Introduction

On Tuesday, we studied another representation of degree three curves using Hermite Polynomials.
We went over how to convert between the Bezier Curve Representation and the Hermite Representations of curves.
We then began looking at Bezier curves of general degree and discussed some of their properties.
Today, we will show how to extend De Casteljau's method and Recursive subdivision to Bezier curves of general degree.
Then we will look at Bezier surfaces and their properties.

De Casteljau Revisited

Recall in the degree three case, that to find the point corresponding to `u` along a Bezier curve `\vec q(u)` with control points `\vec p_0`, `\vec p_1`, `\vec p_2`, `\vec p_3`, we linear interpolated (lerp'd) `u` along each pair of points to get three new points `\vec r_0`, `\vec r_1`, `\vec r_2`. We then repeated this process between each successive pair of `\vec r_i`'s to get points `\vec s_0`, `\vec s_1`. One last iteration of this process got us a point `t_0(u)`, which we claimed was `u` along the Bezier curve.
To generalize this to degree `k` Bezier curves define `\vec p_i^0(u) = \vec p_i`. Then for `r > 0` and `0\leq i \leq k-r`, let
`\vec p_i^r(u) = (1 - u) \vec p_i^(r-1)(u) + u\vec p_(i+1)^(r-1)(u)`.

Correctness of De Casteljau's Method

Theorem Let `\vec q(u)` and `\vec p_i^r` be as above. Then, for all `u`, `\vec q(u) = \vec p_0^k(u)`.

The theorem follows from the `r=k` case of the following claim:

Claim Let `0\leq r \leq k` and `0\leq i \leq k-r`. Then:
`\vec p_i^r(u) = \sum_(j=0)^r B_j^r(u)\vec p_(i+j)`.

The proof is by induction on `r`. For `r=0`, the equation above reduces to `\vec p_i = \vec p_i^0(u) = B_0^0(u)\vec p_i^0 = 1\cdot p_i^0` as desired. Assume the result holds of `r`. To see the `r+1` we calculate:
`\vec p_i^(r+1)(u) = (1 - u) \vec p_i^r(u) + u\vec p_(i+1)^r(u)`
Applying the induction hypothesis to `p_i^r(u)` and `p_(i+1)^r(u)`
`\vec p_i^(r+1)(u) = (1 - u) \sum_(j=0)^r B_j^r(u)\vec p_(i+j) + u\sum_(j=0)^r B_j^r(u)\vec p_(i+j+1)`.
Combining these two sums gives
`\vec p_i^(r+1)(u) = \sum_(j=0)^(r+1) ((1 - u)B_j^r(u) + u B_(j-1)^r(u) )p_(i+j).
Finally, from Pascal's Triangle we know that `((r),(j)) + ((r),(j-1)) = ((r+1),(j))`, from which it follows that
`(1-u)B_j^r(u) + uB_(j-1)^r(u) = B_j^(r+1)(u)`,
which establishes the claim.

Recursive Subdivision Revisited

Recall to split a degree three Bezier `\vec q(u)` at the point `u_0`, we calculated the `\vec r_i(u_0)`'s, `\vec s_i(u_0)`'s and `\vec t_0(u_0)`. Then we used control points `\vec p_0, \vec r_0, \vec s_0, \vec t_0` for the first curve and `\vec p_3, \vec r_2, \vec s_1, \vec t_0` for the second curve.
This generalizes to higher degree where we now use the points `p_i^r(u_0)`.
Namely, if we have a degree `k` Bezier curve `\vec q(u)` and we split it into two curves `\vec q_1(u) = \vec q(u_0u)` and `\vec q_2(u) = = \vec q(u_0 +(1-u_0)u)`, then both of these curves will be Bezier curves of degree `k`. Furthermore, the control points of `\vec q_1(u)` are given by `p_0^0, p_0^1, p_0^2, \ldots, p_0^k` and the control points of `\vec q_2(u)` are given by `p_0^k, p_1^(k-1), p_2^(k-2), \ldots, p_k^0`.
As you can see, this matches with our earlier result when `k=3`.
The proof of this fact is not too hard, but grundgy so I am skipping it, although it is in the book.

Degree Elevation

Sometimes it is useful to take a Bezier curve of lower degree and raise it to one of higher degree.
For example, you might express a circle using degree two Bezier curves and want to draw it using Postscript which takes as input degree three curves.
Given a polynomial of degree `k`, you could always add a lead term of the form `0\cdot x^(k+1)` to make it of degree `k+1`; however, we are interested in defining curves in terms of control points.
As a special case suppose we start with a degree two Bezier curve `\vec q(u)` defined via control points `\vec p_0, \vec p_1, \vec p_2`.
Since this curve start and ends at `\vec p_0` and `\vec p_2`, if our degree three Bezier curve `\vec Q(u)` is specified by points `\vec P_0`, `\vec P_1`, `\vec P_2`, `\vec P_3`, we know we should have the control points: `\vec P_0 = \vec p_0` and `\vec P_3 = \vec p_2`.
Using our equations for the derivative of a Bezier curve at its start points from last day, we know
`\vec q'(0) = 2(\vec p_1 - \vec p_0)`
`\vec q'(1) = 2(\vec p_2 - \vec p_1)`
But we also have conditions on our degree Bezier curve that
`\vec Q'(0) = 3(\vec P_1 - \vec P_0)`
`\vec Q'(1) = 3(\vec P_3 - \vec P_2)`
Since we want `\vec Q'(u) = \vec q'(u)` as the two curves are supposed to be the same, we get
`2(\vec p_1 - \vec p_0) = 3(\vec P_1 - \vec p_0)` and
`2(\vec p_1 - \vec p_0) = 3(\vec p_2 - \vec P_2)`
from which we can solve `\vec P_1 = 1/3\vec p_0 +2/3\vec p_1` and `\vec P_2 = 2/3\vec p_1 + 1/3 \vec p_2`.

Generalizing Degree Elevation

Suppose now `\vec q(u)` is a degree `k` with control points `\vec p_0, \vec p_1, \vec p_2, \ldots, \vec p_k`. We want to find the control points `\vec P_0, \vec P_1, \vec P_2, \ldots, \vec P_(k+1)` for degree `k+1` curve `\vec Q(u) = \vec q(u)`.
It turns out the following equations will work (proof is in the book):
`\vec P_0 = \vec p_0` and `\vec P_(k+1) = \vec p_k` and
`\vec P_i = \frac(i)(k+1)\vec p_(i-1) + \frac(k - i +1)(k+1)\vec p_i`.

Bezier Surface Patches

A picture of the degree three Bezier patch and its sixteen control points

A Bezier curve is a one-dimensional parametric curve; a Bezier patch is a two-dimensional surface.
Such a patch is parametrized using two variables `u` and `v` each ranging over `[0, 1]`.
The patch is then the parametric surface `\vec q(u,v)` where `\vec q` is a vector-valued function defined on the unit square.
A degree three Bezier patch is given by a 4 x 4 array of control points `\vec p_(i,j)`, where `i`, `j` take on values `0, 1, 2, 3`.
The patch defined by these control points is given by the formula:
`\vec q(u,v) = \sum_(i=0)^3\sum_(j=0)^3B_i(u)B_j(v)\vec p_(i,j)`.

More on Bezier Patches

Notice if you hold `u` fixed in the equation for `\vec q` and let `v` vary, the resulting curve is a Bezier curve with control points
`\vec s_j = \sum_(i=0)^3B_i(u)\vec p_(i,j)`.
You also get Bezier curves if you hold `v` fixed and let `u` vary.
Since we eventually want to be able to join patches together in order to take complex surfaces, knowing the partial derivatives of a Bezier patch at its control points is useful.
The holding one parameter fixed property we just noticed allows us to figure out the partial derivatives for Bezier patches using our knowledge of the derivatives of Bezier curves.
Namely, we have:
`\frac{\partial \vec q}{\partial v}(u, 0) = \sum_(i=0)^3 3B_i(u)(\vec p_(i,1) - \vec p_(i,0))`
`\frac{\partial \vec q}{\partial v}(u, 1) = \sum_(i=0)^3 3B_i(u)(\vec p_(i,3) - \vec p_(i,2))`
`\frac{\partial \vec q}{\partial u}(0, v) = \sum_(j=0)^3 3B_j(v)(\vec p_(1,j) - \vec p_(0,j))`
`\frac{\partial \vec q}{\partial u}(1, v) = \sum_(j=0)^3 3B_j(v)(\vec p_(3,j) - \vec p_(2,j))`
To get corners of patches to join up correctly, we are also going to need to know the second-order partial derivatives. These can be calculated from the equation above as:
`\frac{\partial^2 \vec q}{\partial u \partial v}(0, 0) = 9\cdot (\vec p_(1,1) - \vec p_(0,1) - \vec p_(1, 0) - \vec p_(0,0))`
`\frac{\partial^2 \vec q}{\partial u \partial v}(0, 1) = 9\cdot (\vec p_(1,3) - \vec p_(0,3) - \vec p_(1, 2) - \vec p_(0,2))`
`\frac{\partial^2 \vec q}{\partial u \partial v}(1, 0) = 9\cdot (\vec p_(2,1) - \vec p_(2,1) - \vec p_(3, 0) - \vec p_(2,0))`
`\frac{\partial^2 \vec q}{\partial u \partial v}(1, 1) = 9\cdot (\vec p_(3,3) - \vec p_(2,3) - \vec p_(3, 2) - \vec p_(2,2))`

Bezier Curves and Patches

Outline