### 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 up these fragments in all possible ways. Easy to write, but hard to do - one way is shown below for [3, 3, 2, 2, 1]:

The two possible fragment lists are shown at the top, and below is one way to connect each list. It so happens that there are two ways to connect each list to give different molecules. The molecules are on the right of the figure.

The rules to follow here are : 1) divide the fragments into two lists as close as possible to the same number of 'spare bonds' (clear circles), 2) connect between the lists (dark circles + lines), 3) leave spare circles unless it's the final step.

Using this procedure should lead to all molecules with the same formula, with some trial and error. Hope this helps! Oh, and as a poster on reddit/r/chemistry pointed out, remember to count stereoisomers...

chanaka lakshan said…
Thanks

### 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…

### Common Vertex Matrices of Graphs

There is an interesting set of papers out this year by Milan Randic et al (sorry about the accents - blogger seems to have a problem with accented 'c'...). I've looked at his work before here.

[1] Common vertex matrix: A novel characterization of molecular graphs by counting
[2] On the centrality of vertices of molecular graphs

and one still in publication to do with fullerenes. The central idea here (ho ho) is a graph descriptor a bit like path lengths called 'centrality'. Briefly, it is the count of neighbourhood intersections between pairs of vertices. Roughly this is illustrated here:

For the selected pair of vertices, the common vertices are those at the same distance from each - one at a distance of two and one at a distance of three. The matrix element for this pair will be the sum - 2 - and this is repeated for all pairs in the graph. Naturally, this is symmetric:

At the right of the matrix is the row sum (∑) which can be ordered to provide a graph invarian…

### Signatures with user-defined edge colors

A bug in the CDK implementation of my signature library turned out to be due to the fact that the bond colors were hard coded to just recognise the labels {"-", "=", "#" }. The relevant code section even had an XXX above it!

Poor show, but it's finally fixed now. So that means I can handle user-defined edge colors/labels - consider the complete graph (K5) below:

So the red/blue colors here are simply those of a chessboard imposed on top of the adjacency matrix - shown here on the right. You might expect there to be at least two vertex signature classes here : {0, 2, 4} and {1, 3} where the first class has vertices with two blue and two red edges, and the second has three blue and two red.

Indeed, here's what happens for K4 to K7:

Clearly even-numbered complete graphs have just one vertex class, while odd-numbered ones have two (at least?). There is a similar situation for complete bipartite graphs:

Although I haven't explored any more of these…