Barnsley Fern is a fractal set that looks like a shape of a fern. If you plot this and magnify any one part (e.g, the tip of a leaf) you will see the similar shape of the whole leafs wherever you magnify.

The Barnsley Fern is a well-known example of a fractal, a complex geometric shape that exhibits self-similarity. This means that the fern's structure looks similar at various levels of magnification. It closely resembles the shape of an actual fern plant, with its intricate and repetitive patterns.

The Barnsley Fern is generated using a set of mathematical equations, specifically an iterative process known as an iterated function system (IFS). By repeatedly applying these functions to points on a plane, the image of the fern gradually emerges. Each transformation contributes a specific part to the overall structure, such as the stem, the large fronds, and the smaller leaflets.

One of the fascinating properties of the Barnsley Fern is its self-similarity. If you zoom into a specific part of the fern, for instance, the tip of a leaf, you will find that this small section has the same general shape as the entire fern. This recursive pattern repeats infinitely, making the fern a classic example of fractal geometry.

The Barnsley Fern not only demonstrates the beauty of mathematical constructs but also has practical applications in fields such as computer graphics, where fractals are used to generate natural-looking structures and landscapes efficiently.

The procedure to plot this as follows.

i) define four sets of 2D transformation equation as follows. (p stands for the coordinate for the current point, m is a 2x2 matrix that is specially created for this transformation, v is 2x1 vector)

tf1 = m1 . p + v1

tf2 = m2 . p + v2

tf3 = m2 . p + v3

tf4 = m2 . p + v4

ii) pick any arbitray points (let's say this point is named as p)

iii) pick one transformation equation out of the 4 equation defined in step i). The probability of the random selection for each transformation equation should be predefined).

iv) transform p with the transformation equation selected at step iii).

v) move the current point to the transformed location and set the new position to the current point (p).

vi) repeat the process iii) ~ v) and plot the point p on the coordinate at each iterations.

Following is the Octave/Matlab code that I wrote as per the procedure described above.

m1 = [0.85 0.04;-0.04 0.85]; v1 = [0.0;1.60];

m2 = [0.2 -0.26;0.23 0.22]; v2 = [0.0;1.60];

m3 = [-0.15 0.28;0.26 0.24]; v3 = [0.0;0.44];

m4 = [0.0 0.0;0.0 0.16]; v4 = [0.0;0.0];

 

prob = [0.85 0.07 0.07 0.01];

 

Np = 50000;

 

p_old = [0.5;0.5];

 

pList = [p_old'];

 

for i = 1:Np

    r = rand();

    if (r < prob(1))

       p_new = m1 * p_old + v1;

    elseif (r < (prob(1)+prob(2)))

       p_new = m2 * p_old + v2;

    elseif (r < (prob(1)+prob(2)+prob(3)))

       p_new = m3 * p_old + v3;

    else

       p_new = m4 * p_old + v4;

    endif;

    pList = [pList;p_new'];

    p_old = p_new;

end;

Note : the syntax of if statement of Octave may be a little different from Matlab. I tried this only on Octave, so not sure if this code would work in matlab.