The document explains an algorithm to determine if a line intersects a polyhedron by examining the edges and faces of the polyhedron. It details how the intersection is evaluated based on whether the line intersects an odd or even number of edges of the faces. Additionally, it discusses the implications of adding a diagonal to the polyhedron, showcasing how the Euler characteristic (v - e + f) remains unchanged.

1. Estimate the time needed for the Point in Polyhedron test for a polyhedron with v vertices, e
edges, and f faces? Please explain in details.
THIS IS WHAT I KNOW.
The line L intersects P if and only if there is a face F that has a point q in common with L. On the
plane containing the face F, + defines a ray beginning at q. This ray intersects F in an odd
number of sides, since q lies inside F. It follows that + intersects an odd number of edges
bounding F.
If, however, the line and a face do not intersect, the half-plane can intersect only an even number
of edges of this face. This proves the theorem.
This theorem provides the basis for the following algorithm which solves the intersection of a
line and a polyhedron problem.
Consider an edge of P and test for intersection with +. If the edge and the half-plane have only
one point in common, then the counters for the two faces containing the edge are incremented by
1 (the counters for the faces of P are initially zeroed.) If after all of the edges have been tested at
least one of the counters has an odd value, then L and P intersect.
This algorithm is based on a simple test for the intersection of an edge (line segment) and a half-
plane.
Consider a point p and a polyhedron P. Take an arbitrary line L passing through p and an
arbitrary plane passing through L. Let L+ denote one of the parts of L divided by p, and let + be
one of the half-planes obtained by dividing with L. We say that an edge e of P is marked if +
intersects e and only one of the two planes of faces of P that share e intersects L+. Suppose +
does not pass through vertices of P.
Then p belongs to P if and only if the number of marked edges is odd, and p lies outside P if this
number is even.
Solution
and look at the network we get after performing Step 1 just once. Now, by drawing a diagonal
we added one edge. Our original face has become two faces, so we have added one to the
number of faces. We haven't changed the number of vertices. The network now has V vertices,
E + 1 edges and F + 1 faces. So how has V - E + F changed after we performed Step 1 once?
Using what we know about the changes in V, E and F we can see that V - E + F has become V -
(E + 1) + (F + 1). Now we have
V - (E + 1) + (F + 1) = V - E - 1 + F + 1 = V - E + F.