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docs/chapters/canonical/content.en-GB.md
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While quadratic curves are relatively simple curves to analyze, the same cannot be said of the cubic curve. As a curvature is controlled by more than one control point, it exhibits all kinds of features like loops, cusps, odd colinear features, and as many as two inflection points because the curvature can change direction up to three times. Now, knowing what kind of curve we're dealing with means that some algorithms can be run more efficiently than if we have to implement them as generic solvers, so is there a way to determine the curve type without lots of work?
As it so happens, the answer is yes, and the solution we're going to look at was presented by Maureen C. Stone from Xerox PARC and Tony D. deRose from the University of Washington in their joint paper ["A Geometric Characterization of Parametric Cubic curves"](http://graphics.pixar.com/people/derose/publications/CubicClassification/paper.pdf). It was published in 1989, and defines curves as having a "canonical" form (i.e. a form that all curves can be reduced to) from which we can immediately tell what features a curve will have. So how does it work?
As it so happens, the answer is yes, and the solution we're going to look at was presented by Maureen C. Stone from Xerox PARC and Tony D. deRose from the University of Washington in their joint paper ["A Geometric Characterization of Parametric Cubic curves"](https://graphics.pixar.com/people/derose/publications/CubicClassification/paper.pdf). It was published in 1989, and defines curves as having a "canonical" form (i.e. a form that all curves can be reduced to) from which we can immediately tell what features a curve will have. So how does it work?
The first observation that makes things work is that if we have a cubic curve with four points, we can apply a linear transformation to these points such that three of the points end up on (0,0), (0,1) and (1,1), with the last point then being "somewhere". After applying that transformation, the location of that last point can then tell us what kind of curve we're dealing with. Specifically, we see the following breakdown:
@@ -287,9 +287,9 @@ That's kind of super-simple to write out in code, I think you'll agree. Coding m
### How do you track all that?
Doing maths can be a pain, so whenever possible, I like to make computers do the work for me. Especially for things like this, I simply use [Mathematica](http://www.wolfram.com/mathematica). Tracking all this math by hand is insane, and we invented computers, literally, to do this for us. I have no reason to use pen and paper when I can write out what I want to do in a program, and have the program do the math for me. And real math, too, with symbols, not with numbers. In fact, [here's](http://pomax.github.io/gh-weblog-2/downloads/canonical-curve.nb) the Mathematica notebook if you want to see how this works for yourself.
Doing maths can be a pain, so whenever possible, I like to make computers do the work for me. Especially for things like this, I simply use [Mathematica](https://www.wolfram.com/mathematica/). Tracking all this math by hand is insane, and we invented computers, literally, to do this for us. I have no reason to use pen and paper when I can write out what I want to do in a program, and have the program do the math for me. And real math, too, with symbols, not with numbers. In fact, [here's](https://pomax.github.io/gh-weblog-2/downloads/canonical-curve.nb) the Mathematica notebook if you want to see how this works for yourself.
Now, I know, you're thinking "but Mathematica is super expensive!" and that's true, it's [$295 for home use](http://www.wolfram.com/mathematica-home-edition), but it's **also** [free when you buy a $35 raspberry pi](http://www.wolfram.com/raspberry-pi). Obviously, I bought a raspberry pi, and I encourage you to do the same. With that, as long as you know what you want to *do*, Mathematica can just do it for you. And we don't have to be geniuses to work out what the maths looks like. That's what we have computers for.
Now, I know, you're thinking "but Mathematica is super expensive!" and that's true, it's [$344 for home use, up from $295 when I original wrote this](https://www.wolfram.com/mathematica-home-edition/), but it's **also** [free when you buy a $35 raspberry pi](https://www.wolfram.com/raspberry-pi/). Obviously, I bought a raspberry pi, and I encourage you to do the same. With that, as long as you know what you want to *do*, Mathematica can just do it for you. And we don't have to be geniuses to work out what the maths looks like. That's what we have computers for.
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