### Millions of Graphs : Slow Yet Correct Generation

My newest version of canonical path augmentation code for generating graphs has reached a new high point - generating 11,716,571 graphs on ten vertices. Of course, it also gets the number of nines (261,080) and the number of eights (11,117) correct as well ... which is great, but I'm cautious about declaring it 'correct'. Especially given the last version did not get the sevens and eights right. See, for example these past failures:

So how does it get the right answer? Well, it now properly uses the method mentioned in this post to only pick canonical deletions that are not cut-vertices. That turns out only to be necessary for graphs on 8 vertices, but you still have to check this for all augmentations, which seems expensive. However, there was a more fundamental problem; consider the example below (basically nicked from Derick Stolee's blog post):

Obviously A and B are isomorphic, yet how do we properly distinguish them? Well, the key is the set of vertices added to - on the image, these are the labels on the edges between graphs : {0}, {1, 3}, etc. When a new graph is created, a vertex is chosen - using canonical labelling, in my case - and the vertices attached to it must be the ones we used to make that augmented graph. I was checking the set of augmented vertices in the automorphism group of the parent, not the child.

So, the canonical checking is now better. I seem to have written a thousand of these methods, but this one (I think!) finally does it right. What I was getting wrong was checking the orbit of the canonical deletion vertex, and not the orbit of the set of vertices it was being connected to. Great! Now what? How long does it take? See this, where the purple line is the new code, and the others are older attempts:

Clearly the problem now is that of verifying the results - it's quite slow to generate these large datasets, and storing them (uncompressed) takes a lot of space. The nines took minutes and megabytes of space, while the tens took hours and over a gigabyte. At this rate, the 11s would take days and 10s of gigabytes. In any case - where do you stop?

Tobias Kind said…
Hi,
I tried to download the code but the old URL gives the good ole 404.
https://github.com/gilleain/generate/blob/master/src/test/scheme3/TimingTests.java

Also for those people interested, but not able to install all JAVA dependencies,
a compiled JAR with *.sh *.bat *.exe would be good (that would be me and others :-).

For these benchmarks it would be also useful to have the machine, CPU
and disk/ramdisk specifications.

Good to read some new stuff (heavy github contributions in April/May 2015)
Cheers
Tobias

### 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…

### The Gale-Ryser Theorem

This is a small aside. While reading a paper by Grüner, Laue, and Meringer on generation by homomorphism they mentioned the Gale-Ryser (GR) theorem. As it turns out, this is a nice small theorem closely related to the better known Erdős-Gallai (EG).

So, GR says that given two partitions of an integer (p and q) there exists a (0, 1) matrixA iff p*dominatesq such that the row sum vector r(A) = p and the column sum vector c(A) = q.

As with most mathematics, that's quite terse and full of terminology like 'dominates' : but it's relatively simple. Here is an example:

The partitions p and q are at the top left, they both sum to 10. Next, p is transposed to get p* = [5, 4, 1] and this is compared to q at the bottom left. Since the sum at each point in the sequence is greater (or equal) for p* than q, the former dominates. One possible matrix is at the top left with the row sum vector to the right, and the column sum vector below.

Finally, the matrix can be interpreted as a bi…

### 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…