### Fullerene Layout with Spokes and Arches

Having tried (and failed) to layout fullerene structures using various optimisation methods, I thought I would try direct positioning of the atoms. In other words, 'logical' placement rather than 'physics' based layout. For example:

These are two regular fullerenes that work very well. The algorithm is simple in principle:

1) Given a planar embedding G, calculate the inner dual id(G) and the 'face layers'.
2) The innermost layer is the 'core' which is one of: a single vertex, a connected pair, or a cycle.
3) Layout the core, and then each layer outwards, by spoke and arch.

So, to explain some of this; a 'face layer' is a set of faces all at the same distance from the outer cycle, measured by graph distance on id(G). So the faces adjacent to the outer cycle are the first layer, and the second layer is adjacent to that, and so on. This is roughly illustrated here:

The concentric circles represent the layers of faces, with the innermost being the core. On the right is a cartoon of two spokes on an outer path, connected by a dashed line to show the arch. The outer path starts off as the edges around the core, and is replaced by the arches of the next level.

This method does work - it creates diagrams that are not too horrible - but does not give as good results for the less-symmetric examples. Something like Bojan Mohar's circle-packing method would probably be better, or even a properly implemented spring layout...

Here are some final examples, all with paired-cores:

(The colors are signature classes, by the way. All images made using CDK's new classes. Code here).

### How many isomers of C4H11N are there?

One of the most popular queries that lands people at this blog is about the isomers of C4H11N - which I suspect may be some kind of organic chemistry question on student homework. In any case, this post will describe how to find all members of a small space like this by hand rather than using software.

Firstly, lets connect all the hydrogens to the heavy atoms (C and N, in this case). For example:

Now eleven hydrogens can be distributed among these five heavy atoms in various ways. In fact this is the problem of partitioning a number into a list of other numbers which I've talked about before. These partitions and (possible) fragment lists are shown here:

One thing to notice is that all partitions have to have 5 parts - even if one of those parts is 0. That's not strictly a partition anymore, but never mind. The other important point is that some of the partitions lead to multiple fragment lists - [3, 3, 2, 2, 1] could have a CH+NH2 or an NH+CH2.

The final step is to connect u…

### Havel-Hakimi Algorithm for Generating Graphs from Degree Sequences

A degree sequence is an ordered list of degrees for the vertices of a graph. For example, here are some graphs and their degree sequences:

Clearly, each graph has only one degree sequence, but the reverse is not true - one degree sequence can correspond to many graphs. Finally, an ordered sequence of numbers (d1 >= d2 >= ... >= dn > 0) may not be the degree sequence of a graph - in other words, it is not graphical.

The Havel-Hakimi (HH) theorem gives us a way to test a degree sequence to see if it is graphical or not. As a side-effect, a graph is produced that realises the sequence. Note that it only produces one graph, not all of them. It proceeds by attaching the first vertex of highest degree to the next set of high-degree vertices. If there are none left to attach to, it has either used up all the sequence to produce a graph, or the sequence was not graphical.

The image above shows the HH algorithm at work on the sequence [3, 3, 2, 2, 1, 1]. Unfortunately, this produce…

### Generating Trees

Tree generation is a well known (and solved!) problem in computer science. On the other hand, it's pretty important for various problems - in my case, making tree-like fusanes. I'll describe here the slightly tortuous route I took to make trees.

Firstly, there is a famous theorem due to Cayley that the number of (labelled) trees on n vertices is nn - 2 which can be proved by using PrÃ¼fer sequences. That's all very well, you might well say - but what does all this mean?

Well, it's not all that important, since there is a fundamental problem with this approach : the difference between a labelled tree and an unlabelled tree. There are many more labeled trees than unlabeled :

There is only one unlabeled tree on 3 vertices, but 3 labeled ones
this is easy to check using the two OEIS sequences for this : A000272 (labeled) and A000055 (unlabeled). For n ranging from 3 to 8 we have [3, 16, 125, 1296, 16807, 262144] labeled trees and [1, 2, 3, 6, 11, 23] unlabeled ones. Only 23 …