### Dodecahedrane has 12 faces, right?

So there was this guy called Euler, and he had a formula that goes something like F = E - V + 2. Well, actually it is χ = V - E + F, where χ is the Euler characteristic, and this is equal to 2 for polyhedra. Anyway, the point is that dodecahedrane has 12 faces (cycles).

For the SSSRFinder, however, it has only 11; which is annoying. Moreover the ring equivalence class method only distinguishes based on the underlying simple graph - in other words it ignores bond order. In some applications this might be exactly what is needed, but I'm glad that my method gives a more detailed result:

So, apart from being a ridiculously detailed image, the above shows the face (ring, cycle) equivalence classes for dodecahedrane with a particular double bond network. Clearly any face could be 'glued' to another along one of the edges, following the vertex classes. All possible combinations of faces are shown in the 'face quotient graph' at the bottom right.

12 faces = 11 bonds to cut to remove all cycles... I think the SSSR is build around the latter concept. alpha-pinene has three unique cycles, but only two bonds to cut, and two a SSSR of size two?
gilleain said…
I really should read the papers that the code references. What I suspect is happening is that the SSSRFinder is properly implemented, but that the algorithm doesn't work the way I expect it to. Oh, and I checked and yes alpha-pinene does have 2 rings in its SSSR.

### 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 Prüfer 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 …