Signature bond compatibility

So, given my previous posts on what Faulon's signatures are, here is an explanation of how they are used in the structure enumeration algorithm that I am almost finished implementing.

The core test in this algorithm is for compatible bonds. Two atoms are only joined if : a) they have compatible target signatures and b) there are less than the target number of bonds already. A target signature here is just a signature that is set on the atom for it to match, like a pattern.

The first of these tests is illustrated here:

Another (overly) complex diagram! But the formula here is a bit difficult to interpret otherwise. In the top left corner is a graph G (slightly resembling hexane without the hydrogens) which is, by convention, composed of vertices (V) and edges (E).

The equation to the right of the graph defines part of the condition for a compatible bond. The tau terms are just target signatures, as shown on the upper right. The tricky term is ^h-1σ_τ(y)(z) which means 'a signature starting from the neighbour z of y in the subgraph defined by τ(y)'. This requires using a target signature (τ(y)) as if it was a subgraph - shown on the bottom left for the target b and the neighbour n.

The same process is repeated on the bottom right of the figure for b and m - which matches the height - 1 target signature for c. This should make sense, since the atoms labelled with b in the graph are attached to both a and c - so the signature for b must be compatible with both. It is easy to check that a and c are not compatible, and cannot therefore be bonded.

Comments

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...

Some Stuff

Search This Blog