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Take a square paper of size S X S units, and thickness of paper 1 unit.

How many half folds should we do to get form a cube out it?

Consider infinite foldable paper.

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  • 2
    $\begingroup$ Fun question! Brought me to try folding an origami cube :D ... (there are various manuals online of someone) $\endgroup$ – Nick Mar 7 at 9:59
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    $\begingroup$ Full cube, or hollow cube? $\endgroup$ – Laurent LA RIZZA Mar 7 at 11:39
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    $\begingroup$ Actually it was a diamond window, hollow cube. I should have counted the number of foldings though.. $\endgroup$ – Nick Mar 7 at 13:28
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Firstly there must be an even number of folds for the folded paper to remain square

Consider the volume of the paper which is $S \times S \times 1 = S^2$

Therefore the resultant cube side is $\sqrt[3]{S^2}$

After $N$ pairs of folds the side length is $S / (2^N)$

So $S / (2^N) = \sqrt[3]{S^2}$

So $S = 2^{3N}$

So $N = \log S / (3 \times \log 2) $
Or $N = \log S / \log 8 $

Now suppose $ S = 512$ thickness $T = 1$
The computation gives $ N = 3 $ pairs of folds

Worked example:
After 1st pair of folds $S = 256$ with $T = 4$
After 2nd pair of folds $S = 128$ with $T = 16$
After 3rd pair of folds $S = 64$ with $T = 64$ which is a cube

The question asks how many half folds?
Answer:
Half folds = $ 2 \times \log S / \log 8 $

Obviously the paper can only be folded thus if $N$ is an integer

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Consider what happens by doing $2$ folds, one in each direction.

You now have a square that is half the size, but $4$ times the thickness.

So after $2k$ folds

the square has edge length $S/2^k$ and thickness $4^k$.

For this to be a cube we need:

$$\frac{S}{2^k} = 4^k\\ S=8^k \\ k = \log_8{S} = \frac{\log S}{\log 8}$$

So the number of folds we need is

$$2k = \frac{2\log S}{\log 8}$$ It will only be a cube if this result is an even whole number, i.e. if S is a power of $8$.

In reality, this will only work if you cut the paper in half and stack the pieces, instead of folding the paper. I have ignored the amount of paper connecting the different layers of the folded cube, which is quite substantial, and the round folded edges also keep it from being in the shape of a cube.

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I think it should be...

$N = 2 \log_2{\sqrt[\leftroot{-2}\uproot{2}3]{S}}$, where $N$ is the number of folds.
As a caveat, $S$ can only be $2^{3x}$, where $x \in \mathbb{N}$.

For example:
$S=4096$ ($=2^{12})$
$\sqrt[\leftroot{-2}\uproot{2}3]{S} = 16$; $\log_2{16} = 4$; $4 \times 2=8$ folds.
$4096 \div 2^4$ (I fold each side $4$ times) $= 256$.
Thickness is $2^8$ ($8$ folds) = $256$.

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It is simple.

The volume of paper is $S^2$.
So the side of the cube we want is $\sqrt[3]{S^2} = S^\frac{2}{3}$.
Each fold doubles the thickness. It starts with $1$. After $N$ folds the thickness is $2^N$.
To get the correct thickness for the cube we need to have $2^N = S^\frac{2}{3}$.
Taking $log_2$ on both sides we get $N = \frac{2}{3} log_2 S$.
Of course, it only works if N is an even integer.

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