Hidden Simplicity (Maybe-Ferns 5)

Mathematicians like to do things a little differently. An excellent example was the Mathematische Arbeitstagung, a yearly event held in Bonn, where the (mathematical) audience was asked to publicly suggest speakers.

Hirzebruch

Friedrich Hirzebruch would write the suggested names on the board (he sometimes misheard…), and then create a list of speakers on the fly. Sometimes they ended up with unexpected results. One year, Michael Barnsley was suggested, who had been working on a new fractal image compression method.

Nonfern4

His talk was exciting for us graduate students, because we for once could understand something. The idea was to use special types of iterated function systems: Take a few linear maps that are all contractions, and use them to map a subset  of the plane to the union of the images of that set under all the linear maps. This becomes a contraction of the space of closed subsets of the plane to itself with respect to the Hausdorff distance, and hence has a fixed point, which is again a subset of the plane.

Nonfern3

It turns out that these subsets are highly complicated fractals, encoded just by a few numbers. For instance, all images on this page (except for the photo of Hirzebruch at the top) were made with just two linear maps, requiring 12 decimal numbers.

Nonfern2

Barnsley claimed that he could reverse engineer this: Start with an image, and find a small collection of linear maps that would produce the given image very accurately. If true, this would revolutionize image compression.  We went home and tried it out on our Atari ST computers and the likes. All we could produce were ferns, twigs, and leaves.

Nonfern1

Paul Bourke has a nice web site where he explains how one can design some simple fractals, and has also some very impressive images of ferns using four and more linear maps. Below are the two simple maps used to create the polypodiopsida psychedelica above.

Formula

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Carpets (Foldables 4)

The last (for now) example in this series of bifoldable designs is a woven carpet. Will create a doubly periodic polyhedron that consists of the Miura tubes below (which are almost 50 years old!).MiuraTube

We begin with a corner type we call Double L

MiuraWeave 1

Four copies of it (using reflecions) can be combined into a translationa fundamental piece like so:

MiuraWeave 2

The tubes (of double length) emerge when we replicate this piece several times in both directions:

MiuraWeave 3

Above is its most symmetric state. This carpet does not need to be rolled, it can be squeezed in both of its translational directions, as below:MiuraWeave 4

So you can push this Miura Carpet to any of the four sides of a room.

Dos Equis (Foldables 3)

Whenever you show a mathematician two examples, s(he) wants to know them all. So, after the introductory examples of Butterfly and Fractal it’s time to make something more complicated. Jiangmei and I started by classifying all possible vertex types that can occur when you build polyhedra using only translations of four of the six types of faces of the rhombic dodecahedron (and make sure they attach to each other as they do it there). We found 14 different ones, and a particularly intriguing one is what we called the X:

TripleX1The central vertex has valency 8, and we were wondering whether we could use it to build a triply periodic bifoldable polyhedron. It is easy to combine two such Xs to a Double X:

TripleX2

One can then put a second such Double X (with the order of the Xs switched) in front. Note that these are still polyhedra. Below are two deformation states of these quadruple Xs. We see that they are quite different.

Xdef

So far, the construction can be periodically continued up/down and forward/backward. It is also possible to extend to the left/right, and there are in fact two such possibilities, allowing for infinite variations, because one has this choice for every left/right extension. They are indicated by the arrays below. 

TripleX3carrowIf you don’t have the time to build your own model, here again is a movie showing the unfolding/folding of a rotating Dos Equis.

 

The Fractal (Foldables 2)

The second bifoldable object Jiangmei showed me was this:

Fractal 1

You can find a movie showing how this folds together in two ways here. To understand how and why this works, let’s first look at a simple saddle:

Saddle

This is a polyhedron with a non-planar 8-gon as boundary. Its faces are precisely the four types of faces that are allowed in our polyhedra: All others have to be parallel to these four. The four edges that meet at the center of this saddle constitute the star I talked about the last time. Again, all edges that can occur must be parallel to one of these four. One can fold the saddle by moving the upwards pointing star edges further up (or down), and the downwards pointing edges further down (or up), thereby keeping the faces congruent. This works locally everywhere and therefore allows a global folding of anything built that way. Fractal 0

For instance, the hollow rhombic dodecahedron above can be bi-folded. Now note that this piece is also a polyhedron with boundary. In fact, its boundary is exactly the same octagon as the boundary of the saddle. 

Observe also that at the center of this piece we have a vertex in saddle form. This suggests to subdivide all rhombi into four smaller rhombi, remove the saddle an the middle vertex of the doubled hollow dodecahedron, and replace it by a copy of the standard hollow dodecahedron. This gives you Jiangmei’s fractal. Repeating this is now easy. Below is the generation 2 fractal (animation):

Fractal 2

And, just for fun, the generation 10 fractal:

 Fractal 10 colorgradient

You can see it being bifolded here. So far, the two completely folded states of our polyhedra looked very much the same. We will see next week that this doesn’t need to be the case.

The Butterfly (Foldables 1)

We all know that cardboard cubes are rigid, which is why we get our packages in boxes. We also all know that if we remove two opposite faces from a cube, we can fold it together. This started to interest me when I noticed that the polyhedral approximation of the Schwarz P surface is surprisingly flexible. This summer, I showed this to our local Origami and Paper Folding expert, Jiangmei Wu from our School of Art and Design, and she became interested. A few days later she came with a paper model that looked like this:

ButterflyTriply

She called it a simple variation of the polyhedral P-surface. Hmm. This is a triply periodic polyhedral surfaces tiled with rhombi. To understand it, we build it out of smaller units (which we called butterflies):

Butterbuild

The really cool thing about it is that it can be folded together in two different ways, like so:

Butterdeform

You can find an animation showing the continuous deformation here. We stared at this for a (long) while, until we realized that this has to do with rhombic dodecahedra. The structures up above are composed of the rhomboids from last week that tile a rhombic dodecahedron. The latter has, as the name hints, 12 faces, which occur in opposite pairs. Like the cube, it is rigid per se, but becomes foldable if we remove two pairs of parallel faces, leaving us with four faces to use, which are distinguished by color up above.Fractal0b

Above you can see the four hollow parallelepipeds (which we called hollowpeds). The almost trivial but nevertheless mind bending realization is that everything you build out of these hollowpeds becomes a structure foldable in two different ways. Next week I’ll show Jiangmei’s second model, a foldable fractal… If you can’t wait, check out this.

 

 

 

Unpacking the Hypercube

Last week we learned how Rototiler moves can unpack a cube.  As a warmup, below are the moves for a 2x2x2 cube projected parallel along a cube diagonal onto hexagons:

HexaSwap2

We start at the left, remove the frontmost cube, and keep going. The solution is far from being unique, but not too complicated. Today, we do the same with a  hypercube. The projection of a 1x1x1x1 into 3-space along a main diagonal is a rhombic dodecahedron, tiled by four rhomboids. These rhombic dodecahedra have 8 obtuse, 3-valent vertices at the corners of a cube, and 6 acute, 4-valent vertices at the corners of an octahedron.

Rhomb1

There are two ways to tile the rhombic dodecahedron with these rhomboids, and changing one to the other corresponds to a rototiler move in space. Let’s do this with the 2x2x2x2 hypercube, whose projection is a rhombic dodecahedron tiled by 32 rhomboids.

Rhomb2

At first it seems as if there is no swappable rhombic dodecahedron available, but if we remove three rhomboids and  look inside (which is the direction of the fourth dimension, after all), we can see it. After swapping it, we also remove the frontmost rhomboid of the  swapped dodecahedron.

Rhomb3

 

 

We then see that the four removed rhomboids together make up another swappable dodecahedron. We replace it by its swap. The same can be done at three other places.

Rhomb4

The next thing to do is to swap 6 more dodecahedra. One of them is the one which shows yellow and purple rhomboids in the right figure above, sitting between the red and blue “vertex”. All these dodecahedra correspond to the edges of the tetrahedron whose vertices are the already swapped four peripheral dodecahedra. Doing these six swaps leads to a tiling very much like the one above to the right, where now the other four obtuse vertices mark swappable dodecahedra. Swapping these and finally the hidden central dodecahedron  completely unpacks the hypercube. It took us 1+4+6+4+6+1 = 32 swaps, as expected.

 

Next week we’ll see what this is good for…

 

Roto-Tiler

Today we look at a puzzle invented by Alan Schoen that he calls Roto-Tiler. He explained this to me a few years ago, and when I showed him notes I made for a class, he denied that this is the puzzle he described. I insist it is, and it is quite certainly not mine.

Roto0

Things happen on a hexagonal board like the one above (it can but doesn’t need to be regular), tiled by hexagonal rhombi of equal size. The acute angles are marked by 1/3-circles, which occasionally happen to close up when three acute angles meet. In that case, a move consists of rotating the three involved rhombi by 120º either way.

Roto3

Above you can see the possible four moves from the central position. At this point it is not clear at all that a move is always possible. The puzzle consists of transforming one given tiling by rhombi to another given tiling of the same hexagon. For instance, a simple example asks to find the smallest number of moves that takes the left tiling to the right tiling.

Roto2

The clue to solve this puzzle is to view the hexagons as the parallel projection of a box subdivided into smaller cubes, and the rhombi as the projections of the faces of the smaller cubes. This becomes visually more intuitive if we color the rhombi by their orientation so that parallel cube faces have the same color:

Roto1

Then the hexagon above becomes the projection of a box partially filled with cubes, and a move consists of adding or removing a frontmost cube. This step into the third dimension explains everything: We see that we can solve every Roto-Tiler puzzle by emptying and filling boxes with cubes. Last week’s first example was a 1-dimensional version of this, next week we will try to grasp a 3-dimensional version and practice our 4-dimensional intuition.