### 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.

### 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…

### The Gale-Ryser Theorem

This is a small aside. While reading a paper by Grüner, Laue, and Meringer on generation by homomorphism they mentioned the Gale-Ryser (GR) theorem. As it turns out, this is a nice small theorem closely related to the better known Erdős-Gallai (EG).

So, GR says that given two partitions of an integer (p and q) there exists a (0, 1) matrixA iff p*dominatesq such that the row sum vector r(A) = p and the column sum vector c(A) = q.

As with most mathematics, that's quite terse and full of terminology like 'dominates' : but it's relatively simple. Here is an example:

The partitions p and q are at the top left, they both sum to 10. Next, p is transposed to get p* = [5, 4, 1] and this is compared to q at the bottom left. Since the sum at each point in the sequence is greater (or equal) for p* than q, the former dominates. One possible matrix is at the top left with the row sum vector to the right, and the column sum vector below.

Finally, the matrix can be interpreted as a bi…

### Havel-Hakimi Algorithm for Generating Graphs from Degree Sequences

A degree sequence is an ordered list of degrees for the vertices of a graph. For example, here are some graphs and their degree sequences:

Clearly, each graph has only one degree sequence, but the reverse is not true - one degree sequence can correspond to many graphs. Finally, an ordered sequence of numbers (d1 >= d2 >= ... >= dn > 0) may not be the degree sequence of a graph - in other words, it is not graphical.

The Havel-Hakimi (HH) theorem gives us a way to test a degree sequence to see if it is graphical or not. As a side-effect, a graph is produced that realises the sequence. Note that it only produces one graph, not all of them. It proceeds by attaching the first vertex of highest degree to the next set of high-degree vertices. If there are none left to attach to, it has either used up all the sequence to produce a graph, or the sequence was not graphical.

The image above shows the HH algorithm at work on the sequence [3, 3, 2, 2, 1, 1]. Unfortunately, this produce…