A few years ago after the birth of our second daughter and in anticipation of our third, I started designing a one-room addition for our house. One of the problems I faced was figuring out the shape of the roof. I learned of the concept of the straight skeleton of a polygon, first defined by Oswin Aichholzer and Franz Aurenhammer in a book chapter “Straight Skeletons for General Polygonal Figures in the Plane” in 1996. Wikipedia provides an intuitive definition:

“The straight skeleton of a polygon is defined by a continuous shrinking process in which the edges of the polygon are moved inwards parallel to themselves at a constant speed. As the edges move in this way, the vertices where pairs of edges meet also move, at speeds that depend on the angle of the vertex. If one of these moving vertices collides with a nonadjacent edge, the polygon is split in two by the collision, and the process continues in each part. The straight skeleton is the set of curves traced out by the moving vertices in this process. In the illustration the top figure shows the shrinking process and the middle figure depicts the straight skeleton in blue.”

but the concept is best understood by picture:

The fact that straight skeletons fit “symmetrically” into the polygons that generated them, made me think about whether they could constitute aesthetic representations of phylogenetic trees. So I asked the inverse question: given a phylogenetic tree, i.e. a graph that is a tree with weighted edges, together with a cyclic orientation on its vertices, is there a convex polygon such that the tree is the straight skeleton of that polygon? A few google searches didn’t reveal anything, but fortunately and coincidentally, Satyan Devadoss, who is a topologist and computational geometer was visiting my group on his sabbatical (2009–2010).

Now, a few years later, Satyan and coauthors have written a paper providing a partial answer to my question. Their paper is about to appear in the next issue of Discrete Applied Mathematics:

- Howard Cheng, Satyan L. Devadoss, Brian Li, and Andrej Risteski, Skeletal configurations of ribbon trees,
*Discrete Applied Mathematics*170, 19 June 2014, 46–54.

The main theorem is formally about *ribbon trees**:*

**Definition**. A ribbon tree is a tree (a connected graph with no cycles) for which each edge is assigned a nonnegative length, each internal vertex has degree at least three, and the edges incident to each vertex are cyclically ordered.

The authors prove the interesting result that there exists only a finite set of planar embeddings of a tree appearing as straight skeletons of convex polygons. Specifically, they show that:

**Theorem. **A ribbon tree with *n* leaves has at most *2n−5* suitable convex polygons.

Its fun to work out by hand the case of a star tree with three leaves:

The algebra works out to solving a cubic system of three equations that can be seen to have one unique positive solution (Lemma 5 in the paper).

The proof of the main theorem relies on some elementary trigonometry and algebra, as well as an interesting analogy of Cauchy’s “arm lemma” (if one increases one of the angles of a convex polygonal chain, then the distance between the endpoints will only increase). Furthermore, a few interesting cases and connections are pointed out along the way. For example, some ribbons, even caterpillar ribbons, are not realized by any polygon. There is also a related conference publication by many of the same authors, and in addition including Aichholzer himself, that provides some interesting constructions in special cases:

- Oswin Aichholzer, Howard Cheng, Satyan L. Devadoss, Thomas Hackl, Stefan Huber, Brian Li, Andrej Risteski, What makes a tree a straight skeleton?, Canadian Conference on Computational Geometry, 2012.

Straight skeletons may yet find application in drawing phylogenetic trees, but for now the best out there are radial or circular representations optimizing various layout considerations.

My addition is now complete and the roof is absolutely beautiful.

## 4 comments

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March 25, 2014 at 2:03 pm

Allen KnutsonHuh! So the graph is oriented (which way the vertex was moving). Am I right in thinking that modulo degeneracies, each internal vertex is trivalent, with out-degree 2 or 0?

March 25, 2014 at 2:39 pm

Lior PachterYes, but I think you mean in-degree 2 or 3. Said differently, the tree is rooted.

March 26, 2014 at 12:08 pm

Saheli Datta (@SaheliDatta)Just curious–was there an architect involved?

March 26, 2014 at 12:13 pm

Lior PachterAn architect

hadto get involved. We live in Berkeley.