diagrams-contrib-0.6: Collection of user contributions to diagrams EDSL

Maintainer byorgey@cis.upenn.edu None

Diagrams.TwoD.Apollonian

Description

Generation of Apollonian gaskets. Any three mutually tangent circles uniquely determine exactly two others which are mutually tangent to all three. This process can be repeated, generating a fractal circle packing.

See J. Lagarias, C. Mallows, and A. Wilks, "Beyond the Descartes circle theorem", Amer. Math. Monthly 109 (2002), 338--361. http://arxiv.org/abs/math/0101066.

Synopsis

# Circles

data Circle Source

Representation for circles that lets us quickly compute an Apollonian gasket.

Constructors

 Circle Fieldsbend :: DoubleThe bend is the reciprocal of signed radius: a negative radius means the outside and inside of the circle are switched. The bends of any four mutually tangent circles satisfy Descartes' Theorem. cb :: Complex DoubleProduct of bend and center represented as a complex number. Amazingly, these products also satisfy the equation of Descartes' Theorem.

Instances

 Eq Circle Floating Circle The `Num`, `Fractional`, and `Floating` instances for `Circle` (all simply lifted elementwise over `Circle`'s fields) let us use Descartes' Theorem directly on circles. Fractional Circle Num Circle Show Circle

Arguments

 :: Double signed radius -> P2 center -> Circle

Create a `Circle` given a signed radius and a location for its center.

Get the center of a circle.

Get the (unsigned) radius of a circle.

# Descartes' Theorem

descartes :: Floating a => [a] -> [a]Source

Descartes' Theorem states that if `b1`, `b2`, `b3` and `b4` are the bends of four mutually tangent circles, then

```     b1^2 + b2^2 + b3^2 + b4^2 = 1/2 * (b1 + b2 + b3 + b4)^2.
```

Surprisingly, if we replace each of the `bi` with the product of `bi` and the center of the corresponding circle (represented as a complex number), the equation continues to hold! (See the paper referenced at the top of the module.)

`descartes [b1,b2,b3]` solves for `b4`, returning both solutions. Notably, `descartes` works for any instance of `Floating`, which includes both `Double` (for bends), `Complex Double` (for bend/center product), and `Circle` (for both at once).

other :: Num a => [a] -> a -> aSource

If we have four mutually tangent circles we can choose one of them to replace; the remaining three determine exactly one other circle which is mutually tangent. However, in this situation there is no need to apply `descartes` again, since the two solutions `b4` and `b4'` satisfy

```     b4 + b4' = 2 * (b1 + b2 + b3)
```

Hence, to replace `b4` with its dual, we need only sum the other three, multiply by two, and subtract `b4`. Again, this works for bends as well as bend/center products.

initialConfig :: Double -> Double -> Double -> [Circle]Source

Generate an initial configuration of four mutually tangent circles, given just the signed bends of three of them.

apollonian :: Double -> [Circle] -> [Circle]Source

Given a threshold radius and a list of four mutually tangent circles, generate the Apollonian gasket containing those circles. Stop the recursion when encountering a circle with an (unsigned) radius smaller than the threshold.

# Diagram generation

drawCircle :: Renderable (Path R2) b => Double -> Circle -> Diagram b R2Source

Draw a circle.

drawGasket :: Renderable (Path R2) b => [Circle] -> Diagram b R2Source

Draw a generated gasket, using a line width 0.003 times the radius of the largest circle.

apollonianGasket :: Renderable (Path R2) b => Double -> Double -> Double -> Double -> Diagram b R2Source

Draw an Apollonian gasket: the first argument is the threshold; the recursion will stop upon reaching circles with radii less than it. The next three arguments are bends of three circles.