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Let A,B,C,D be four consecutive vertices of a regular hexagon. The curve is based loosely on the line segment AD. It sits entirely within the trapezium ABCD.
Start with the line AD. Replace it with three lines, half its length; namely AB, BC and CD.
The middle line, BC, remains a straight line segment, and forms part of the final fractal.
Replace AB with three lines, each half the length of AB, so they form a trapezium, half the dimensions of ABCD. Replace CD with three lines, each half the length of CD, in the same way.
At each stage, the middle third of each group of three remains a straight line segment, and the outer two get replaced.
On a smaller length-scale, there are two copies of the fractal, based on AB and CD, connected by the line segment BC. The copies are each half the size of the original.
So $L(d/2) = 2\times (L(d)/2) + d/2 = L(d) + d/2$. On a smaller length-scale, it is $L(d/4) = 4\times(L(d)/4) + d/2 + 2(d/4)$. In general, $$L(d/2^n)=L(d)+nd/2$$ The curve is infinitely long, but increases with the logarithm of the length-scale, rather than with a power of the length-scale. Is its dimension one, or something else?
In the image, the red lines get replaced by the following recursive step, and the blue lines are the straight line-segments that join fractal pieces together. enter image description here

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    $\begingroup$ Could you describe the recursive construction in a clear way, and before arguing about lengths? $\endgroup$ – Christian Blatter Dec 10 '17 at 10:10
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Your argument is correct. Because of the logarithmic (in the covering ball radius) growth of the "effective" length of the curve (i. e., the number of balls times their radius), the dimension is 1.

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