Two Different Ways to Handle Disconnected Graphs when Generating Graphs with Signatures

Disconnected graphs were mentioned in this post on OMG's algorithm. I also said that signatures could not handle such graphs, but this is not totally true. Here are a couple of approaches to handle them.

Firstly, why is this a problem at all? Well, a signature of a graph is a bit like a spanning tree - and the canonical signature is the maximal string form of the set of trees. However, for disconnected graphs you get a spanning forest - which is just a set of spanning trees. The maximum tree from the forest will no longer span the whole graph.

The image shows an example of this for a very simple graph. The upper panel shows a disconnected graph and its corresponding forest. The lower panel shows two different labelings of the same graph (A, B) and their canonical form. The algorithm for canonicalization simply has to 'paste' together canonical labels or signature strings, and has to make sure that a canonical signature is generated for each component. The components are ordered by their canonical strings.

An alternative solution to the problem for generating graphs is mentioned in a blog post by Derrick Stolee. Simply choose the canonical deleted edge to be one that will not disconnect the graph. For example, with an edge-addition canonical path algorithm, any edge that is a bridge (as discussed yesterday) would disconnect the graph if deleted. So the canonical edge is no longer the last edge in the canonically labelled graph but the last non-bridge edge.

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