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.

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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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