/code/txt/ Apollonian Gaskets in Python
by 5xa
I implemented this in python as a command line program that saves those as svg images. Below I will explain the math behind it and show some images.
1 How it works
The process of generating an Apollonian gasket roughly works like this:
We start with a triple of circles, each of which touches the other two from the outside. Now we try to find a fourth circle that is also tangent to each of the three. It’s easy to see that there are two possibilities for this: Either it lies in the middle of the three given circles (externally tangent) or it encloses them (internally tangent).
Left: Three circles with different radii. Each is tangent to the other two.
Right: Two possibilities for a fourth circle that is tangent to the first three. Externally tangent (pink) or internally tangent (green).
For now we want take the enclosing one.
There are two equations relating the curvatures and centers of four mutually tangent circles. That means, knowing our three circles from the start, we just plug in their values and get our enclosing circle.
Now comes the clever bit: The equations are quadratic polynomials, so we can use Vieta’s formulas on them. This is very useful, because quadratic polynomials have two solutions (in this case: the inner and outer circle). With Vieta’s formulas it is trivially easy to get the second solution if we already know the first.
Applied to our four circles that means:
 Considering our three circles form the start as “fixed” and the outer one as the first solution, Vieta gives us the small circle inside the three.
 Considering one of the starting circles as the first solution and keeping the rest fixed, Vieta gives another new circle. Because we started out with three circles, there are three candidates to plug into Vieta as the first solution, so we also get three new circles.
The grey circles are fixed, the green circle is the first solution we already know about.
Pink is the second solution given by Vieta.
In total we have added four new circles.
This can be continued with our shiny new circles now added to the gasket: Find four mutually tangent ones, in turn consider each of them as the first of Vieta’s solutions and get the second solution. Rinse and repeat.
When we think the gasket is full enough, we fill in some nice colors and save it to a file.
Now that you have an idea what this is about, you might want to read the rest of the article which details all the maths and programming aspects (which are not that bad, actually) or you may rather skip to the end of this article and try the Apollonian Gasket Generator for yourself: Try it!.
2 The Math
The math stuff can be broken into some smaller individual problems.
2.1 Why it works: Vieta revisited
If you are rusty on Vieta’s formulas, here is a quick recap:
Consider the quadratic polynomial f(x)=x2+px+q with the solutions x1 and x2 of f(x)=0. Vieta’s formulas state that
x1+x2=−p(1)
x1⋅x2=q(2)
If we happen to know x1 already, x2 can be trivially computed as x2=−p−x1.
Note that we do not use x2=q/x1 as x1 might be 0 and the whole thing could explode in our faces with a “division by zero” error.
2.2 Representing Circles
There are (among others) two ways to represent a circle:
 Center and radius
 Center and curvature
The center is straightforward and can be thought of as a point in the 2Dplane or a complex number.
The radius as half the diameter is not even remotely difficult, only thing you might not already have heard is the curvature. Curvature is simply the reciprocal of the radius.
A circle with infinite radius has the curvature 0 and is, in fact, a straight line. On the other hand, infinite curvature means a radius of zero — a point.
For us, it will be convenient to think of a circle as a complex number for the center and a curvature.
2.3 Descartes and his Disciples
Although the problem of the tangent circles was first described (and solved) by Apollonius of Perga (about 200 BC), doing this by hand with a compass and a straightedge is quite tedious still today.
The first step to an algorithmic solution was made by Descartes in 1643. He proved that the curvatures b1,b2,b3,b4 of four mutually tangent circles satisfy
b21+b22+b23+b24=12(b1+b2+b3+b4)2(3)
This is known as Descartes’ theorem.
Note that for this to work an enclosing (internally tangent) circle has to have a negative curvature.
Let’s pretend we already know three circles’ curvatures b1,b2,b3 and need to calculate b4. For this solve for b4, and after some rearranging of ugly long lines we get
b4=b1+b2+b3±2b1b2+b2b3+b3b1‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√(4)
But this only gives us the curvatures. Luckily, in 2001 Wilks et al discovered a very similar equation for the centers of Apollonian circles.
Let b1,b2,b3,b4 be the curvatures and z1,z2,z3,z4 the centers (as complex numbers) of four mutually tangent circles. They satisfy
(b1z1)2+(b2z2)2+(b3z3)2+(b4z4)2=12(b1z1+b2z2+b3z3+b4z4)2(5)
Like Descartes’ theorem, this can be solved for b4z4, and with b4 from equation (4) it is a simple division to compute z4.
b4z4=b1z1+b2z2+b3z3±2b1b2z1z2+b2b3z2z3+b3b1z3z1‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√(6)
This does not look really elegant (square roots tend to evaluate to ugly irrational numbers, not to speak of square roots of complex numbers), but these equations are only needed one time: When we produce an enclosing circle for the three starting circles.
2.4 Vieta to the rescue!
Now that we have our four mutually tangent circles we can begin using one of Vieta’s formulas (1) on them.
Let’s first look at the curvatures:
Suppose we know values for bi with i=1…4 and are looking for the second solution for b1 that satisfies Descartes’ theorem. We can rewrite it in a form more resembling the quadratic polynomial from (1).
First note that Descartes’ therorem (3) can be written as 2(b21+b22+b23+b24)=(b1+b2+b3+b4)2, i.e. multiplied by two. From there we have:
2(b21+b22+b23+b24)===(b1+b2+b3+b4)2b21+b22+b23+b24+2b1b2+2b1b3+2b1b4+2b2b3+2b2b4+2b3b4(b21+b22+b23+b24)+2b1(b2+b3+b4)+2(b2b3+b2b4+b3b4)
Rearranging stuff to the left side we get
b21+b22+b23+b24−2b1(b2+b3+b4)−2(b2b3+b2b4+b3b4)b21+(−2(b2+b3+b4))pb1+(−2(b2b3+b2b4+b3b4)+b22+b23+b24)q==00
Hooray, we’re nearly there! Now we have something looking like a quadratic polynomial that we can feed to Vieta. Let b1,old,b1,new be the solutions to b21+pb1+q=0. Then we see from (1) that b1,old+b1,new=−p.
b1,new=−p−b1,old=2(b2+b3+b4)−b1,old(7)
And that’s it. With this equation, we can calculate the alternative fourth circle’s curvature from the known four.
With the other equation, for the center points, it works the same, just replace bi with bizi. In the end we get
z1,new=2(b2z2+b3z3+b4z4)−b1,oldz1,oldb1,old(8)
2.5 How to find starting triples
Now that we know how to proceed from four known circles to new ones we’re only missing a method to come up with these three in the first place.
Finding appropriate circles is not that difficult either. Our goal is a method that, when given three curvatures b1,b2,b3, returns the curvature of a fourth, enclosing, circle and the centers of all of them.

We will do geometry with the radii of the circles, so let ri=1/bi.

The first circle is easy. We stick it in the origin at (0,0).

No worries about the second circle either, just place it to the right of the first at (r1+r2,0).

The third one is a bit more tricky. We want to place it below the first and second, but how do we get the coordinates of (mx,my)?
To solve this, we take a look just at the triangle defined by the three radii.
We can say that (mx,my)=(x,y) in the sketch above. We can further say that by Pythagoras’ theorem the following must be the case:
y2+x2y2+(r1+r2−x)2==(r1+r3)2(r2+r3)2
Solving for x and y we get:
x=r21+r1r3+r1r2−r2r3r1+r2y=(r1+r3)2−x2‾‾‾‾‾‾‾‾‾‾‾‾‾‾√(9)(10)

The fourth circle is easy again, we just feed everything about circles #1 to #3 into our equations (4) and (6) from earlier and get our last circle. Note that in place of the ± we take a −, because we want an enclosing circle, which has a negative curvature.
2.6 When it breaks
Note that it is not possible to generate Apollonian gaskets from all starting curvatures, for example, when the enclosing circle would need to have zero curvature to be tangent to the three.
One combination with this property is b1=1,b2=1,b3=4, or more generally any combination that satisfies b3=2b1⋅b2‾‾‾‾‾‾√+b1+b2. In these cases equation (4) gives b4=0.
3 The Code
You can find the code on github.
First of all, we need to store the information about the circles. Unsurprisingly, the Circle
class takes care of this.
1 class Circle(object): 2 """ 3 A circle represented by center point as complex number and radius. 4 """ 5 def __init__ ( self, mx, my, r ): 6 """ 7 @param mx: x center coordinate 8 @param my: y center coordinate 9 @param r: radius 10 """ 11 self.r = r 12 self.m = (mx +my*1j) 13 14 def curvature (self): 15 return 1/self.r
Next, we want to get the enclosing circle for three mutually tangent circles. This is a straightforward implementation of equation (4) and (6).
1 def outerTangentCircle( circle1, circle2, circle3 ): 2 """ 3 Takes three externally tangent circles and calculates the fourth one enclosing them. 4 """ 5 cur1 = circle1.curvature() 6 cur2 = circle2.curvature() 7 cur3 = circle3.curvature() 8 m1 = circle1.m 9 m2 = circle2.m 10 m3 = circle3.m 11 cur4 = 2 * sqrt( cur1*cur2 + cur2*cur3 + cur1 * cur3 ) + cur1 + cur2 + cur3 12 m4 = ( 2 * sqrt( cur1*m1*cur2*m2 + cur2*m2*cur3*m3 + cur1*m1*cur3*m3 ) + cur1*m1 + cur2*m2 + cur3*m3 ) / cur4 13 circle4 = Circle( m4.real, m4.imag, 1/cur4 ) 14 15 return circle4
We can directly use this if we comfortably want to generate the first four circles from three given radii. And while we are at it, we also implement equations (9) and (10) for the center of circle #3/4 (slightly other numbering in the code than in the equations).
1 def tangentCirclesFromRadii( r2, r3, r4 ): 2 """ 3 Takes three radii and calculates the corresponding externally 4 tangent circles as well as a fourth one enclosing them. The 5 enclosing circle is the first one. 6 """ 7 8 circle2 = Circle( 0, 0, r2 ) 9 circle3 = Circle( r2 + r3, 0, r3 ) 10 m4x = (r2*r2 + r2*r4 + r2*r3  r3*r4) / (r2 + r3) 11 m4y = sqrt( (r2 + r4) * (r2 + r4)  m4x*m4x ) 12 circle4 = Circle( m4x, m4y, r4 ) 13 circle1 = outerTangentCircle( circle2, circle3, circle4 ) 14 return ( circle1, circle2, circle3, circle4 )
The more interesting parts are next: Using Vieta to jump from circles we already know to new solutions:
1 def secondSolution( fixed, c1, c2, c3 ): 2 """ 3 If given four tangent circles, calculate the other one that is tangent 4 to the last three. 5 6 @param fixed: The fixed circle touches the other three, but not 7 the one to be calculated. 8 9 @param c1, c2, c3: Three circles to which the other tangent circle 10 is to be calculated. 11 """ 12 13 curf = fixed.curvature() 14 cur1 = c1.curvature() 15 cur2 = c2.curvature() 16 cur3 = c3.curvature() 17 18 curn = 2 * (cur1 + cur2 + cur3)  curf 19 mn = (2 * (cur1*c1.m + cur2*c2.m + cur3*c3.m)  curf*fixed.m ) / curn 20 return Circle( mn.real, mn.imag, 1/curn )
The ApollonianGasket
class holds the starting circles and a list of all generated circles.
1 class ApollonianGasket(object): 2 """ 3 Container for an Apollonian Gasket. 4 """ 5 def __init__(self, c1, c2, c3): 6 self.start = tangentCirclesFromRadii( 1/c1, 1/c2, 1/c3 ) 7 self.genCircles = list(self.start)
And finally the most interesting part: Recursively generating the new circles from the starting set.
1 def recurse(self, circles, depth, maxDepth): 2 """Recursively calculate the smaller circles of the AG up to the 3 given depth. Note that for depth n we get 2*3^{n+1} circles. 4 5 @param maxDepth: Maximal depth of the recursion. 6 @type maxDepth: int 7 8 @param circles: 4Tuple of circles for which the second 9 solutions are calculated 10 @type circles: (L{Circle}, L{Circle}, L{Circle}, L{Circle}) 11 12 @param depth: Current depth 13 @type depth: int 14 """ 15 if( depth == maxDepth ): 16 return 17 (c1, c2, c3, c4) = circles 18 if( depth == 0 ): 19 # First recursive step, this is the only time we need to 20 # calculate 4 new circles. 21 del self.genCircles[4:] 22 cspecial = secondSolution( c1, c2, c3, c4 ) 23 self.genCircles.append( cspecial ) 24 self.recurse( (cspecial, c2, c3, c4), 1, maxDepth ) 25 26 cn2 = secondSolution( c2, c1, c3, c4 ) 27 self.genCircles.append( cn2 ) 28 cn3 = secondSolution( c3, c1, c2, c4 ) 29 self.genCircles.append( cn3 ) 30 cn4 = secondSolution( c4, c1, c2, c3 ) 31 self.genCircles.append( cn4 ) 32 33 self.recurse( (cn2, c1, c3, c4), depth+1, maxDepth ) 34 self.recurse( (cn3, c1, c2, c4), depth+1, maxDepth ) 35 self.recurse( (cn4, c1, c2, c3), depth+1, maxDepth )
For each circle we construct the second solution by Vieta’s formula and recurse with the original circle replaced by the new one. By sticking it in the first position of the circles
quadruple and not recursing for the first circle (except in the first call of the function) we avoid that the second solution of the second solution is calculated, as this would again be the “grandparent” circle.
The remainder of the code just wraps all of this stuff in a command line program and adds some helper functions for svg output (ugly) and coloring.
4 Try it!
In addition to the command line program I wrote a small online version. You can play around with different starting curvatures, recursion depths and colours.
You can find it at http://lsandig.org/cgibin/apollon/index.cgi
source
link http://lsandig.org/blog/2014/08/apollonpython/en/
post https://www.facebook.com/groups/ALGOinDSGN/permalink/1238389939511828/