### 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 right is the starting degree sequence - [3, 2, 2, 2, 1] - and there are two graphs at the bottom that realise this sequence. The 'tree of trees' in between is the recursive search through sets of neighbours for vertices in the graph. So the top tree shows the possible choices for neighbours of the last (4th) vertex; the next level shows them for the 3rd vertex, and so on.

The key point here is that only red leaves of a particular tree are valid choices, and these pass through a path of red and black edges. A red edge in the tree represents an edge in the graph, while a black edge indicates no edge from this vertex. So the left hand graph is [{0} : 4, {0, 1} : 3, {0, 1} : 2], using the notation {V0, V1, ..., Vn} : Vm for a set of edges {V0:Vm, V1:Vm, ..., Vn:Vm}. The final edge for the right hand graph (0:1) is not shown as a tree, since the degree sequence is [1, 1, 0, 0, 0] at that point - leaving only one choice.

Also not shown are colors on the internal nodes of the tree. Király's paper describes how to color these nodes so that the algorithm never visits any black leaf. This is vital for efficient output, but I have not ye implemented that part. However, cross-checking the results against the graphs output by McKay's method is promising so far (up to 7 vertices). I should note that Király's method seems to produce isomorphic solutions.

Code for this is here, although it is a somewhat naïve implementation.

Anonymous said…
very Good blogThank you!
Anonymous said…
Thank you for explaining the method. But seems the link for the code does not work.
gilleain said…
Sorry about that. Link is now fixed. Let me know if you have any problems with the code or suggestions to improve it.
lea336 said…
Great post, thank you! I believe the image you posted has a mistake, though: in the right-hand depth 3 tree, the leaf associated with {0,1} is colored red, but the degree sequence [2,0,2] is not graphic. Same is true for the leaf {1,2} and degree sequence [3,0,1]. If I have misunderstood the algorithm, please explain. Thank you
Anonymous said…
Thank you for explaining the method. But code run error selection does not contain a main type. Can you help me. Thanks you so much.

### 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…