Symmetric Generations

I've been trying to work on an implementation of a structure generator algorithm due to Faulon (its a paper in this list somewhere). One problem I have difficulty with is reduction of the number of isomers generated. For example:

It may be hard to read, but the idea is that the full tree of possible ways to attach {Br, Cl, F, I} to c1ccc1 is 4 * 3 * 2 * 1 (you can attach Bromine to all four of the carbons, leaving 3 places to attach chlorine, and so on).

Of course c1(Br)c(Cl)cc1 is the same as c1(Cl)c(Br)cc1 [never mind that C1(Br)=C(Cl)C=C1 is not the same as C1=C(Br)C(Cl)=C1]. Or, in other words, there is a high degree of symmetry in the tree.

There is a way to solve this, perhaps even described in the paper - if I could just understand them...

Comments

Egon Willighagen said…

Symmetry is indeed the main blocker in structure generators. The only way to get performance is to detect symmetry as soon as possible... the deterministic generator we had in CDK 1.0 was doing this, as there is a trick to detect this in the connectivity matrix, but it needs to be normalized. I never had/took the time to fully grasp the math behind it to isolate the bugs in the implementation. But please do look up the literature; it's a must read for your work.

13 March 2009 at 11:55

gilleain said…

Finding the eigenvalues of the matrix, perhaps?

Reading the literature is one thing, understanding it is another. I was reading "Isomorphism, Automorphism Partitioning, and Canonical
Labeling Can Be Solved in Polynomial-Time for Molecular Graphs" (Faulon, 1998) and it's heavy going :(

13 March 2009 at 12:04

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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