A Visit to Martin State Forrest

DSC 0794

State Forests are an ambivalent thing. The designation means they are in possession of the state, and managed as such. This means logging, hunting, fishing, and depending on the extent of the above, the availability of roads and trails for public use.
DSC 0899

There are beautiful spots like Pine Lake. The fire tower offers a view of the gypsum mine near Shoals. There is a lot going on underground here.

DSC 0886

Abandoned places

DSC 0836

offer nested views in the past.

DSC 0854

The Tank Spring Trail leads to another abondoned place, an old water reservoir.

DSC 0803

The sink holes are ready.

DSC 0824

Advertisements

Stellating the Icosidodecahedron in Black and White

DSC 1737

It is also this time of the year to chase away the dark hours by making presents. As in previous years,
we will make a stellation out of paper without glue. This year, we are going to stellate the Icosidodecahedron, one of the fancier Archimedean solids.

Icosidodecahedron

The stellation is quite simple, it is also a compound of the dodecahedron and icosahedron. The simpler compounds of a Platonic solid with its dual are also doable, see the post from two years ago.

Compound

To make it, we will need 20 triangles and 12 pentagons, so printing and cutting two of the templates below will do. I suggest to print four templates in two different colors and to make two models.

StellaDodecaIcosa

Then we start by sliding five triangles into one pentagon like so:

DSC 1739

Then we add five pentagons between two adjacent triangles.

DSC 1740

Next another five triangles:

DSC 1742

Now we have finished one half of the model. This already would make a nice dome for the backyard.

You can make another half and try to attach them, but I think it is easier to just keep going.
This next step is a little tricky, because to prevent the polygons from falling out, it is best to add a ring of alternating pentagons and triangles. When done, it looks like this:

DSC 1746

The last two steps (add five more triangles and one more pentagon) are then pretty clear, but still tricky because you have to insert the new polygons in four or five slits essentially at once.

DSC 1750

Hunan and Yunnan

It’s this time of the year when to look back to see what has been memorable, in the good way. I discovered some wonderful teas, black teas from China. New for me were two teas from the Hunan province, both distributed in the US by reliable Harney & Sons.

DSC 1725

The simpler of them is the Hunan Forest Twist that comes with long leaves that slowly uncurl in the pot while the tea steeps.

DSC 1732

The taste is similar to that of a full-bodied Assam but without any bitterness.
The second tea from Hunan is the Wild Hunan Gold.

DSC 1761

The leaves have wonderful golden tips, and the cup is a wild ride through intense chocolate notes. This tea clearly deserves its name.

DSC 1767

Even more golden are the leaves of the Yunnan Golden Ring tea from the Yunnan province, purchased through Imperial Tea Court. I usually have one or two good quality Yunnan teas around because I like their soft and distinctive flavor, but this tea is something else.

DSC 1751

The rolled golden leaves are amazing, and the tea, while clearly Yunnan, is more complex than the standard Yunnan, with subtle hints of chocolate.

DSC 1757

Instant Insanity

This post is about a mathematical puzzle and not the current state of daily affairs. The puzzle consists of four cubes, with faces colored in four colors:

Cubes

If you want to make your own, you can print out the nets below and fold.

Insanenets

The goal is to stack the four cubes together so that each of the four long faces of the 4 x 1 x 1 tower show all four colors.

According to Jerry Slocum, this puzzle goes back to 1900 when Frederick A. Schossow marketed a version of it under the name Katzenjammer Puzzle. Katzenjammer is German for hangover…

It has since appeared in many variations. Its latest incarnation under the current name Instant Insanity was discovered by Frank Armbruster and has been marketed by Parker Brothers since 1967.

We will solve he puzzle using graph theory. Each cube will get encoded in a graph. To do so, we use as vertices the colors (yellow, orange, purple, and green), and connect two colors by an edge if the two colors occur on opposite faces of the cube. Thus we obtain for each colored cube a graph with four vertices (colors) and three edges (opposite faces).

Singlenets

Before we continue, let’s see why these arbitrary looking graphs contains all the information about the cubes that is necessary to play the puzzle: If the graph of a cube is given, we know all colors of pairs of opposite faces for that cube. Using a blank cube, we can first color the front and back face with a pair of these colors. It doesn’t matter which color we pick for the front, because we could turn the cube over. Using a second edge of the graph, we then color the left and right face with the two colors of the end points of the edge. Again it doesn’t matter which color we use for the left side, because there is a rotation fixing front and back that flips left and right. Finally, there are two possibilities remaining for coloring the top and bottom face of the cube. These lead to truly differently colored cubes, but both are equivalent for solving the puzzle, as the top and bottom colors of the tower don’t matter for the puzzle.

Coming back to solving the puzzle, we combine the four graphs we have created for each cube into a single graph with multiple edges. The edges are now labeled from 1 through 4 to indicate from which cube they come. How does this graph help us to solve the puzzle?

InstantInsanity

Suppose we have stacked the cubes together so that both the front and back side of the tower show all four difference colors. The four front and back sides of each cube represent each an edge in our graph, which we give a direction so that they always point to the color of the back edge. Let’s mark these edges say blue. Thus we get four blue edges that begin at four different colors and end at four different colors. As there are just four colors, each vertex has an edge ending and an edge beginning there. This means that following the blue edges, we have found a doubly Hamiltonian circuit – a cycle or collection of cycles that pass through all four vertices (colors) of the graph, and uses edges with each of the four labels (cubes).

Vice versa, any such path can be used to stack the cubes so that front and back side of the tower show all four colors.

Next we will show that any such system of circuits must go once around the square marked by the four colors.

If this were not the case, it would decompose into several components of lengths 1+3, 1+1+2, or 2+2.
While in each case there are Hamiltonian circuits, none of them are doubly Hamiltonian.

Thus we only need to look for Hamiltonian paths that go around the square, and can ignore the loops at orange and green as well as the diagonal connection from purple to orange.

Consider the top and bottom edge, which both have an edge labeled 4. As each edge label can only occur once, one of these two edges must use an edge with a label other than 4. We make a case distinction: First, let’s assume the bottom edge uses label 3. Necessarily then, the right edge then must use label 2, the left edge label 1, and the top edge label 4, and we get the Hamiltonian path marked red.

Now let’s assume that the top edge is using label 1. Thus we have to use for the left either label 2 or 3, and the other label 3 or two for the right edge. Either choice leaves us with label 4 for the bottom edge. One of the choices is the path marked blue, the other one has labels 2 and 3 exchanged in the left and right connections. Thus there are only three different possible Hamiltonian paths.

Solution

To finally solve the puzzle completely, we will need to take care of the left and right faces of our tower. By associating a directed edge for each cube from every left face to every right face, we get a second doubly Hamiltonian path in our graph.
This second path must be edge disjoint with the first, as we cannot use a pair of opposite faces of the same cube twice.

As the second and third of our three Hamiltonian paths are not disjoint, one of the paths we seek must be the red path. This uses edge 2 on the right hand side, so that the second Hamiltonian path can only be the blue one. They are indeed edge disjoint, and thus solve the puzzle.

The Unruly

One of the fascinating aspects of the DePauw Nature Park is that one can observe who takes possession of this devastated landscape.
A contender (my favorite) is the Sycamore tree.

DSC 7742

Even while still little, they make gigantic leaves.

DSC 7703

A little older, they begin to show their unruly temperament. This is not a pretty tree for an English Garden. But they show character.

DSC 7617

When mature, they become imposing. Their distinctively peeling bark makes me think of ghosts.

DSC 7658

Right now, they stand mostly isolated, or against the backdrop of the quarry walls.

DSC 7713

But hundreds of little ones are growing, hidden between the shrubs. Let them have this place. They will cause no harm.

DSC 7777

Reflections on the Letters r,s,t (Groups II)

Continuing the discussion from last week, let’s consider the 3-letter alphabet {r,s,t}. We are allowed to form all possible words in these letters (and their inverses, if you want to), but we agree that rr=ss=tt=1 and (rs)^2=(st)^3=(tr)^6=1. This defines the Coxeter group G(2,3,6). Last time we saw that the very similar group G(2,3,3) is finite, today we will see that G(2,3,6) is infinite. Below is the beginning of its Cayley graph.

Cayley236 01

We travel from one word to the next by appending r,s, or t. This looks much more complicated than what we saw for G(2,3,3), but things become clearer when we look at another group. Consider a yellow triangle with 30, 60, 90 degree angles (writing this as π/2, π/3,π/6 makes 2-3-6 reappear), and let ρ, σ, τ be the reflections at the lines extending its edges.

Reflectiongroup 01

These three reflections generate a group Γ(2,3,6) of Euclidean symmetries which has the yellow triangle as its fundamental domain. The clue is that Γ(2,3,6) = G(2,3,6). We can easily map G(2,3,6) to Γ(2,3,6) by sending R to ρ, s to &sigma, t to τ. This works because ρ, σ, τ satisfy the same relations as r, s, t. It doesn’t work as easily the other way because ρ, σ, &tau could also satisfy other, hidden relations.

Reflpath 01

Let’s look at the word tsrtst. Reading it from left to right gives us a path on the Cayley graph from the initial triangle to a target triangle. Translating from Latin tsrtst to Greek τσρτστ gives a composition of reflections that takes the initial triangle to the same target triangle. This is not completely trivial, you prove it by induction. Remember that the composition is applied from the right to the left, so we also change reading direction.

This observation can be used to show that the translation map G(2,3,6) to Γ(2,3,6) is injective. If a word in G is the identity in $γ, its path in the Cayley graph must be a closed loop. As the Euclidean plane where the tiling lives is simply connected, we can homotope it to a constant path, using elementary operations: Backtracking an edge, or shrinking a loop around a vertex to a point. The former is the accomplished using the relations rr=ss=tt=1, the latter using the other relations. This shows that the geometric homotopy can be realized using the relations of the group, and thus we can reduce the word to the trivial word 1.

Cayley
This is essentially the proof of a famous theorem by Walther Franz Anton von Dyck: The group G(a,b,c) is finite if and only if 1/a+1/b+1/c>1. We have seen the relevant examples in the case
1/a+1/b+1/c>1 and 1/a+1/b+1/c=1. If 1/a+1/b+1/c <1, we need hyoperbolic geometry. Above is a picture of the Cayley graph of G(2,3,7) within the tiling of the hyperbolic plane by (π/2,π/3,π/7)-triangles.

Reflections on the Letter T

Almost a year ago, when there was still hope, I posted a few Fall themed images with the title Yellow. The third image shows a view of McCormick’s Creek with a tree trunk that looked in 2008 like this:

DSC0848

The perspective of the two images is not quite identical, but you will see that in the older image above there are two prominent trees, the right of which has become the trunk in the second image of the image of last year’s post.

DSC 1986

Above is another image, from 2009, tree still standing, again from a slightly shifted perspective. The view has always tantalized me, because it looked promising, but I could never turn it into a picture I was happy with. Below is a view from the other side, another year later.

DSC0838

Then, in 2015, this unexpected change:

DSC 0691

With the tree reduced to a stump,

DSC 7530

the place has become more balanced and serene.

DSC 7533

Sometimes it is worth the wait.