### Dodecahedrane has 12 faces, right?

So there was this guy called Euler, and he had a formula that goes something like F = E - V + 2. Well, actually it is χ = V - E + F, where χ is the Euler characteristic, and this is equal to 2 for polyhedra. Anyway, the point is that dodecahedrane has 12 faces (cycles).

For the SSSRFinder, however, it has only 11; which is annoying. Moreover the ring equivalence class method only distinguishes based on the underlying simple graph - in other words it ignores bond order. In some applications this might be exactly what is needed, but I'm glad that my method gives a more detailed result:

So, apart from being a ridiculously detailed image, the above shows the face (ring, cycle) equivalence classes for dodecahedrane with a particular double bond network. Clearly any face could be 'glued' to another along one of the edges, following the vertex classes. All possible combinations of faces are shown in the 'face quotient graph' at the bottom right.

12 faces = 11 bonds to cut to remove all cycles... I think the SSSR is build around the latter concept. alpha-pinene has three unique cycles, but only two bonds to cut, and two a SSSR of size two?
gilleain said…
I really should read the papers that the code references. What I suspect is happening is that the SSSRFinder is properly implemented, but that the algorithm doesn't work the way I expect it to. Oh, and I checked and yes alpha-pinene does have 2 rings in its SSSR.

### 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) …