User blog:Edwin Shade/A Mengol

A Mengol  is the minimum number of square tiles needed to construct an iteration-$$10^{100}$$ Menger Sponge.

To explain how we would go about deriving such a figure, we must first understand what a Menger Sponge is. A Menger sponge is a type of fractal that consists of a cube which is then subdivided into 27 equal portions, with the center most cube and six center exterior cubes being removed. This process is repeated for these smaller cubes, and iterated an infinite number of time.

The following explanation demonstrates this graphically.

This is an example of an iteration-0 Menger fractal, it is a singular cube which consists of 6 squares.



Next we subdivide this cube into 27 equal cubes, as shown below.



Next we remove the innermost cube, that is, the one in the center, and the six cubes in the center of the cube's faces. The resultant shape has a volume of 20 cubes, and a surface area of 72 squares



We reiterate this process on these smaller cubes and end up with this shape, or an iteration-2 Menger sponge. This shape has a volume of 400 cubes, and 1,056 squares.



Let us call this $$M_2$$ for brevity's sake.

The following is a picture of $$M_3$$, which has 8,000 cubes and a surface area of 18,048 squares.



We can easily go on doing this, but it begs the question, how many squares are in each successive iteration ? Well, the answer may not be initially obvious, but a few years ago I found a solution to this problem, in the form of a recursive equation.

Put simply, we know that for any sponge $$M_n$$, it will contain at most 20 times the surface area of $$M_{n-1}$$, since $$M_n$$ is composed of twenty copies of $$M_{n-1}$$. We also know that for each subsequent iteration of the Menger sponge the faces of $$M_{n-1}$$ cover each other to form $$M_n$$.

The number of squares on the faces of a sponge $$M_n$$ can be given by the expression $$8^n$$. This follows from the fact that a Menger sponge has as it's faces Sierpinski carpets, (pictured below along with their construction).



There are 48 carpets that overlap and cancel out in the joining together of $$M_{n-1}$$'s to create one $$M_n$$, hence the surface area in square of $$M_n$$ will be $$20{M_{n-1}}-48{8^{n-1}}$$, where $$M_1=72$$.

This gives us the recursive equation $$S_n=20{S_{n-1}}-48{8^{n-1}},S_1=72$$, where $$S_n$$ refers to the surface area in squares of an iteration-n Menger sponge.

Therefore, a Mengol is equal to $$S_{10^{100}}$$, as defined by the above recursive equation. It is most likely above a googolplex, though much smaller than a googolduplex.