Last Thursday, we were developing a basic toolkit of algorithms which are useful for geometric modeling.
We went over the minimax method for determining if two bounding boxes intersect.
We developed the theory of generalized bounding boxes and gave a minimax method for them.
Then we talked about surrounding tests. i.e., tests to see if a set contains a point.
We have described two tests so far which work with linear polyhedra: normal tests and equation tests.
Today, we continue to develop our toolkit beginning with more surrounding tests.
Barycentric Coordinate Test
This test only works with polyhedra in the plane.
Suppose such a polyhedra `Q` is described as `\vec p_0\vec p_1\cdots\vec p_k`, a fan of triangles `\vec p_0\vec p_i\vec p_(i+1)`.
To check if `\vec p` is in `Q` it suffices to check whether it belongs to any of these triangles.
To see if `\vec p` belongs to a triangle such as `Δ = \vec p_0\vec p_1\vec p_2` one writes `\vec p` in barycentric coordinates with respect to the triangle i.e., `\vec p - \vec p_0 = a(\vec p_1 -\vec p_0) + b(\vec p_2 -\vec p_0)` and then we check if
`a,b \ge 0` and `a+b \le 1`.
Wedge Test
This is another test that works in the plane.
Given `Q=\vec p_0\vec p_1\cdots\vec p_k`, we add a central point `\vec q` (computed as the centroid of the other points).
The rays through the central point and vertices of `Q` then divide the plane into wedges, where each wedge is cut in half by one edge of `Q`.
One can find the wedge that contains `\vec p` by doing a binary search for the angle of the `\vec p - \vec q` among the angles of the `\vec p_i - \vec q`.
Finally one checks where `\vec p` lies with respect to the edge that cuts through this wedge to determine membership.
Parity or Crossing Test
For this test one checks how many times any ray that starts at `\vec p` will intersect the boundary of `Q`.
If it intersects an even number of times, `\vec p` is outside `Q`.
If the ray is tangent to boundary of `Q` we count it as having intersected twice.
The polyhedron does not have to be convex for this test to work.
This test also works in three dimensions.
Quiz
Which of the following statements is true?
The regular operations starting from a finite set of halfspaces can generate an infinite collection of sets.
The way we learned in class to test line-plane intersections involved using cross-products.
A slab in `\mathbb{R}^n` is the region between two parallel hyperplanes.
Convexity Tests
We next look at the question of how to determine if a polygon `P` is convex. Recall a set is convex if given any points in the set the line segment between them is also in the set.
To be more precise on the question: "When does a sequence of points in a plane define a convex polygon?"
One test is to check if the segments which make up `P` keep turning in one direction. Definition. Two linearly independent vectors `\vec u` and `\vec v` in `\mathbb{R}^2` are said to determine a left turn
if the ordered basis `(\vec u, \vec v)` determines the standard orientation. Otherwise, we say they determine a right turn. If
`\vec u` and `\vec v` are linearly dependent we say they determine both a left and right turn.
Using this notion yields the following convexity test: Theorem. (Convexity test for polygons) Assume that a planar polygon `P` is defined by a sequence of points `\vec p_0,\vec p_1,\ldots,\vec p_(n),\vec p_(n+1)=\vec p_0`. The polygon `P` will be convex iff the vectors `\vec p_i\vec p_(i+1)` and `\vec p_(i+1)\vec p_(i+2)` for each `i` all determine left turns or all right turns.
The `\vec e_i` denote the `i`th unit vector in the standard coordinate basis. Then one can also check for a left turn using cross products by verifying that
`det((\vec u),(\vec v)) = \vec e_3 \cdot (\vec u \times \vec v) \ge 0.`
More on Convexity
Suppose that the polygon `F` is the face of a solid `S` in `\mathbb{R}^3` and that `\vec p` is a point in the interior of `F`. Definition. Let `\vec n` be any normal for `F`. We say that `\vec n` is an inward-pointing normal to `F` with respect to solid S,
if the segment `[\vec p, \vec p +\epsilon \vec n]` is entirely contained in `S` for some `\epsilon > 0`. In that case, `-\vec n` is called an outward-pointing normal for `F` with respect to `S`.
If `\vec q` is any point in the interior of S, then `\vec n` will be an outward-pointing normal for `F` if `\vec n \cdot (\vec q - \vec p) \ge 0`. Otherwise it will be inward-pointing.
Given a normal unit normal `\vec n`vector to a face `F`, this vector determine an orientation on the face: Pick two independent vectors `\vec u` and `\vec v` in the face and order them so that the standard basis can be rotated onto the basis one get by applying Gram-Schmidt to `\vec u`, `\vec v`, and then the vector `\vec n`.
Similarly, given `\vec u`, `\vec v`, there is a unique unit normal `\vec n` such that standard basis rotates onto Gram-Schmidt applied to `\vec u`, `\vec v` and the vector `\vec n`. Namely, `\vec n= \frac{1}{|\vec u \times \vec v|}\vec u \times \vec v`
Assume that a faces is defined by listing its vertices in some order. This order thus defines an orientation and normal vector in a unique way.
To test for convexity one can take successive edges from a face and for each edge cross it with the normal. Finally, check that the signed of each such product are all the same.
More Intersection Tests
We have already looked at how to check if a line intersects a plane. Let's briefly look at some other simple intersection tests.
To begin let's see how to determine the point `I` (if it exists) that is the intersection of two planar segments `[A,B]` and `[P,Q]`.
Let's assume that these segments determine the lines `L` and `L'` respectively. Let `N` be a normal for `L` and `N'` a normal for `L'`.
The point `I` occurs when `A + s(B-A) = P + t(Q-P)`.
It follows that
`A\cdot N = (A+ s(B-A))\cdot N = (P + t(Q-P))\cdot N = P\cdot N + t(Q-P)\cdot N`.
Solving for `t` gives: `t = \frac{(P-A)\cdot N}{(Q-P)\cdot N}`
If one solves for `s` in a similar fashion one gets `s= \frac{(P-A)\cdot N'}{(B-A)\cdot N'}`
If the denominator in these equations is zero, then `L` and `L'` are parallel. They are the same line if `(P-A)\cdotN = 0` otherwise the two segments definitely don't intersect. If they are the same line one checks whether the real segments `[0, |B-A|]` and `[|P-A|, |Q-A|]` intersect.
