
In analogy with the 1D situation, if you knew the value of s at some vertex
v 0 and at all but one of its neighbors, and you knew the Laplacian of s at every
vertex, then you could compute the value of s at the last neighbor. And that might
give you enough information to compute the value of s at another vertex, etc.
There's a difference from the discrete-signal situation, however: The Lapla-
cian values are not all independent.
|
N ( v )
|
Inline Exercise 25.11: Draw a tetrahedron, and write the numbers 1, 3, 0, and
5 at the four vertices, thus defining a “signal” s on the tetrahedron. Compute
the Laplacian L ( s )( v ) at each of these vertices, using C = 1 as the constant.
What do you notice about the sum of these values?
Thus, although the Laplacian of a signal once again represents the part that's
independent of the addition of a constant at every vertex, and some other kind
of alteration similar to shearing in the 1D discrete-signal case, it's no longer the
case that an arbitrary set of values at vertices
can be treated as
the Laplacian of some signal and “integrated” to recover the original signal. The
{
h ( v ): v
V
}
 
 
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