### Expensive and Exhaustive Ring Finding

As the title states - this is a computationally expensive way to get all rings in a graph, but it's fairly simple, and illustrates some nice principles. For a better way to do things, perhaps Rich Apodaca's description of the Hanser, Jauffret, and Kaufmann algorithm would suit.

Anyway, back to the expensive way. The set of cycles in a graph form what is called a 'cycle space' - which I didn't understand at all for a while, but is not actually that hard. For example, here is a basis set for the cycle space on a 3x3 hexagonal lattice:

Looks like a bunch of cycles, really. The important thing is that it is possible to combine any subset of these cycles to get another cycle (or one of the other cycles in the basis). By 'combine' we mean XOR or the symmetric difference of the edge vectors. This sounds more complicated than necessary, so it's useful to consider a simple example. Well, the example is simple - the picture is not:

On the left here is a graph (top left) and a cycle basis (lower left; A-C). Each cycle is just a bit set (vector) of edges. So cycle A is the set of edges {a, b, d, j} and cycle C is {e, f, g, h}. On the right is one example of a combination of two cycles - known as a ring sum - to produce a new cycle D.

It should be clear that A + B is equal to {a, b, d, j} + {c, d, f, i} - {d}. In words, this is the union of the edge sets minus their intersection. One important point is that the ring sum of two disconnected cycles is just those two cycles. So A + C = A + C...

All the cycles in the graph can be generated in this way - for example A + B + C = {a, b, c, e, g, h, i, j} or the outer cycle of the graph. I think I am right in saying that the order of combination does not matter.

So why is this expensive? Well, the problem lies in the fact that there are a large number of subsets of the basis set - 2 to the power |S|, for the set S. Many of these will be the same cycle, so there are undoubtedly ways to cleverly choose subsets; I'm not sure what these are.

Finally, here is an example of a set of (unique) cycles on a lattice. The procedure to get these was ridiculous : a) find all rings as above, b) layout to get the faces, c) construct the inner dual, d) make a signature of dual to filter duplicates.

Obviously, from a drawing point of view, the top row are nicely symmetric ways to layout 18-rings. Probably the top-center and top-right are the best, as they are also more convex.

### Generating Dungeons With BSP Trees or Sliceable Rectangles

So, I admit that the original reason for looking at sliceable rectangles was because of this gaming stackoverflow question about generating dungeon maps. The approach described there uses something called a binary split partition tree (BSP Tree) that's usually used in the context of 3D - notably in the rendering engine of the game Doom. Here is a BSP tree, as an example:

In the image, we have a sliced rectangle on the left, with the final rectangles labelled with letters (A-E) and the slices with numbers (1-4). The corresponding tree is on the right, with the slices as internal nodes labelled with 'h' for horizontal and 'v' for vertical. Naturally, only the leaves correspond to rectangles, and each internal node has two children - it's a binary tree.

So what is the connection between such trees and the sliceable dual graphs? Well, the rectangles are related in exactly the expected way:

Here, the same BSP tree is on the left (without some labels), and the slicea…

### 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…