The.Algorithm.Design.Manual.Springer-Verlag.1998

War Story: Getting the Graph
``Well, it starts out with a list of the 3D-coordinates of the vertices used in the model and then follows
with a list of triangles. Each triangle is described by a list of three indices into the vertex coordinates.
Here is a small example:''
VERTICES 4
0.000000 240.000000 0.000000
204.000000 240.000000 0.000000
204.000000 0.000000 0.000000
0.000000 0.000000 0.000000
TRIANGLES 2
0 1 3
1 2 3
``I see. So the first triangle must use all but the third point, since all the indices start from zero. The two
triangles must share an edge formed by points 1 and 3.''
``Yeah, that's right,'' she confirmed.
``OK. Now tell me how you built your dual graph from this file.''
``Well, I can pretty much ignore the vertex information, once I know how many vertices there will be.
The geometric position of the points doesn't affect the structure of the graph. My dual graph is going to
have as many vertices as the number of triangles. I set up an adjacency list data structure with that many
vertices. As I read in each triangle, I compare it to each of the others to check whether it has two
numbers in common. Whenever it does, I add an edge from the new triangle to this one.''
I started to sputter. ``But that's your problem right there! You are comparing each triangle against every
other triangle, so that constructing the dual graph will be quadratic in the number of triangles. Reading in
the input graph should take linear time!''
``I'm not comparing every triangle against every other triangle. On average, it only tests against half or a
third of the triangles.''
``Swell. But that still leaves us with an algorithm. That is much too slow.''
She stood her ground. ``Well, don't just complain. Help me fix it!''
Fair enough. I started to think. We needed some quick method to screen away most of the triangles that
would not be adjacent to the new triangle (i,j,k). What we really needed was just a list of all the triangles
that go through each of the points i, j, and k. Since each triangle goes through three points, the average
point is incident on three triangles, so this would mean comparing each new triangle against fewer than
ten others, instead of most of them.
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