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!

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Adamantane, Diamantane, Twistane

After cubane, the thought occurred to look at other regular hydrocarbons. If only there was some sort of classification of chemicals that I could use look up similar structures. Oh wate, there is . Anyway, adamantane is not as regular as cubane, but it is highly symmetrical, looking like three cyclohexanes fused together. The vertices fall into two different types when colored by signature: The carbons with three carbon neighbours (degree-3, in the simple graph) have signature (a) and the degree-2 carbons have signature (b). Atoms of one type are only connected to atoms of another - the graph is bipartite . Adamantane connects together to form diamondoids (or, rather, this class have adamantane as a repeating subunit). One such is diamantane , which is no longer bipartite when colored by signature: It has three classes of vertex in the simple graph (a and b), as the set with degree-3 has been split in two. The tree for signature (c) is not shown. The graph is still bipartite accordin

1,2-dichlorocyclopropane and a spiran

As I am reading a book called "Symmetry in Chemistry" (H. H. Jaffé and M. Orchin) I thought I would try out a couple of examples that they use. One is 1,2-dichlorocylopropane : which is, apparently, dissymmetric because it has a symmetry element (a C2 axis) but is optically active. Incidentally, wedges can look horrible in small structures - this is why: The box around the hydrogen is shaded in grey, to show the effect of overlap. A possible fix might be to shorten the wedge, but sadly this would require working out the bounds of the text when calculating the wedge, which has to be done at render time. Oh well. Another interesting example is this 'spiran', which I can't find on ChEBI or ChemSpider: Image again courtesy of JChempaint . I guess the problem marker (the red line) on the N suggests that it is not a real compound? In any case, some simple code to determine potential chiral centres (using signatures) finds 2 in the cyclopropane structure, and 4 in the

General Graph Layout : Putting the Parts Together

An essential tool for graph generation is surely the ability to draw graphs. There are, of course, many methods for doing so along with many implementations of them. This post describes one more (or perhaps an existing method - I haven't checked). Firstly, lets divide a graph up into two parts; a) the blocks, also known as ' biconnected components ', and b) trees connecting those blocks. This is illustrated in the following set of examples on 6 vertices: Trees are circled in green, and blocks in red; the vertices in the overlap between two circles are articulation points. Since all trees are planar, a graph need only have planar blocks to be planar overall. The layout then just needs to do a tree layout on the tree bits and some other layout on the embedding of the blocks. One slight wrinkle is shown by the last example in the image above. There are three parts - two blocks and a tree - just like the one to its left, but sharing a single articulation point. I had

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