Skip to main content

Timing Four Augmentation Algorithms

There are many possible ways to canonically augment graphs, but here I'm picking two pairs of possibilities - vertex vs edge augmentation, and filtering duplicates vs picking a representative from symmetrically equivalent positions. So, the four algorithms are vertex/filter (V/Fil), vertex/symmetric (V/Sym), edge/filter (E/Fil), and edge/symmetric (E/Sym).

Here is a graph of log-average timings (in milliseconds) of the four implementations running on graphs of 4-8 vertices. One very important caveat is that the graph counts for 7 and 8 vertices are not 100% correct.


The rows in blue on the table (4, 5, 6, 7, 8) are the log-averages of rows abov; so 4 = log(average(4a, 4b, 4c)), etc. The full spreadsheet is available here on github (as a .numbers file), or the code is here. I'm not particularly confident in the crude System.getTimeMilliseconds() as a timing method.

However, the striking thing to me is that these numbers suggest that for larger input sizes (n > 6), V/Sym is consistently better than E/Fil. In other words, augmenting by vertex at non-equivalent positions may be faster than augmenting by edges and then filtering duplicates.

One final point that is relevant for molecule generation is that these graphs have no degree-limit apart from n-1 edges for a graph on n. So the behaviour could be different for low-degree graphs (deg <= 4) especially since there will be less symmetries (I think?) for each parent.

Comments

Popular posts from this blog

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) …