[math-fun] Folding flat figures with holes
How many folds are needed to turn a square piece of paper into a flat figure with k holes? As an example, you can fold a square up into a long skinny rectangle like the wrapper a drinking straw comes in, and then bend that long skinny rectangle in lots of places to create lots of crossings (and lots of holes between the crossings). I can show that asymptotically you can get k holes with something like constant times sqrt(k) folds, but I don't know if this is best possible. Jim Propp
What happens when the intersection of a fold line with the paper is disconnected? Does it count as multiple folds? Rich ------- Quoting James Propp <jamespropp@gmail.com>:
How many folds are needed to turn a square piece of paper into a flat figure with k holes?
As an example, you can fold a square up into a long skinny rectangle like the wrapper a drinking straw comes in, and then bend that long skinny rectangle in lots of places to create lots of crossings (and lots of holes between the crossings).
I can show that asymptotically you can get k holes with something like constant times sqrt(k) folds, but I don't know if this is best possible.
Jim Propp _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
That’s a good question. Maybe we should bifurcate the problem till we know which branch is more fun. I’m guessing that one or both versions of the problem have been considered before, especially in the mathematical origami community. The question occurred to me as an offshoot of Jeannine Mosely’s upcoming talk at https://www.gathering4gardner.org/g4gs-celebration-of-mind-2020/. Jim Propp On Mon, Oct 19, 2020 at 1:24 AM <rcs@xmission.com> wrote:
What happens when the intersection of a fold line with the paper is disconnected? Does it count as multiple folds?
Rich
------- Quoting James Propp <jamespropp@gmail.com>:
How many folds are needed to turn a square piece of paper into a flat figure with k holes?
As an example, you can fold a square up into a long skinny rectangle like the wrapper a drinking straw comes in, and then bend that long skinny rectangle in lots of places to create lots of crossings (and lots of holes between the crossings).
I can show that asymptotically you can get k holes with something like constant times sqrt(k) folds, but I don't know if this is best possible.
Jim Propp _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Another variant would be to lift the restriction that the folded figure be flat. There may be some topological restrictions on the final figure that one would want to impose. If you folded it into a hollow cube, with no holes, would that count as having *removed* a hole? Tom James Propp writes:
That’s a good question. Maybe we should bifurcate the problem till we know which branch is more fun.
I’m guessing that one or both versions of the problem have been considered before, especially in the mathematical origami community. The question occurred to me as an offshoot of Jeannine Mosely’s upcoming talk at https://www.gathering4gardner.org/g4gs-celebration-of-mind-2020/.
Jim Propp
On Mon, Oct 19, 2020 at 1:24 AM <rcs@xmission.com> wrote:
What happens when the intersection of a fold line with the paper is disconnected? Does it count as multiple folds?
Rich
------- Quoting James Propp <jamespropp@gmail.com>:
How many folds are needed to turn a square piece of paper into a flat figure with k holes?
As an example, you can fold a square up into a long skinny rectangle like the wrapper a drinking straw comes in, and then bend that long skinny rectangle in lots of places to create lots of crossings (and lots of holes between the crossings).
I can show that asymptotically you can get k holes with something like constant times sqrt(k) folds, but I don't know if this is best possible.
Jim Propp
If folding along a single line counts as just one fold even when the intersection of line and paper is disconnected, then you only need O(log(k)) folds. Fold a skinny rectangle into a triangle, then fold a vertex onto the opposite side to get two triangles, and keep folding along lines parallel to the base, doubling the count of triangles each time. The original rectangle has to have width O(1/k), which is another log(k) folds. Of course zero-thickness paper can only exist in a frictionless vacuum. On Mon, Oct 19, 2020 at 7:10 AM Tom Karzes <karzes@sonic.net> wrote:
Another variant would be to lift the restriction that the folded figure be flat. There may be some topological restrictions on the final figure that one would want to impose.
If you folded it into a hollow cube, with no holes, would that count as having *removed* a hole?
Tom
James Propp writes:
That’s a good question. Maybe we should bifurcate the problem till we know which branch is more fun.
I’m guessing that one or both versions of the problem have been considered before, especially in the mathematical origami community. The question occurred to me as an offshoot of Jeannine Mosely’s upcoming talk at https://www.gathering4gardner.org/g4gs-celebration-of-mind-2020/.
Jim Propp
On Mon, Oct 19, 2020 at 1:24 AM <rcs@xmission.com> wrote:
What happens when the intersection of a fold line with the paper is disconnected? Does it count as multiple folds?
Rich
------- Quoting James Propp <jamespropp@gmail.com>:
How many folds are needed to turn a square piece of paper into a flat figure with k holes?
As an example, you can fold a square up into a long skinny rectangle like the wrapper a drinking straw comes in, and then bend that long skinny rectangle in lots of places to create lots of crossings (and lots of holes between the crossings).
I can show that asymptotically you can get k holes with something like constant times sqrt(k) folds, but I don't know if this is best possible.
Jim Propp
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