### Colored Tree Paths to Represent Bond Order Assignements

A couple of different sources pointed me to a paper by Dehof et al (Bioinformatics, 2011; doi: 10.1093/bioinformatics/btq718) about bond order assignment by various methods. One of them is the A*-algorithm, which finds paths. Not in the molecule graph, but in a kind of tree of possible bond combinations. Precisely, the tree has a layer for each decision, and a leaf for each combination of decisions.

For example, here is the usual example of a square four-cycle (G):

with the tree shown below. If we pick a particular path through the tree - say to leaf 5 - we get the sequence of colors in the box in the upper center of the image. This corresponds to the assignment on the upper right. Clearly, it has too many double bonds to be carbon skeleton, but still.

So, it's a nice, simple way to represent the set of solutions. Well, it could be even simpler but representing it as a tree lets the paper authors assign weights to the edges based on the chemical rules outlined in work they reference (Wang et al, 2006). What happens if you don't have any weights? Here are all the assignments from the tree:

where the assignments are labelled by the leaf number (0-15) and have a letter label (A-F). These letters stand for 'equivalence classes' - really just isomorphism classes. Oh, and the grey vertices have unlikely atom types for carbons.

### 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) …