# Sierpinski triangle

(Redirected from Sierpiński Triangle)
Sierpinski triangle
Generated using a random algorithm
Sierpinski triangle in logic: The first 16 conjunctions of lexicographically ordered arguments
The columns interpreted as binary numbers give 1, 3, 5, 15, 17, 51... (sequence A001317 in OEIS)

The Sierpinski triangle (also with the original orthography Sierpiński), also called the Sierpinski gasket or the Sierpinski Sieve, is a fractal and attractive fixed set named after the Polish mathematician Wacław Sierpiński who described it in 1915. However, similar patterns appear already in the 13th-century Cosmati mosaics in the cathedral of Anagni, Italy,[1] and other places, such as in the nave of the Roman Basilica of Santa Maria in Cosmedin.[2]

Originally constructed as a curve, this is one of the basic examples of self-similar sets, i.e. it is a mathematically generated pattern that can be reproducible at any magnification or reduction.

Comparing the Sierpinski triangle or the Sierpinski carpet to equivalent repetitive tiling arrangements, it is evident that similar structures can be built into any rep-tile arrangements.

## Construction

An algorithm for obtaining arbitrarily close approximations to the Sierpinski triangle is as follows:

Note: each removed triangle (a trema) is topologically an open set.[3]

1. Start with any triangle in a plane (any closed, bounded region in the plane will actually work). The canonical Sierpinski triangle uses an equilateral triangle with a base parallel to the horizontal axis (first image).
2. Shrink the triangle to ½ height and ½ width, make three copies, and position the three shrunken triangles so that each triangle touches the two other triangles at a corner (image 2). Note the emergence of the central hole - because the three shrunken triangles can between them cover only 3/4 of the area of the original. (Holes are an important feature of Sierpinski's triangle.)
3. Repeat step 2 with each of the smaller triangles (image 3 and so on).

This process of recursively removing triangles is an example of a finite subdivision rule.

Note that this infinite process is not dependent upon the starting shape being a triangle—it is just clearer that way. The first few steps starting, for example, from a square also tend towards a Sierpinski triangle. Michael Barnsley used an image of a fish to illustrate this in his paper "V-variable fractals and superfractals."[4]

The actual fractal is what would be obtained after an infinite number of iterations. More formally, one describes it in terms of functions on closed sets of points. If we let $d_a$ note the dilation by a factor of ½ about a point a, then the Sierpinski triangle with corners a, b, and c is the fixed set of the transformation $d_a$ U $d_b$ U $d_c$.

This is an attractive fixed set, so that when the operation is applied to any other set repeatedly, the images converge on the Sierpinski triangle. This is what is happening with the triangle above, but any other set would suffice.

If one takes a point and applies each of the transformations $d_a$, $d_b$, and $d_c$ to it randomly, the resulting points will be dense in the Sierpinski triangle, so the following algorithm will again generate arbitrarily close approximations to it:

Start by labeling p1, p2 and p3 as the corners of the Sierpinski triangle, and a random point v1. Set vn+1 = ½ ( vn + prn ), where rn is a random number 1, 2 or 3. Draw the points v1 to v. If the first point v1 was a point on the Sierpiński triangle, then all the points vn lie on the Sierpinski triangle. If the first point v1 to lie within the perimeter of the triangle is not a point on the Sierpinski triangle, none of the points vn will lie on the Sierpinski triangle, however they will converge on the triangle. If v1 is outside the triangle, the only way vn will land on the actual triangle, is if vn is on what would be part of the triangle, if the triangle was infinitely large.

Animated creation of a Sierpinski triangle using the chaos game
Animated construction of a Sierpinski triangle

Or more simply:

1. Take 3 points in a plane to form a triangle, you need not draw it.
2. Randomly select any point inside the triangle and consider that your current position.
3. Randomly select any one of the 3 vertex points.
4. Move half the distance from your current position to the selected vertex.
5. Plot the current position.
6. Repeat from step 3.

Note: This method is also called the Chaos game. You can start from any point outside or inside the triangle, and it would eventually form the Sierpinski Gasket with a few leftover points. It is interesting to do this with pencil and paper. A brief outline is formed after placing approximately one hundred points, and detail begins to appear after a few hundred.

Sierpinski triangle using IFS

Or using an Iterated function system

An alternative way of computing the Sierpinski triangle uses an Iterated function system and starts by a point at the origin (x0 = 0, y0 = 0). The new points are iteratively computed by randomly applying (with equal probability) one of the following three coordinate transformations (using the so-called chaos game):
xn+1 = 0.5 xn
yn+1 = 0.5 yn; a half-size copy
This coordinate transformation is drawn in yellow in the figure.

xn+1 = 0.5 xn + 0.25
yn+1 = 0.5 yn + 0.5 $\sqrt{3}\over 2$; a half-size copy shifted right and up
This coordinate transformation is drawn using red color in the figure.

xn+1 = 0.5 xn + 0.5
yn+1 = 0.5 yn; a half-size copy doubled shifted to the right

This coordinate transformation is drawn using blue color in the figure.

Or using an L-system — The Sierpinski triangle drawn using an L-system.

bitwise AND - The 2D AND function, z=AND(x,y) can also produce a white on black right angled Sierpinski triangle if all pixels of which z=0 are white, and all other values of z are black.

bitwise XOR - The values of the discrete, 2D XOR function, z=XOR(x,y) also exhibit structures related to the Sierpinski triangle. For example, one could generate the Sierpinski triangle by setting up a 2 dimensional matrix, [rows][columns] placing the uppermost point on [1][n/2], then cycling through the remaining cells row by row the value of the cell being XOR([i-1][j-1],[i-1][j+1])

Other means — The Sierpinski triangle also appears in certain cellular automata (such as Rule 90), including those relating to Conway's Game of Life. The automaton "12/1" when applied to a single cell will generate four approximations of the Sierpinski triangle.

If one takes Pascal's triangle with 2n rows and colors the even numbers white, and the odd numbers black, the result is an approximation to the Sierpinski triangle. More precisely, the limit as n approaches infinity of this parity-colored 2n-row Pascal triangle is the Sierpinski triangle.

## Properties

For integer number of dimensions d, when doubling a side of an object, 2 d copies of it are created, i.e. 2 copies for 1 dimensional object, 4 copies for 2 dimensional object and 8 copies for 3 dimensional object. For Sierpinski triangle doubling its side creates 3 copies of itself. Thus Sierpinski triangle has Hausdorff dimension log(3)/log(2) ≈ 1.585, which follows from solving 2 d = 3 for d.[5]

The area of a Sierpinski triangle is zero (in Lebesgue measure). The area remaining after each iteration is clearly 3/4 of the area from the previous iteration, and an infinite number of iterations results in zero.[citation needed]

The points of a Sierpinski triangle have a simple characterization in Barycentric coordinates.[6] If a point has coordinates (0.u1u2u3…,0.v1v2v3…,0.w1w2w3…), expressed as Binary numbers, then the point is in Sierpinski's triangle if and only if ui+vi+wi=1 for all i.

## Analogues in higher dimensions

A Sierpinski square-based pyramid and its 'inverse'
A Sierpiński triangle-based pyramid as seen from above (4 main sections highlighted). Note the self-similarity in this 2-dimensional projected view, so that the resulting triangle could be a 2D fractal in itself.

The tetrix is the three-dimensional analogue of the Sierpinski triangle, formed by repeatedly shrinking a regular tetrahedron to one half its original height, putting together four copies of this tetrahedron with corners touching, and then repeating the process. This can also be done with a square pyramid and five copies instead. A tetrix constructed from an initial tetrahedron of side-length L has the property that the total surface area remains constant with each iteration.

The initial surface area of the (iteration-0) tetrahedron of side-length L is $L^2 \sqrt{3}$. At the next iteration, the side-length is halved

$L \rightarrow { L \over 2 }$

and there are 4 such smaller tetrahedra. Therefore, the total surface area after the first iteration is:

$4 \left( \left( {L \over 2} \right)^2 \sqrt{3} \right) = 4 { {L^2} \over 4 } \sqrt{3} = L^2 \sqrt{3}.$

This remains the case after each iteration. Though the surface area of each subsequent tetrahedron is 1/4 that of the tetrahedron in the previous iteration, there are 4 times as many—thus maintaining a constant total surface area.

The total enclosed volume, however, is geometrically decreasing (factor of 0.5) with each iteration and asymptotically approaches 0 as the number of iterations increases. In fact, it can be shown that, while having fixed area, it has no 3-dimensional character. The Hausdorff dimension of such a construction is $\textstyle\frac{\ln 4}{\ln 2}=2$ which agrees with the finite area of the figure. (A Hausdorff dimension strictly between 2 and 3 would indicate 0 volume and infinite area.)