### 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…

### 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) …