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### Double Bonds and Edge Colorings

There is an effort going on to improve the double bond assignment machinery in the CDK, which is great. Of interest to me, however, is how many possible arrangements of double bonds can you have in fused ring systems. This was mentioned in at least one previous post - or perhaps two.

However, lets get a very rough upper bound; how many ways are there to color the N edges of a graph with two colors? This is 2 to the power N, or the set of all subsets of the edges. Of course, many of these are chemically meaningless, where atoms have too high a valence. So filter out those where adjacent edges have the same color - or more exactly, where adjacent edges are colored with the 'double bond' color (let's call it '2').

The image shows a sketch of the simple procedure (above) and a slightly better approach (below). The better way of doing things is similar to the k-independent chessboard solution (sorry to link to my own pages so much - but it is relevant!). The idea is to use the symmetries of the graph (or chessboard) to prune solutions that must have already been tried.

This does seem to work, but there are a huge number of solutions - even for relatively small graphs. For example, fusanes with just 4 rings have thousands of partial solutions. For these four examples, there is quite a variation:

The most symmetric (green) example has the least - of course - but the large numbers of solutions of size 5 for the orange example is odd. It's a bit difficult to look through these to see why, unfortunately. What is much easier is the 'full' solutions of size 9. For example, for the green graph:

A bit difficult to distinguish these drawings - actual double lines are clearer, it seems. Anyway, below each is a kind of 'name' based on the bond equivalence classes (a-e) in the lower center. So, "b3.d3" means three 'b' bonds and three 'd' bonds are colored.

### 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 …