Hausdorff Dimension for Koch-like Fractal

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I'm trying to define is a simple fractal: detach the middle third of a horizontal line, push it upwards by one third the length of the original segment and then add in vertical lines to keep it all connected. I'd be surprised if this fractal doesn't already exist but wouldn't know what it's called so can't find it via a search. Let's try to define the fractal formally.

A line segment in $\mathbb{R}^2$ can be denoted by: $[(x_1, y_1), (x_2, y_2)]$. Let's abbreviate horizontal and vertical line segments as follows:

  • $H_y[a, b] = [(a, y), (b, y)]$
  • $V_x[a, b] = [(x, a), (x, b)]$

Define the following sets:

  • $H = \{H_y[a, b] | a, b, y \in \mathbb{R}\}$
  • $V = \{V_x[a, b] | a, b, x \in \mathbb{R}\}$
  • $S = H \cup V$

Define the following function $f : S \to \mathcal{P}(S)$ as:

$$f(H_y[a, a + d]) = \{(H_y[a, a + \frac13d]), H_{y + \frac13d}[a + \frac13d, a + \frac23d]), H_{y}[a + \frac23d, a + d]), V_{a + \frac13d}[y, y + \frac13d], V_{a + \frac23d}[y, y + \frac13d]\}$$

and

$$f(V_x[a, b]) = \{V_x[a, b]\}$$

Define $F : \mathcal{P}(S) \to \mathcal{P}(S)$ as:

$F(U) = \bigcup f(U)$

Finally then we can define a fractal $K$ as the limit of the iteration $F$ on the line segment $H_0[0, 1]$. I'm not sure exactly how to do that formally (maybe a least fixed point?).

Phew... I Hope I got all that formalism right. My question is: Does the Hausdorff dimension of this thing exist? And if so, what is it?

My Attempt at Calculating the Hausdorff dimension

Let $d$ be the Hausdorff dimension. I noticed that the vertical line segments make an argument based on self-similarity difficult, because no further iterations happen on those. So I ended up reasoning like this. If we start with a line segment of length $3l$, we end up with $3$ copies of the fractal of length $l$, plus an extra two vertical line segments of length $l$. So if there is a Hausdorff dimension, I figured it would satisfy, for any $l$:

$$3^dl^d - 3l^d = 2l$$

So for $l = 1$, we have:

$$d = \frac{\ln5}{\ln3}$$

But that was just a specific $l$. The Hausdorff dimension, if it exists, should be independent of $l$ I assume. And the equation becomes hard to solve when $l \neq 1$. So I'm not really satisfied with the above.

Infinite Length

It's easy to show that the sum of the length of all the line segments in this thing diverges. At each iteration, it can be shown that we add a constant length to the length of the previous iteration, namely $\frac23$. So my intuition is telling me that it has some sort of "dimension" that's greater than $1$.

Image of First Few Iterations

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