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631
components/sections/matrixsplit/index.js
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631
components/sections/matrixsplit/index.js
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var React = require("react");
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var Graphic = require("../../Graphic.jsx");
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var SectionHeader = require("../../SectionHeader.jsx");
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var MatrixSplit = React.createClass({
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statics: {
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title: "Splitting curves using matrices"
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},
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render: function() {
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return (
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<section>
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<SectionHeader {...this.props}>{ MatrixSplit.title }</SectionHeader>
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<p>Another way to split curves is to exploit the matrix representation of
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a Bézier curve. In <a href="#matrix">the section on matrices</a> we saw that
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we can represent curves as matrix multiplications. Specifically, we saw these
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two forms for the quadratic, and cubic curves, respectively (using the reversed
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Bézier coefficients vector for legibility):</p>
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<p>\[
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B(t) = \begin{bmatrix}
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1 & t & t^2
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\end{bmatrix}
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\cdot
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\begin{bmatrix}
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1 & 0 & 0 \\
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-2 & 2 & 0 \\
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1 & -2 & 1
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\end{bmatrix}
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\cdot
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\begin{bmatrix}
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P_1 \\ P_2 \\ P_3
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\end{bmatrix}
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\]</p>
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<p>and</p>
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<p>\[
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B(t) = \begin{bmatrix}
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1 & t & t^2 & t^3
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\end{bmatrix}
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\cdot
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\begin{bmatrix}
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1 & 0 & 0 & 0\\
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-3 & 3 & 0 & 0\\
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3 & -6 & 3 & 0\\
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-1 & 3 & -3 & 1
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\end{bmatrix}
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\cdot
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\begin{bmatrix}
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P_1 \\ P_2 \\ P_3 \\ P_4
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\end{bmatrix}
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\]</p>
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<p>Let's say we want to split the curve at some point <em>t = z</em>, forming
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two new (obviously smaller) Bézier curves. To find the coordinates for these
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two Bézier curves, we can use the matrix representation and some linear algebra.
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First, we split out the the actual "point on the curve" information as a new matrix
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multiplication:</p>
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<p>\[
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B(t) =
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\begin{bmatrix}
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1 & (z \cdot t) & (z \cdot t)^2
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\end{bmatrix}
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\cdot
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\begin{bmatrix}
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1 & 0 & 0 \\
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-2 & 2 & 0 \\
|
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1 & -2 & 1
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\end{bmatrix}
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\cdot
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\begin{bmatrix}
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P_1 \\ P_2 \\ P_3
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\end{bmatrix}
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=
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\begin{bmatrix}
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1 & t & t^2
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\end{bmatrix}
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\cdot
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\begin{bmatrix}
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1 & 0 & 0 \\
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0 & z & 0 \\
|
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0 & 0 & z^2
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\end{bmatrix}
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\cdot
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\begin{bmatrix}
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1 & 0 & 0 \\
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-2 & 2 & 0 \\
|
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1 & -2 & 1
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\end{bmatrix}
|
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\cdot
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\begin{bmatrix}
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P_1 \\ P_2 \\ P_3
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\end{bmatrix}
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\]</p>
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<p>and</p>
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<p>\[
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B(t) =
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\begin{bmatrix}
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1 & (z \cdot t) & (z \cdot t)^2 & (z \cdot t)^3
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\end{bmatrix}
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\cdot
|
||||
\begin{bmatrix}
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1 & 0 & 0 & 0 \\
|
||||
-3 & 3 & 0 & 0 \\
|
||||
3 & -6 & 3 & 0 \\
|
||||
-1 & 3 & -3 & 1
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||||
\end{bmatrix}
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\cdot
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\begin{bmatrix}
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P_1 \\ P_2 \\ P_3 \\ P_4
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\end{bmatrix}
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=
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\begin{bmatrix}
|
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1 & t & t^2 & t^3
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\end{bmatrix}
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\cdot
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\begin{bmatrix}
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1 & 0 & 0 & 0\\
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0 & z & 0 & 0\\
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0 & 0 & z^2 & 0\\
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0 & 0 & 0 & z^3
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\end{bmatrix}
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\cdot
|
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\begin{bmatrix}
|
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1 & 0 & 0 & 0 \\
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-3 & 3 & 0 & 0 \\
|
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3 & -6 & 3 & 0 \\
|
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-1 & 3 & -3 & 1
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||||
\end{bmatrix}
|
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\cdot
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\begin{bmatrix}
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P_1 \\ P_2 \\ P_3 \\ P_4
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\end{bmatrix}
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\]</p>
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<p>If we could compact these matrices back to a form <strong>[t values] · [bezier matrix] · [column matrix]</strong>,
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with the first two staying the same, then that column matrix on the right would be the coordinates
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of a new Bézier curve that describes the first segment, from <em>t = 0</em> to <em>t = z</em>.
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As it turns out, we can do this quite easily, by exploiting some simple rules of linear algebra
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(and if you don't care about the derivations, just skip to the end of the box for the results!).</p>
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<div className="note">
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<h2>Deriving new hull coordinates</h2>
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<p>Deriving the two segments upon splitting a curve takes a few steps, and the higher
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the curve order, the more work it is, so let's look at the quadratic curve first:</p>
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<p>\[
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B(t) =
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\begin{bmatrix}
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1 & t & t^2
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\end{bmatrix}
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\cdot
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\begin{bmatrix}
|
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1 & 0 & 0 \\
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0 & z & 0 \\
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0 & 0 & z^2
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||||
\end{bmatrix}
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\cdot
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\begin{bmatrix}
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1 & 0 & 0 \\
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-2 & 2 & 0 \\
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1 & -2 & 1
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||||
\end{bmatrix}
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\cdot
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||||
\begin{bmatrix}
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P_1 \\ P_2 \\ P_3
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\end{bmatrix}
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\]</p>
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<p>\[
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=
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\begin{bmatrix}
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1 & t & t^2
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\end{bmatrix}
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\cdot
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\underset{we\ turn\ this...}{\underbrace{Z \cdot M}}
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\cdot
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||||
\begin{bmatrix}
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P_1 \\ P_2 \\ P_3
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\end{bmatrix}
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\]</p>
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<p>\[
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=
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||||
\begin{bmatrix}
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1 & t & t^2
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\end{bmatrix}
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\cdot
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\underset{...into\ this!}{\underbrace{ M \cdot M^{-1} \cdot Z \cdot M }}
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\cdot
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\begin{bmatrix}
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P_1 \\ P_2 \\ P_3
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\end{bmatrix}
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\]</p>
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<p>\[
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=
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||||
\begin{bmatrix}
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1 & t & t^2
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||||
\end{bmatrix}
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\cdot
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||||
M
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||||
\cdot
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Q
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\cdot
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||||
\begin{bmatrix}
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P_1 \\ P_2 \\ P_3
|
||||
\end{bmatrix}
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\]</p>
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<p>We do this, because [<em>M · M<sup>-1</sup></em>] is the identity matrix (a bit like
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multiplying something by x/x in calculus. It doesn't do anything to the function, but it
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does allow you to rewrite it to something that may be easier to work with, or can be
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broken up differently). Adding that as matrix multiplication has no effect on the total
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formula, but it does allow us to change the matrix sequence [<em>something · M</em>] to
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a sequence [<em>M · something</em>], and that makes a world of difference: if we know
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what [<em>M<sup>-1</sup> · Z · M</em>] is, we can apply that to our coordinates, and be
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left with a proper matrix representation of a quadratic Bézier curve (which is
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[<em>T · M · P</em>]), with a new set of coordinates that represent the curve from
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<em>t = 0</em> to <em>t = z</em>. So let's get computing:</p>
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<p>\[
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Q = M^{-1} \cdot Z \cdot M =
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\begin{bmatrix}
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1 & 0 & 0 \\
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1 & \frac{1}{2} & 0 \\
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||||
1 & 1 & 1
|
||||
\end{bmatrix}
|
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\cdot
|
||||
\begin{bmatrix}
|
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1 & 0 & 0 \\
|
||||
0 & z & 0 \\
|
||||
0 & 0 & z^2
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||||
\end{bmatrix}
|
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\cdot
|
||||
\begin{bmatrix}
|
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1 & 0 & 0 \\
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-2 & 2 & 0 \\
|
||||
1 & -2 & 1
|
||||
\end{bmatrix}
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=
|
||||
\begin{bmatrix}
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1 & 0 & 0 \\
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-(z-1) & z & 0 \\
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(z - 1)^2 & -2 \cdot (z-1) \cdot z & z^2
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\end{bmatrix}
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\]</p>
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<p>Excellent! Now we can form our new quadratic curve:</p>
|
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||||
<p>\[
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B(t) =
|
||||
\begin{bmatrix}
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||||
1 & t & t^2
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||||
\end{bmatrix}
|
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\cdot M \cdot Q \cdot
|
||||
\begin{bmatrix}
|
||||
P_1 \\ P_2 \\ P_3
|
||||
\end{bmatrix}
|
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=
|
||||
\begin{bmatrix}
|
||||
1 & t & t^2
|
||||
\end{bmatrix}
|
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\cdot
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M
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\cdot
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\left (
|
||||
Q
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||||
\cdot
|
||||
\begin{bmatrix}
|
||||
P_1 \\ P_2 \\ P_3
|
||||
\end{bmatrix}
|
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\right )
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||||
\]</p>
|
||||
|
||||
<p>\[
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||||
=
|
||||
\begin{bmatrix}
|
||||
1 & t & t^2
|
||||
\end{bmatrix}
|
||||
\cdot
|
||||
\begin{bmatrix}
|
||||
1 & 0 & 0 \\
|
||||
-2 & 2 & 0 \\
|
||||
1 & -2 & 1
|
||||
\end{bmatrix}
|
||||
\cdot
|
||||
\left (
|
||||
\begin{bmatrix}
|
||||
1 & 0 & 0 \\
|
||||
-(z-1) & z & 0 \\
|
||||
(z - 1)^2 & -2 \cdot (z-1) \cdot z & z^2
|
||||
\end{bmatrix}
|
||||
\cdot
|
||||
\begin{bmatrix}
|
||||
P_1 \\ P_2 \\ P_3
|
||||
\end{bmatrix}
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||||
\right )
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||||
\]</p>
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||||
|
||||
<p>\[
|
||||
=
|
||||
\begin{bmatrix}
|
||||
1 & t & t^2
|
||||
\end{bmatrix}
|
||||
\cdot
|
||||
\begin{bmatrix}
|
||||
1 & 0 & 0 \\
|
||||
-2 & 2 & 0 \\
|
||||
1 & -2 & 1
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||||
\end{bmatrix}
|
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\cdot
|
||||
\begin{bmatrix}
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P_1 \\
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z \cdot P_2 - (z-1) \cdot P_1 \\
|
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z^2 \cdot P_3 - 2 \cdot z \cdot (z-1) \cdot P_2 + (z - 1)^2 \cdot P_1
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||||
\end{bmatrix}
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\]</p>
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<p><strong><em>Brilliant</em></strong>: if we want a subcurve from <em>t = 0</em>
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to <em>t = z</em>, we can keep the first coordinate the same (which makes sense),
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our control point becomes a z-ratio mixture of the original control point and the start
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point, and the new end point is a mixture that looks oddly similar to a bernstein
|
||||
polynomial of degree two, except it uses (z-1) rather than (1-z)... These new
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||||
coordinates are actually really easy to compute directly!</p>
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<p>Of course, that's only one of the two curves. Getting the section from <em>t = z</em>
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to <em>t = 1</em> requires doing this again. We first observe what what we just did is
|
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actually evaluate the general interval [0,<em>z</em>], which we wrote down simplified
|
||||
becuase of that zero, but we actually evaluated this:</p>
|
||||
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||||
<p>\[
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||||
B(t) =
|
||||
\begin{bmatrix}
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||||
1 & ( 0 + z \cdot t) & ( 0 + z \cdot t)^2
|
||||
\end{bmatrix}
|
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\cdot
|
||||
\begin{bmatrix}
|
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1 & 0 & 0 \\
|
||||
-2 & 2 & 0 \\
|
||||
1 & -2 & 1
|
||||
\end{bmatrix}
|
||||
\cdot
|
||||
\begin{bmatrix}
|
||||
P_1 \\ P_2 \\ P_3
|
||||
\end{bmatrix}
|
||||
\]</p>
|
||||
|
||||
<p>\[
|
||||
=
|
||||
\begin{bmatrix}
|
||||
1 & t & t^2
|
||||
\end{bmatrix}
|
||||
\cdot
|
||||
\begin{bmatrix}
|
||||
1 & 0 & 0 \\
|
||||
0 & z & 0 \\
|
||||
0 & 0 & z^2
|
||||
\end{bmatrix}
|
||||
\cdot
|
||||
\begin{bmatrix}
|
||||
1 & 0 & 0 \\
|
||||
-2 & 2 & 0 \\
|
||||
1 & -2 & 1
|
||||
\end{bmatrix}
|
||||
\cdot
|
||||
\begin{bmatrix}
|
||||
P_1 \\ P_2 \\ P_3
|
||||
\end{bmatrix}
|
||||
\]</p>
|
||||
|
||||
<p>If we want the interval [<em>z</em>,1], we will be evaluating this instead:</p>
|
||||
|
||||
<p>\[
|
||||
B(t) =
|
||||
\begin{bmatrix}
|
||||
1 & ( z + (1-z) \cdot t) & ( z + (1-z) \cdot t)^2
|
||||
\end{bmatrix}
|
||||
\cdot
|
||||
\begin{bmatrix}
|
||||
1 & 0 & 0 \\
|
||||
-2 & 2 & 0 \\
|
||||
1 & -2 & 1
|
||||
\end{bmatrix}
|
||||
\cdot
|
||||
\begin{bmatrix}
|
||||
P_1 \\ P_2 \\ P_3
|
||||
\end{bmatrix}
|
||||
\]</p>
|
||||
|
||||
<p>\[
|
||||
=
|
||||
\begin{bmatrix}
|
||||
1 & t & t^2
|
||||
\end{bmatrix}
|
||||
\cdot
|
||||
\begin{bmatrix}
|
||||
1 & z & z^2 \\
|
||||
0 & 1-z & 2 \cdot z \cdot (1-z) \\
|
||||
0 & 0 & (1-z)^2
|
||||
\end{bmatrix}
|
||||
\cdot
|
||||
\begin{bmatrix}
|
||||
1 & 0 & 0 \\
|
||||
-2 & 2 & 0 \\
|
||||
1 & -2 & 1
|
||||
\end{bmatrix}
|
||||
\cdot
|
||||
\begin{bmatrix}
|
||||
P_1 \\ P_2 \\ P_3
|
||||
\end{bmatrix}
|
||||
\]</p>
|
||||
|
||||
<p>We're going to do the same trick, to turn <em>[something · M]</em> into <em>[M · something]</em>:</p>
|
||||
|
||||
<p>\[
|
||||
Q' = M^{-1} \cdot Z' \cdot M =
|
||||
\begin{bmatrix}
|
||||
1 & 0 & 0 \\
|
||||
1 & \frac{1}{2} & 0 \\
|
||||
1 & 1 & 1
|
||||
\end{bmatrix}
|
||||
\cdot
|
||||
\begin{bmatrix}
|
||||
1 & z & z^2 \\
|
||||
0 & 1-z & 2 \cdot z \cdot (1-z) \\
|
||||
0 & 0 & (1-z)^2
|
||||
\end{bmatrix}
|
||||
\cdot
|
||||
\begin{bmatrix}
|
||||
1 & 0 & 0 \\
|
||||
-2 & 2 & 0 \\
|
||||
1 & -2 & 1
|
||||
\end{bmatrix}
|
||||
=
|
||||
\begin{bmatrix}
|
||||
(z-1)^2 & -2 \cdot z \cdot (z-1) & z^2 \\
|
||||
0 & -(z-1) & z \\
|
||||
0 & 0 & 1
|
||||
\end{bmatrix}
|
||||
\]</p>
|
||||
|
||||
<p>So, our final second curve looks like:</p>
|
||||
|
||||
<p>\[
|
||||
B(t) =
|
||||
\begin{bmatrix}
|
||||
1 & t & t^2
|
||||
\end{bmatrix}
|
||||
\cdot M \cdot Q \cdot
|
||||
\begin{bmatrix}
|
||||
P_1 \\ P_2 \\ P_3
|
||||
\end{bmatrix}
|
||||
=
|
||||
\begin{bmatrix}
|
||||
1 & t & t^2
|
||||
\end{bmatrix}
|
||||
\cdot
|
||||
M
|
||||
\cdot
|
||||
\left (
|
||||
Q'
|
||||
\cdot
|
||||
\begin{bmatrix}
|
||||
P_1 \\ P_2 \\ P_3
|
||||
\end{bmatrix}
|
||||
\right )
|
||||
\]</p>
|
||||
|
||||
<p>\[
|
||||
=
|
||||
\begin{bmatrix}
|
||||
1 & t & t^2
|
||||
\end{bmatrix}
|
||||
\cdot
|
||||
\begin{bmatrix}
|
||||
1 & 0 & 0 \\
|
||||
-2 & 2 & 0 \\
|
||||
1 & -2 & 1
|
||||
\end{bmatrix}
|
||||
\cdot
|
||||
\left (
|
||||
\begin{bmatrix}
|
||||
(z-1)^2 & -2 \cdot z \cdot (z-1) & z^2 \\
|
||||
0 & -(z-1) & z \\
|
||||
0 & 0 & 1
|
||||
\end{bmatrix}
|
||||
\cdot
|
||||
\begin{bmatrix}
|
||||
P_1 \\ P_2 \\ P_3
|
||||
\end{bmatrix}
|
||||
\right )
|
||||
\]</p>
|
||||
|
||||
<p>\[
|
||||
=
|
||||
\begin{bmatrix}
|
||||
1 & t & t^2
|
||||
\end{bmatrix}
|
||||
\cdot
|
||||
\begin{bmatrix}
|
||||
1 & 0 & 0 \\
|
||||
-2 & 2 & 0 \\
|
||||
1 & -2 & 1
|
||||
\end{bmatrix}
|
||||
\cdot
|
||||
\begin{bmatrix}
|
||||
z^2 \cdot P_3 - 2 \cdot z \cdot (z-1) \cdot P_2 + (z-1)^2 \cdot P_1 \\
|
||||
z \cdot P_3 - (z-1) \cdot P_2 \\
|
||||
P_3
|
||||
\end{bmatrix}
|
||||
\]</p>
|
||||
|
||||
<p><strong><em>Nice</em></strong>: we see the same as before; can keep the last
|
||||
coordinate the same (which makes sense), our control point becomes a z-ratio
|
||||
mixture of the original control point and the end point, and the new start point
|
||||
is a mixture that looks oddly similar to a bernstein polynomial of degree two,
|
||||
except it uses (z-1) rather than (1-z). These new coordinates are <em>also</em>
|
||||
really easy to compute directly!</p>
|
||||
</div>
|
||||
|
||||
<p>So, using linear algebra rather than de Casteljau's algorithm, we have determined
|
||||
that for any quadratic curve split at some value <em>t = z</em>, we get two subcurves
|
||||
that are described as Bézier curves with simple-to-derive coordinates.</p>
|
||||
|
||||
<p>\[
|
||||
\begin{bmatrix}
|
||||
1 & 0 & 0 \\
|
||||
-(z-1) & z & 0 \\
|
||||
(z - 1)^2 & -2 \cdot (z-1) \cdot z & z^2
|
||||
\end{bmatrix}
|
||||
\cdot
|
||||
\begin{bmatrix}
|
||||
P_1 \\ P_2 \\ P_3
|
||||
\end{bmatrix}
|
||||
=
|
||||
\begin{bmatrix}
|
||||
P_1 \\
|
||||
z \cdot P_2 - (z-1) \cdot P_1 \\
|
||||
z^2 \cdot P_3 - 2 \cdot z \cdot (z-1) \cdot P_2 + (z - 1)^2 \cdot P_1
|
||||
\end{bmatrix}
|
||||
\]</p>
|
||||
|
||||
<p>and</p>
|
||||
|
||||
<p>\[
|
||||
\begin{bmatrix}
|
||||
(z-1)^2 & -2 \cdot z \cdot (z-1) & z^2 \\
|
||||
0 & -(z-1) & z \\
|
||||
0 & 0 & 1
|
||||
\end{bmatrix}
|
||||
\cdot
|
||||
\begin{bmatrix}
|
||||
P_1 \\ P_2 \\ P_3
|
||||
\end{bmatrix}
|
||||
=
|
||||
\begin{bmatrix}
|
||||
z^2 \cdot P_3 - 2 \cdot z \cdot (z-1) \cdot P_2 + (z-1)^2 \cdot P_1 \\
|
||||
z \cdot P_3 - (z-1) \cdot P_2 \\
|
||||
P_3
|
||||
\end{bmatrix}
|
||||
\]</p>
|
||||
|
||||
<p>We can do the same for cubic curves. However, I'll spare you the actual derivation
|
||||
(don't let that stop you from writing that out yourself, though) and simply show you
|
||||
the resulting new coordinate sets:</p>
|
||||
|
||||
<p>\[
|
||||
\begin{bmatrix}
|
||||
1 & 0 & 0 & 0 \\
|
||||
-(z-1) & z & 0 & 0 \\
|
||||
(z-1)^2 & -2 \cdot (z-1) \cdot z & z^2 & 0 \\
|
||||
-(z-1)^3 & 3 \cdot (z-1)^2 \cdot z & -3 \cdot (z-1) \cdot z^2 & z^3
|
||||
\end{bmatrix}
|
||||
\cdot
|
||||
\begin{bmatrix}
|
||||
P_1 \\ P_2 \\ P_3 \\ P_4
|
||||
\end{bmatrix}
|
||||
=
|
||||
\begin{bmatrix}
|
||||
P_1 \\
|
||||
z \cdot P_2 - (z-1) \cdot P_1 \\
|
||||
z^2 \cdot P_3 - 2 \cdot z \cdot (z-1) \cdot P_2 + (z-1)^2 \cdot P_1 \\
|
||||
z^3 \cdot P_4 - 3 \cdot z^2 \cdot (z-1) \cdot P_3 + 3 \cdot z \cdot (z-1)^2 \cdot P_2 - (z-1)^3 \cdot P_1
|
||||
\end{bmatrix}
|
||||
\]</p>
|
||||
|
||||
<p>and</p>
|
||||
|
||||
<p>\[
|
||||
\begin{bmatrix}
|
||||
-(z-1)^3 & 3 \cdot (z-1)^2 \cdot z & -3 \cdot (z-1)^3 \cdot z^2 & z^3 \\
|
||||
0 & (z-1)^2 & -2 \cdot (z-1) \cdot z & z^2 \\
|
||||
0 & 0 & -(z-1) & z \\
|
||||
0 & 0 & 0 & 1
|
||||
\end{bmatrix}
|
||||
\cdot
|
||||
\begin{bmatrix}
|
||||
P_1 \\ P_2 \\ P_3 \\ P_4
|
||||
\end{bmatrix}
|
||||
=
|
||||
\begin{bmatrix}
|
||||
z^3 \cdot P_4 - 3 \cdot z^2 \cdot (z-1) \cdot P_3 + 3 \cdot z \cdot (z-1)^2 \cdot P_2 - (z-1)^3 \cdot P_1 \\
|
||||
z^2 \cdot P_4 - 2 \cdot z \cdot (z-1) \cdot P_3 + (z-1)^2 \cdot P_2 \\
|
||||
z \cdot P_4 - (z-1) \cdot P_3 \\
|
||||
P_4
|
||||
\end{bmatrix}
|
||||
\]</p>
|
||||
|
||||
<p>So, looking at our matrices, did we really need to compute the second segment matrix?
|
||||
No, we didn't. Actually having one segment's matrix means we implicitly have the other:
|
||||
push the values of each row in the matrix <strong><em>Q</em></strong> to the right, with
|
||||
zeroes getting pushed off the right edge and appearing back on the left, and then flip
|
||||
the matrix vertically. Presto, you just "calculated" <strong><em>Q'</em></strong>.</p>
|
||||
|
||||
<p>Implementing curve splitting this way requires less recursion, and is just straight
|
||||
arithmetic with cached values, so can be cheaper on systems were recursion is expensive.
|
||||
If you're doing computation with devices that are good at matrix multiplication, chopping
|
||||
up a Bézier curve with this method will be a lot faster than applying de Casteljau.</p>
|
||||
</section>
|
||||
);
|
||||
}
|
||||
});
|
||||
|
||||
module.exports = MatrixSplit;
|
||||
Reference in New Issue
Block a user