### 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) matrix A iff p* dominates q 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 bipartite graph as shown in the lower right panel. There is an edge between vertices (i, j) if there is a 1 in the cell for (i, j) in the matrix. It's quite hard to see the connections on a graph drawn like this, but it should be clear that there are only edges between the two sets {0, 1, 2, 3} and {4, 5, 6, 7, 8} and not within these.

There's a nice proof by Manfred Krause of the theorem, and a very readable slideshow on the relationship between GR and EG here called "Variations on the Erdos-Gallai" by Grant Cairns.

Prueba said…
Very nice post and references. Just a little issue: the lower left box in the example should say sum(q) instead of sum(q*), right?
gilleain said…
Ah! Good spot - now changed. Thank you for pointing it out.
Rodrigo said…
Hi! Nice explanation!
I'm not sure about how this theorem will deal with a matrix like the sums are p = [0,0,3], q = [0,0,3]. I know that is not a valid (binary) matrix , but it seems that the theorem is telling me otherwise.
Can you help me with that?
gilleain said…
Hi Rodrigo,

Well, [0,0,3] is not a valid partition either. Another way to think about it is that such a sequence could never be the degree sequence of a simple connected graph, since two of the vertices would necessarily be disconnected since they have degree 0.

### 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…

### 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…