The triangle of Sierpiński , also called by Mandelbrot the cylinder head gasket of Sierpiński , is a Fractale, name of Wacław Sierpiński.
A algorithm to obtain approximations arbitrarily close to the triangle of Sierpiński can be written in the following way:
The fractale is obtained after an infinite number of Itération S. With each stage, the surface of the unit decreases, it is multiplied by 3/4.
; Algorithm 2 With algorithm 1, the number of triangles to be treated multiplies in an exponential way (X 3 with each stage: 1, then 3, then 9, then 27, etc: 3n after the nième stage). One can obtain the same result by an equivalent method (but easier to formalize and calculate): to apply three Homothety S of report/ratio 1/2 and center each top of the figure, and to trace the new figure like union of the three results. In this way, with each stage, there are only three operations to make.
Moreover, one can characterize the triangle of Sierpiński as being the fixed unit left by this transformation, which one can note ha U hb U hc, where hx the Homothétie of report/ratio 1/2 of center X , and where has , B and C is the tops of the initial triangle.
That implies that if one applies the infinite iteration of the operation ha U hb U hc to an unspecified finished unit, and has, B and C three distinct points, one obtains (images “convergent” towards) the triangle of Sierpiński of tops has, B and C. the triangle of Sierpiński is a Attracteur of three homotheties of report/ratio 1/2 centered at the tops.
; Algorithm 3 Another interesting property appears if an unspecified point is considered. This point constitutes a unit to which one can apply the operations ha, hb and hc. Thus, the continuation of the points obtained form a dense whole in the triangle of Sierpiński. One can be even satisfied to apply only one of the operations ha, hb and hc, by chance selected with each stage. Thus the following algorithm will generate approximations arbitrarily close to the triangle of Sierpiński:
One considers a point taken randomly v 1. One poses v n + 1= 1/2 ( v n + p n), where pn is a random point equal to has, B or C. One places the points v 1 until v ∞. If the initial point v 1 is a point of the triangle of Sierpiński, then all the points v n belong to the triangle of Sierpinski. If the first point v 1 is in the perimeter of the triangle and is not a point of the triangle of Sierpiński, then none the points v n belongs to the triangle of Sierpiński, however the continuation of the points v n converges towards a point of the triangle of Sierpiński. Even if v 1 is out of the triangle, then the probability that there exists a point following leaving of which all points are in the triangle has, B, C, is equal to 1, and they all will be then either in the triangle of Sierpiński, or all outwards but while approaching some as arbitrarily as desired.
; Algorithm 4 If one builds it starting from a Triangle of Pascal with 2 N lines and that one colors the even numbers in white and the odd numbers in black, then the result is an approximation of the triangle of Sierpiński.
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