Cycles on Lattices

So, first a small correction to the last post : a 'honeycomb' layout is a cycle on a hexagonal lattice, while the other layouts there are not on-lattice. I've renamed them 'flower' layouts, as I'm not sure if they have a name - they're not at all new, I just haven't looked! To be clear:

these are two different layouts of the same 12-cycle. As it happens, the honeycomb layout is also a [5, 5, 5]-flower layout with straight edges on the first edge of the outer cycle. However, not every honeycomb is a flower.

There might be many ways to do it, but one way to make honeycomb layouts is to find cycles on a hexagonal lattice. Assuming, of course, that you have such a lattice already - it's not terribly hard to make one, but connecting it up properly is a bit fiddly. Luckily, it turned out that making the dual of a triangular lattice is slightly easier. To see how this works, consider the three possible (regular) planar lattices:

The grey dots are actually the dual lattice of the black lattice. Click-for-bigger to see the lines connecting the dual points. Of course, a square lattice (on the left) is self-dual, while triangular and hexagonal lattices are duals of each other.

Fine, so given such a hex-lattice, how to get cycles of the right size? Well, the simplest way might be to find all cycles in a grid of some size, and just take ones of whatever size. This ... works - but not very well. Here are the 10-cycles from the hex-grid above:

clearly they are all the same! In fact, this is one of the smaller 'families' (orbits) of cycles. There are 78 cycles of size 12, and 2,082 of size 22. Here is a table:
Clearly, this is not the right approach.

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…

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…

Generating Trees

Tree generation is a well known (and solved!) problem in computer science. On the other hand, it's pretty important for various problems - in my case, making tree-like fusanes. I'll describe here the slightly tortuous route I took to make trees.

Firstly, there is a famous theorem due to Cayley that the number of (labelled) trees on n vertices is nn - 2 which can be proved by using Prüfer sequences. That's all very well, you might well say - but what does all this mean?

Well, it's not all that important, since there is a fundamental problem with this approach : the difference between a labelled tree and an unlabelled tree. There are many more labeled trees than unlabeled :

There is only one unlabeled tree on 3 vertices, but 3 labeled ones
this is easy to check using the two OEIS sequences for this : A000272 (labeled) and A000055 (unlabeled). For n ranging from 3 to 8 we have [3, 16, 125, 1296, 16807, 262144] labeled trees and [1, 2, 3, 6, 11, 23] unlabeled ones. Only 23 …