Average triangle area in the Cantor dust

We have,

\begin{align} \int_0^1\int_0^1 |x-y|dF(y)dF(x) & = 2 \int_0^1\int_0^x (x - y)dF(y)dF(x), \end{align}

where \(F\) denotes distribution function of \(\mu\), \(F(x) := \mu((-\infty, x])\) for \(x\in\mathbb{R}\). \(F\) is also the Cantor-Lebesgue function of \(\mathcal{C}\) (see §1.5 of [3] for more) and in probabilistic terms, it is the cumulative distribution function of \(\mu\).

We may now perform integration by parts on the inner Lebesgue-Stieltjes integral (see Theorem 3.36 of [3]), to obtain:

\begin{align} \int_0^x (x-y)dF(y) = \left.(x-y)F(y)\right|_0^x + \int_0^x F(y)dy. \end{align}

The first term vanishes and so we're left with:

\begin{align} \mathbb{E}(|X - Y|) & = 2\int_0^1\int_0^x F(y)dydF(x) \end{align}

One more productive use of integration by parts, this time on the outer integral, yields:

\begin{align} \label{eq:moments} \mathbb{E}(|X - Y|) & = 2\int_0^1 F(y)dy - 2\int_0^1 F(x)^2dx, \end{align}

where we've made use of \(d(f(x)) = f'(x)dx\) to get the second term.

The terms in \eqref{eq:moments} have been computed in many ways [4,5].

One straightforward computation is to note that \(F(x)\) is constant each of the intervals \(I_{i,j}\) from §2, and they exhaust \([0,1]\) (in the Lebesgue measure), so for any exponent \(M>0\),

\begin{align} \int F^M(x)dx& = \sum_{k=0}^\infty \sum_{j=1}^{2^k} |I_{k,j}| F(I_{k,j})^M. \end{align}

We also have

\begin{align} F(I_{k,j}) = 2^{-k-1}(2j - 1) \end{align}

We can utilize the above and compute the two integrals from:

\begin{align} \int F(x)^Mdx & = \frac{1}{3\cdot 2^M}\sum_{k=0}^\infty (3\cdot 2^M)^{-k}\sum_{j=1}^{2^k}(2j-1)^M, \end{align}

to obtain \(\int F(x)dx = 1/2\) and \(\int F(x)^2dx = 3/10\) which shows \eqref{eq:average-distance}.

Reasonably one expects that a recurring formula may help in the calculation of the \(2/5\) result in \eqref{eq:average-distance}. This would offer an alternative method to compute the result. In §2 of [1], the integral is reduced to calculating, for any \(k\), the sum \(\displaystyle\sum_{i < j} d^k_{ij}\) where \(d^k_{ij}\) denotes the distances of the midpoints of the \(i\)-th and \(j\)-th interval of \(\mathcal{C}_k\), and indeed they recurringly connected to the next set of distances \(d^{k+1}_{m,n}\).

Can all the lines be computed? More explicitly, can all degenerate triangles be computed?

If a line \(L\) is given, how many points does \(|L\cap\mathcal{C}^2|\) contain? (it answers how many degenerate triangles on \(L\) exist)

It may be too difficult to calculate all triangle congruence classes. Can I calculate a subclass such as the pinned triangles? See [6] for an estimate for the integer lattice (my \(r=1/2\)) where it is estimated that the number of congruence classes is \(\sim n^2\) inside \([0,\sqrt{n}]^2\).

I need to figure out exactly what the sums \(\pm r \pm r^2 \pm \cdots \pm r^n\) represent.

Some congruence classes are singular in their area, i.e. there are no two different classes \(C_1, C_2\) of triangles with the same area.

Can these be classified? What makes them unique?

I need to figure out how many operations are taken for each computation. For example, how much memory is required? For instance, here is a count of the congruence classes:

depth	ratio	# of congruence classes
1	1/3	4
2	1/3	67
3	1/3	2533
4	1/3	119388

Note that they change for different classes. We note their stability:

depth	stable ratio	# of congruence classes
1	1/2	4
2	1/5	67
3	1/7	2572

Some remarks on the stability:

• The number of congruence classes increases as the ratio gets smaller.
• The number appears to immediately stabilize once two different ratios give the same number.

Why do these numbers stabilize? Do they really stabilize or not?

• Mention the tables and observations appearing in 4.5.