Consider a polygonal sheet $P$ of area $A$ with $N$ vertices (it material is not stretchable or tearable). Let $n$ be a positive integer >=2.

Question: Let $P$ lie on a flat plane. We need to fold up $P$ so that it now occupies only an area $A/n$ of the plane. It is also needed that the folding is as uniform as possible - ie the number of layers of the material above any given point should be as close to $n$ as possible. We need an algorithm that does it and an estimate of its complexity.

Example: If $P$ is a rectangle of area $A$ and $n$ is an integer, it is easy to see that we can fold it to an area $A/n$ such that it is exactly $n$ layers thick throughout - the 'creases' could simply be $n-1$ equally spaced parallel lines. It appears that no other shape of $P$ has this property of 'perfectly uniform foldability'. Which is the shape(s) of $P$ that causes greatest variation in the number of layers for a given $n$?

Further possibilities: One can further ask: Minimize the perimeter of the area $A/n$ region that is covered by the folded polygon. Alternatively, We could require $P$ to be folded as uniformly as possible so that it can be packed into a rectangular or square box of some specified dimensions - and area not necessarily equal to $A/n$ with n an integer.

  • $\begingroup$ Do you assume that $P$ is convex? $\endgroup$ – M. Winter Jul 15 at 8:23
  • $\begingroup$ Yes A is the area. Made the necessary edit. Thanks. As for P being convex, not necessary. $\endgroup$ – Nandakumar R Jul 15 at 9:48

For every $k\ge 2$ there is a non-rectangular shape allowing a uniform folding for all $n$ that are multiplies of $k$:

The reason is that you can uniformly fold it into a rectangle with $k$ layers.

If you are looking for convex shapes, then $k=2$ above is convex. Here is a non-rectangular convex shape admitting a uniform folding with three layers. If you want to make it rectangular you will get six layers, but then you can proceed in all multiples of 6.

More generally, every regular $n$-gon admits a $2n$-layer folding. And it can be further made into a rectangular $4n$-layer folding (and then every multiple thereof).

The differently colored creases in the pentagon are to be understood as alternatingly folded upwards and downwards.

Or even better, for every $n$ there is a non-rectangular convex shape admitting an $n$-layered folding, or a $2n$-layerd rectangular folding (and then every multiple thereof).

So an interesting question might be whether for every $n$ there is a non-rectangular convex shape admitting a rectangular $n$-layer folding.

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    $\begingroup$ The $(k=3)$-shape folds uniformly into a rectangle with $n$ layers for every $n \geq 2$, not just multiples of $3$. First fold the triangles in. If $n$ is even, then fold the right half along $n/2-1$ creases and the left half along $n-1$ creases; if $n$ is odd, then fold the right half along $(n-1)/2-1$ creases, fold the left half "over it" (i.e. fold in the middle) and fold the "remaining part" (i.e. the part that does not cover the folded right half) of the left half $n-1$ times. $\endgroup$ – Florian Lehner Jul 15 at 10:21
  • $\begingroup$ Then there is the carpet roll approximation. Convert the sheet into a tube of small diameter and squash it. It won't be perfect, but for large n the thicknesses may not vary much. Gerhard "Exercise: Which Shapes Vary Much?" Paseman, 2020.07.15. $\endgroup$ – Gerhard Paseman Jul 15 at 19:26
  • $\begingroup$ Those are quite nice examples! And the naïve guess would be that the answer to your parting question of whether there is some non-rectangular convex shape that admits a uniform n-layer folding for any n might be "no" even if the final folded object is allowed to be non-rectangular. And somehow all smart foldings seem to involve a rectangle at some stage. $\endgroup$ – Nandakumar R Jul 15 at 19:28
  • $\begingroup$ @NandakumarR Do you mind whether I ask this particular instance (the last paragraph of my asnwer) as a separate MO question? You question seems to be also focused on the algorithmic aspect which my asnwer does not touch at all. $\endgroup$ – M. Winter Jul 16 at 19:02
  • $\begingroup$ @M.Winter, Sure... Indeed, happy if the question acquires dimensions I didn't anticipate! $\endgroup$ – Nandakumar R Jul 17 at 5:15

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