## More fractals: Laakso spaces and slit carpets

In early 2000s Tomi Laakso published two papers which demonstrated that metric spaces could behave in very Euclidean ways without looking Euclidean at all. One of his examples became known as the Laakso graph space, since it is the limit of a sequence of graphs. In fact, the best known version of the construction was introduced by Lang and Plaut, who modified Laakso’s original example to make it more symmetric. The building block is this graph with 6 edges:

Each edge is assigned length 1/4, so that the distance between the leftmost and rightmost points is 1. Next, replace each edge with a copy of the building block. This increases the number of edges by the factor of 6; their length goes down by the factor of 4 (so that the leftmost-rightmost distance is still 1). Repeat.

The resulting metric space is doubling: every ball can be covered by a fixed number (namely, 6) balls of half the size. This is the typical behavior of subsets of Euclidean spaces. Yet, the Laakso space does not admit a bi-Lipschitz embedding into any Euclidean space (in fact, even into any uniformly convex Banach space). It remains the simplest known example of a doubling space without such an embedding. Looking back at the building block, one recognizes the cycle $C_4$ as the source of non-embeddability (a single cycle forces a certain amount of distortion; adding cycles withing cycles ad infinitum forces infinite distortion). The extra edges on the left and right are necessary for the doubling condition.

In some sense, the Laakso space is the antipode of the Cantor set: instead of deleting the middle part repeatedly, we duplicate it. But it’s also possible to construct it with a ‘removal’ process very similar to Cantor’s. Begin with the square and slit it horizontally in the center; let the length of the slit be 1/3 of the sidelength. Then repeat as with the Cantor set, except in addition to cutting left and right of the previous cut, we also do it up and down. Like this:

Our metric space if the square minus the slits, equipped with the path metric: the distance between two points is the infimum of the length of curves connecting them within the space. Thus, the slits seriously affect the metric. This is how the set will look after a few more iterations:

I called this a slit pseudo-carpet because it has nonempty interior, unlike true Sierpinski carpets. To better see the similarity with the Laakso space, multiply the vertical coordinate by $\epsilon$ and let $\epsilon\to 0$ (equivalently, redefine the length of curves as $\int |x'|$ instead of $\int \sqrt{|x'|^2+|y'|^2}$). This collapses all vertical segments remaining in the set, leaving us with a version of the Laakso graph space.

Finally, some Scilab code. The Laakso graph was plotted using the Chaos game, calling the function below with parameters laakso(0.7,50000).

function laakso(angle,steps) s=1/(2+2*cos(angle)); xoffset = [0,1-s,s,s,1/2,1/2]; yoffset = [0,0,0,0,s*sin(angle),-s*sin(angle)]; rotation =[0,0,angle,-angle,-angle,angle]; sc=s*cos(rotation); ss=s*sin(rotation); point = zeros(steps,2); vert = grand(1,steps,'uin',1,6); for j = 2:steps point(j,:) = point(j-1,:)*[sc(vert(j)),ss(vert(j));-ss(vert(j)),sc(vert(j))] + [xoffset(vert(j)), yoffset(vert(j))]; end plot(point(:,1),point(:,2),'linestyle','none','markstyle','.','marksize',1); endfunction

## Playing with iterated function systems

After Colin Carroll posted several fractal experiments with Matlab, I decided to do something of the sort. One difference is that I use Scilab, an open-source alternative to Matlab.

The first experiment: drawing the Sierpinski carpet using the Chaos Game. Namely, given a finite family of strict contractions $f_1,\dots,f_r\colon \mathbb R^2\to \mathbb R^2$ and an initial point $p_0$, plot the sequence $p_{n+1}=f_{j_n}(p_n)$, where $j_n \in \{1,\dots,r\}$ is chosen randomly at each step. To simplify matters, let $f_j$ be the similarity transformation with scaling factor $s\in (0,1)$ and the fixed point $v_j$.

A canonical example is: $v_1,v_2,v_3$ are the vertices of equilateral triangle, $s=1/2$. This produces the fractal known as the Sierpinski gasket. For a different example, set $s=1/3$ and let $v_1,\dots,v_8$ be the vertices of square together with midpoints of its sides. The resulting fractal is known as the Sierpinski carpet.

This image was obtained by calling the scilab function given below as Scarpet(1/3, 100000). The function is essentially a translation of Colin’s code to scilab. Caution: if you copy and paste this code, watch out for line breaks and encoding of quote marks.

function Scarpet(scale,steps)
b=1-scale;
x = [1,0,-1,-1,-1,0,1,1];
y = [1,1,1,0,-1,-1,-1,0];
sides=length(x);
point = zeros(steps,2);
vert = grand(1,steps,'uin',1,sides);
for j = 2:steps
point(j,:) = scale*point(j-1,:) + b*[x(vert(j)),y(vert(j))];
end
plot(point(:,1),point(:,2),'linestyle','none','markstyle','.','marksize',1);
endfunction

Regardless of the choice of initial point $p_0$, the set of cluster points of the sequence $(p_n)$ is exactly the invariant set $K$, namely the unique nonempty compact set such that $K=\bigcup_{j=1}^r f_j(K)$. This is proved, for example, in the book Integral, Probability, and Fractal Measures by Gerald Edgar.

The scaling factor $s=1/3$ for the carpet is chosen so that the images of the original square under the eight similarities touch, but do not overlap. With a smaller factor the fractal looks like dust (a totally disconnected set), while with $s\ge 1/2$ it becomes a solid square. The intermediate range $1/3 is tricky: I think that $K$ has measure zero, but can’t even prove that it’s nowhere dense.

It’s also possible to draw $K$ in the opposite way, by removing points rather than adding them. To this end, let $P$ be the convex hull of the set $\{v_1,\dots,v_r\}$; that is, a solid convex polygon. It’s not hard to see that $K\subset P$. Therefore, $\bigcup_{j=1}^r f_j(K)\subset \bigcup_{j=1}^r f_j(P)$, but since the set on the left is $K$ itself, we get $K\subset \bigcup_{j=1}^r f_j(P)$. By induction, $K=\bigcap_{n=1}^{\infty} P_n$ where $P_0=P$ and $P_{n+1}=\bigcup_{j=1}^r f_j(P_n)$.

The above example is ifs(3,3/5,11), calling the Scilab code below.

function ifs(sides,scale,steps)
b=1-scale; t=2*%pi*(1:sides)/sides; x=cos(t); y=sin(t);
xpols=x'; ypols=y';
for j=2:steps
xpols=scale*xpols; ypols=scale*ypols;
xpolsnew=[]; ypolsnew=[];
for k=1:sides
xpolsnew=[xpolsnew xpols+b*x(k)*ones(xpols)];
ypolsnew=[ypolsnew ypols+b*y(k)*ones(ypols)];
end
xpols=xpolsnew; ypols=ypolsnew;
end
a=gca(); a.data_bounds=[-1,-1;1,1];
[m,n]=size(xpols);
xfpolys(xpols,ypols,ones(n,1))
endfunction

The final example is an “upper bound” for the fat pentagonal fractal that Colin created with the Chaos Game: the points $v_1,\dots,v_5$ are the vertices of regular pentagon, and $s=1/2$. The function was called as ifs(5,1/2,8). Again, I think that the invariant set has measure zero, but can’t even prove that the interior is empty. (Or find a reference where this is already done.)