### Generating Chessboards With K-Independant Vertex Sets

After looking at Julio Peironcely's poster on generating chemical structures, where he describes using Canonical Path Augmentation (I think due to Brendan McKay) I went looking for more about it. One thing I found was this talk/slideshow by Mathieu Dutour SikiriÄ‡ - incidentally coauthor of a nice book on Chemical Graphs.

Anyway; that's the context. Now : about chessboards? Well one of the examples given for augmentation (or 'orderly') schemes of independent vertex sets. I'm not sure what made me think of chessboards for this, but I think it's a fairly standard simple toy example. Let me show an example for 3x3 boards:

So, these are all the three by three boards where no two black squares share an edge. In other words, if the board is considered as a grid-shaped graph, then these are the k-independent vertex sets. So the question is : how to generate these?

The simple way, of course, is just to fill in every square and eliminate those boards that have pairs of black squares across an edge. This is the 'brute force' approach, and scales badly : there are 29 boards but we only want 20 of these, or just 3%. For 4x4 boards, this fraction is even smaller - 131 of 216 boards which is only 0.19%. This is 0.0026% for 5x5 boards.

So, the number of boards increases rapidly as the size increases and any way of decreasing this large search space is essential. The approach outlined in Mathieu's talk is quite simple : only try sets (boards) that are the minimal representative in their orbit. Ok, so maybe that doesn't sound so simple :) but look at this:

Here are 3x3 boards, with a numbering (any one will do), the equivalence classes of the cells, and a set of orbits. Lets say we've just generated {0, 4, 6} : how do we check if it is the minimal representative? Well, so long as we have the automorphism group of the board we just apply each permutation in the group to the set of numbers and check to see if any are smaller. In this case, {0, 2, 4} is smaller, so we don't consider {0, 4, 6}.

It's really as simple as that. Of course, we have to check each newly added cell to make sure it is not adjacent, but that can be done without consideration of previously generated boards. This means we don't have to store solutions, which means it can be done in parallel without communication. See how many there are just for 5x5 boards:

I suppose that one last question about all this is : so what? Does this have anything to do with chemistry? Well, actually, it does. Consider the bond order assignment problem : the second image shows all the possible assignments, which you may notice has duplicates (3, 5, 10, and 12 for example). Also consider the line graph approach to double bond systems where again the second image shows a pair of colorings of line graphs. In fact, these are k-independent sets...

Oh, and there is code here

chembioinfo said…
interesting work.

Lemma: When you generate the next solution state, you could ignore previous state.

So this means previous state checks are reduced.

Now the number of solutions are not effected, right?
gilleain said…
Missed this comment, sorry!

Yes, this is exactly right - you can break up the problem into parts, and do each independently.

So, the second post on this topic about graphs should really mention that you can start with a graph on n vertices, and generate all the children on (n+1) vertices.

This can be done for different parents on completely different machines, and you would still get the same number of children.

### 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) …