### Line Graphs and Double Bonding Systems

After looking at a CDK tool for fixing bond orders for aromatic systems (DeduceBondSystemTool in the smiles package) I wondered if there was a more general approach. That is, the problem is to take a molecular graph with no double bonds and generate all possible double bonded systems.

One possibility might be to first convert the graph (G) into a form known as a line graph (lg(G)) where every vertex in lg(G) is an edge in G. If these vertices are labelled to represent the bond order, then an aromatic system has a particular line graph. For example, here is benzene:

The dashed lines show the construction of the line graph, and the labels '-' and '=' mean single and double. Now obviously, the two resulting graphs are essentially the same, so it would be nice to remove this redundancy. An example of two different bonding systems comes from phenanthrene:

Which is great, but how to generate all non-redundant colorings of the line graphs? Since a line graph is just a graph, it can have a signature, and a signature quotient graph. This can then be colored:

However, in this example it is necessary to 'half-color' some of the vertices of the quotient graph ... which doesn't quite seem to work. The numbers in between the colored quotient graphs show how many line graph vertices are in each of the symmetry classes.

In any case, this is a bit of a toy problem, with only a partial solution, but here is a code repository for a sketch of the code. Note that the algorithm is missing!

### 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…

### Generating Dungeons With BSP Trees or Sliceable Rectangles

So, I admit that the original reason for looking at sliceable rectangles was because of this gaming stackoverflow question about generating dungeon maps. The approach described there uses something called a binary split partition tree (BSP Tree) that's usually used in the context of 3D - notably in the rendering engine of the game Doom. Here is a BSP tree, as an example:

In the image, we have a sliced rectangle on the left, with the final rectangles labelled with letters (A-E) and the slices with numbers (1-4). The corresponding tree is on the right, with the slices as internal nodes labelled with 'h' for horizontal and 'v' for vertical. Naturally, only the leaves correspond to rectangles, and each internal node has two children - it's a binary tree.

So what is the connection between such trees and the sliceable dual graphs? Well, the rectangles are related in exactly the expected way:

Here, the same BSP tree is on the left (without some labels), and the slicea…

### Listing Degree Restricted Trees

Although stack overflow is generally just an endless source of questions on the lines of "HALP plz give CODES!? ... NOT homeWORK!! - don't close :(" occasionally you get more interesting ones. For example this one that asks about degree-restricted trees. Also there's some stuff about vertex labelling, but I think I've slightly missed something there.

In any case, lets look at the simpler problem : listing non-isomorphic trees with max degree 3. It's a nice small example of a general approach that I've been thinking about. The idea is to:
Given N vertices, partition 2(N - 1) into N parts of at most 3 -> D = {d0, d1, ... }For each d_i in D, connect the degrees in all possible ways that make trees.Filter out duplicates within each set generated by some d_i. Hmm. Sure would be nice to have maths formatting on blogger....

Anyway, look at this example for partitioning 12 into 7 parts:

At the top are the partitions, in the middle the trees (colored by degree) …