Skip to main content

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 assumed (naively) that the blocks and trees were connected together in a kind of 'meta-tree' - which I called the part-tree. Actually, it is possible to have cycles in this part-graph :


This complicates things. Especially since it means the part-graphs themselves can have blocks and trees of their own! In any case, for a large proportion of graphs, this does not matter. Here are drawings for all the planar graphs on 6 vertices:


Where the only real problems remaining look to be for the block embedding/layout of graphs with large numbers of edges. These are the ones on the lower row - with some examples having line crossings.

Comments

Mbala said…
The benzene ring is listed twice in your chart. A potential bug?
gilleain said…
Well noticed! I'm reasonably sure that is due to a layout bug for the second benzene (or 6-cycle). The underlying set of graphs are non-redundant. Also, due to the way they were generated, the graphs are roughly in order of increasing number of edges, so the it wouldn't make sense for the second 6-cycle to be where it is in the list (increasing left-to-right, top-to-bottom).

Popular posts from this blog

Listing Degree Restricted Trees

Although stack overflow is generally just an endless source of questions on the lines of "HALP plz give CODES!? ... NOT homeWORK!! - don't close :(" occasionally you get more interesting ones. For example this one that asks about degree-restricted trees. Also there's some stuff about vertex labelling, but I think I've slightly missed something there.

In any case, lets look at the simpler problem : listing non-isomorphic trees with max degree 3. It's a nice small example of a general approach that I've been thinking about. The idea is to:
Given N vertices, partition 2(N - 1) into N parts of at most 3 -> D = {d0, d1, ... }For each d_i in D, connect the degrees in all possible ways that make trees.Filter out duplicates within each set generated by some d_i. Hmm. Sure would be nice to have maths formatting on blogger....

Anyway, look at this example for partitioning 12 into 7 parts:

At the top are the partitions, in the middle the trees (colored by degree) …

Common Vertex Matrices of Graphs

There is an interesting set of papers out this year by Milan Randic et al (sorry about the accents - blogger seems to have a problem with accented 'c'...). I've looked at his work before here.

[1] Common vertex matrix: A novel characterization of molecular graphs by counting
[2] On the centrality of vertices of molecular graphs

and one still in publication to do with fullerenes. The central idea here (ho ho) is a graph descriptor a bit like path lengths called 'centrality'. Briefly, it is the count of neighbourhood intersections between pairs of vertices. Roughly this is illustrated here:


For the selected pair of vertices, the common vertices are those at the same distance from each - one at a distance of two and one at a distance of three. The matrix element for this pair will be the sum - 2 - and this is repeated for all pairs in the graph. Naturally, this is symmetric:


At the right of the matrix is the row sum (∑) which can be ordered to provide a graph invarian…

How many isomers of C4H11N are there?

One of the most popular queries that lands people at this blog is about the isomers of C4H11N - which I suspect may be some kind of organic chemistry question on student homework. In any case, this post will describe how to find all members of a small space like this by hand rather than using software.

Firstly, lets connect all the hydrogens to the heavy atoms (C and N, in this case). For example:


Now eleven hydrogens can be distributed among these five heavy atoms in various ways. In fact this is the problem of partitioning a number into a list of other numbers which I've talked about before. These partitions and (possible) fragment lists are shown here:


One thing to notice is that all partitions have to have 5 parts - even if one of those parts is 0. That's not strictly a partition anymore, but never mind. The other important point is that some of the partitions lead to multiple fragment lists - [3, 3, 2, 2, 1] could have a CH+NH2 or an NH+CH2.

The final step is to connect u…