diagrams-contrib- Collection of user contributions to diagrams EDSL

Safe HaskellNone




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.



data Circle Source

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




bend :: Double

The 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 Double

Product of bend and center represented as a complex number. Amazingly, these products also satisfy the equation of Descartes' Theorem.


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 



:: Double

signed radius

-> P2


-> Circle 

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

center :: Circle -> P2Source

Get the center of a circle.

radius :: Circle -> DoubleSource

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 gasket generation

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.