Hi fellows,
Does anyone know the number of holes of a level 2 Menger Sponge ?
3 Answers
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For $n=1$, $g=5$: you drill a vertical hole through the middle and four horizontal holes to meet that vertical hole.
For higher values of $n$, the right way to think of it is in terms of Euler characteristic.
For $n=2$ you start with $20$ copies of a small level $1$ Menger sponge, with Euler characteristic $-8$. You then glue them together a bunch of times on an annulus. The annulus has Euler characteristic $0$, so the final Euler characteristic is $-160$, giving a genus of $81$.
The side of a level $n$ Menger sponge has Euler characteristic $(8-8^n)/7$. In general you glue along $24$ of these, so the recurrence relation for the Euler characteristic is $\chi_{n+1}=20\chi_n + 24 (8^n-8)/7$.
This recurrence relation can produce an explicit formula, but it seems rather awful.
answered May 21, 2012 at 18:34
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What does "hole" mean? All the empty spaces are connected to each other, so probably you should say there is one hole.
answered May 21, 2012 at 15:17
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$\begingroup$ by "holes", I meant topological genus. $\endgroup$– user23855Commented May 21, 2012 at 15:26
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I think that the answer is supposed to be 2^n, where n is the level of the sponge. Not quite so sure though, as I was told so in class very long ago, and I misplaced the paper...
