Skip to main content

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!

Comments

Popular posts from this blog

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) …

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…

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…