Skip to main content

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.

Comments

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.

Popular posts from this blog

Adamantane, Diamantane, Twistane

After cubane, the thought occurred to look at other regular hydrocarbons. If only there was some sort of classification of chemicals that I could use look up similar structures. Oh wate, there is . Anyway, adamantane is not as regular as cubane, but it is highly symmetrical, looking like three cyclohexanes fused together. The vertices fall into two different types when colored by signature: The carbons with three carbon neighbours (degree-3, in the simple graph) have signature (a) and the degree-2 carbons have signature (b). Atoms of one type are only connected to atoms of another - the graph is bipartite . Adamantane connects together to form diamondoids (or, rather, this class have adamantane as a repeating subunit). One such is diamantane , which is no longer bipartite when colored by signature: It has three classes of vertex in the simple graph (a and b), as the set with degree-3 has been split in two. The tree for signature (c) is not shown. The graph is still bipartite accordin

Király's Method for Generating All Graphs from a Degree Sequence

After posting about the Hakimi-Havel  theorem, I received a nice email suggesting various relevant papers. One of these was by Zoltán Király  called " Recognizing Graphic Degree Sequences and Generating All Realizations ". I have now implemented a sketch of the main idea of the paper, which seems to work reasonably well, so I thought I would describe it. See the paper for details, of course. One focus of Király's method is to generate graphs efficiently , by which I mean that it has polynomial delay. In turn, an algorithm with 'polynomial delay' takes a polynomial amount of time between outputs (and to produce the first output). So - roughly - it doesn't take 1s to produce the first graph, 10s for the second, 2s for the third, 300s for the fourth, and so on. Central to the method is the tree that is traversed during the search for graphs that satisfy the input degree sequence. It's a little tricky to draw, but looks something like this: At the top

1,2-dichlorocyclopropane and a spiran

As I am reading a book called "Symmetry in Chemistry" (H. H. Jaffé and M. Orchin) I thought I would try out a couple of examples that they use. One is 1,2-dichlorocylopropane : which is, apparently, dissymmetric because it has a symmetry element (a C2 axis) but is optically active. Incidentally, wedges can look horrible in small structures - this is why: The box around the hydrogen is shaded in grey, to show the effect of overlap. A possible fix might be to shorten the wedge, but sadly this would require working out the bounds of the text when calculating the wedge, which has to be done at render time. Oh well. Another interesting example is this 'spiran', which I can't find on ChEBI or ChemSpider: Image again courtesy of JChempaint . I guess the problem marker (the red line) on the N suggests that it is not a real compound? In any case, some simple code to determine potential chiral centres (using signatures) finds 2 in the cyclopropane structure, and 4 in the