# How To Construct a 4D Menger Sponge

Geometry Level 5

The Problem

The volume of a modified 4D Menger Sponge, which is created by successively subtracting volumes from the interior of a 4D tesseract of side length of 2, can be expressed as

$\dfrac{a}{b} {\pi}^{c},$

where $$a,b$$ and $$c$$ are positive integers with $$a,b$$ coprime.

Find the sum $$a+b+c$$.

The method of successively subtracting volumes from the interior shall now be explained

Preliminaries

Given an arbitrary odd number $$2n+1$$, we have the set of digits $$(0,1,2,3,\ldots , 2n)$$ in base $$2n+1$$. Then we construct all possible distinct sets of the form

$$(n,n,a,b)$$

where $$a,b$$ are digits from that first set, and need not be different, i.e.,$$a$$ may be the same as $$b$$. Then for each such distinct set, we list all possible distinct permutations of it. For example, let the odd number be $$5$$, where $$n=2$$, and we have the set of digits $$(0, 1, 2, 3, 4)$$. We then will have the following distinct sets

$$(2,2,0,0)$$
$$(2,2,0,1)$$
$$(2,2,0,2)$$
$$(2,2,0,3)$$
$$(2,2,0,4)$$
$$(2,2,1,1)$$
$$(2,2,1,2)$$
$$(2,2,1,3)$$
$$(2,2,1,4)$$
$$(2,2,2,2)$$
$$(2,2,2,3)$$
$$(2,2,2,4)$$
$$(2,2,3,3)$$
$$(2,2,3,4)$$
$$(2,2,4,4)$$

and then for example, for the set $$(2,2,0,0)$$, we have the following permutations

$$(2,2,0,0)$$
$$(2,0,2,0)$$
$$(2,0,0,2)$$
$$(0,2,0,2)$$
$$(0,0,2,2)$$
$$(0,2,2,0)$$

Thus, we will have a total of $$113$$ such sets with this example

Subtraction of Volumes

Given a tesseract, for example, we can subdivide it into $${(2n+1)}^{4}={5}^{4}=625$$ smaller tesseracts, of which the 4D coordinates of each can be described as $$(a,b,c,d)$$. We subtract the 113 smaller tesseracts as per the list of 4D coordinates. If we sort the list by the first tuple, that is

$$(0,a, b, c)$$
$$(1, a, b, c)$$
$$(2, a, b, c)$$
$$(3, a, b, c)$$
$$(4, a, b, c)$$

where $$a, b, c$$ are arbitrary digits, then the "layers" $$0, 1, 3, 4$$ will look like this, the green blocks being subtracted

while "layer" $$2$$ will look like this, the green blocks being subtracted

Because there's already so many occurrences of the digit $$2$$, it shouldn't be hard to see that the "middle layer" of $$2$$ is going to have a lot of green blocks being subtracted from it.

Even though these graphics look like 3D graphics, they are still a valid projection of 4D tesseracts, we can imagine each green block as a tesseract each with its own coordinates.

Method of Successive Subtractions

The classic 3D Menger Sponge is created from a 3D cube by letting $$(2n+1)=3$$ for all recursive steps, and the sets are of the form

$$(1, 1, a)$$

where $$a=(0, 1, 2)$$, always. After the first round, $$\dfrac {7}{27}$$ of the volume of a 3D cube is subtracted. Then for the 2nd round, EACH of the remaining cubes has $$\dfrac {7}{27}$$ of their volumes removed. Repeat for the 3rd round and thereafter. We can see that starting with an intial volume of $${2}^{3}=8$$ we are reducing it by a factor of $$\dfrac {20}{27}$$ successively. Since the infinite product of this fixed fraction $$<1$$ has the limit of $$0$$, the classic 3D Menger Sponge has a volume of $$0$$ as the limit. We are going to do things slightly differently here.

Here, instead of letting $$(2n+1)=3$$ for all recursive steps, the value is going to be $$(2n+1)$$ where $$n$$ stands for the $$n$$th recursive step. That is, in the case of the modified 4D Menger Sponge, which has the volume $${2}^{4}=16$$, the first round will reduce it by a factor of

$$\dfrac {81-33}{81}=\dfrac{48}{81}$$.

The 2nd round will reduce it by a factor of

$$\dfrac {625-113}{625}=\dfrac{512}{625}$$

And so on. The infinite product has a limit value which is not zero. What is the final volume?

Note

Here's how a 3D Menger Sponge looks like, after a few recursions, to help with visualizing this problem. Try to generalize from 3D to 4D. This one, as already noted, will have zero volume.

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