### Automorphism groups and fragment graphs

Structure generation involves not just graph theory, but group theory. Or, I should say, it does in some of the papers I have read. For example, in this paper by J.L.Faulon, there is the sentence:
"The two main steps are to compute the orbits of the automorphism group of G and to saturate all the atoms of a chosen orbit
which may well be incomprehensible to many readers, except if the reader is a mathematician.

I am no mathematician, but thanks to some books on groups, I now understand both what an automorphism group is and what an orbit is. On the other hand, I also believe that this definition of how the algorithm works is overly complex. A more simple term might just be "fragment sets" - as it is fairly clear, if not mathematically exact. So, for the fragment graph [CH3, CH3, CH2, CH2, CH, CH] the fragment set is [CH3, CH2, CH].

Anyway, here is a short analysis of the automorphism group of the fragment graph [CH2, CH2]. This first image shows the tiny group of permutations that swaps the two fragments:

The notation is taken from an excellent book called "Visual Group Theory" that is also associated with some software called group explorer on sourceforge. It might be quite general, I suppose (and I hope I'm using it right), but it shows the permutation that swaps the fragments as a circled s. This is an automorphism with respect to the edges - in other words, after the swap, there are still bonds between [1, 2], [2, 3], [4, 5], and [5-6].

Another part of the automorphism group is a 'flip' like:

which is a little more complex, but shows how 'flipping' each fragment separately combines to form four possible permutations. If this does not seem particularly tricky, consider what happens if you take the direct product of these two groups:

Assuming I have done it right, this should show most (all?) of the automorphisms of the fragment graph. It does look pretty cool, but I don't think that it gets me any closer to implementing the cursed algorithm :)

Anonymous said…
Anonymous said…
Give never the wolf the wether to keep.

### 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 …