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### Tree Canonization Simplified

While debugging the methods to make canonical signatures, I learned something about tree isomorphism from various sources, including Prof. Valiente's excellent looking book on trees.

One way of checking isomorphism is canonisation, since two trees are only isomorphic if they have the same canonical form. For simple labelled trees, it looks like there is an almost trivial way to get a canonical string representation. Say we have two trees: The are rooted, labelled trees. So the conversion to a canonised string proceeds as follows; for each node, lexicographically sort the string form of the labels of its children, and return the concatenated string. In python this looks like: which...is unreadable. hmmm. Wish there was a better way to get marked-up code into blog posts. Perhaps there is one, and I don't know of it. Anyway, the point is that it is very short.

edit : the code is...

`def printSorted(node): if len(node.children) > 0:   childStrings = [printSorted(child) for child in node.children]   return node.label + "("+ "".join(sorted(childStrings)) + ")" else:   return node.label`

### Comments  Rich Apodaca said…
Simple and clear. I wonder how can you use this to develop a graph canonicalization algorithm?

For displaying source, you might also check out Gist:

http://gist.github.com/

Or maybe this:

http://stackoverflow.com/questions/687213/what-websites-do-you-use-to-post-temporary-code-snippets gilleain said…
Egon : Thanks - I think I saw that before, but never went to the trouble of finding out how to use it in blogspot. This post was helpful:

http://manuelmoeg.blogspot.com/2009/03/def-hello-pass.html

although it's a bit odd that google themselves have not integrated a prettifier. gilleain said…
Rich : From what I have read (and from what I have understood from that) I don't know if this is possible.

I should point out that the method I describe is not my own, but a simplified version of that described by Tarjan and Hopcroft in a paper on, well, graph isomorphism.

Their method only works for planar graphs, and is fiendishly complex - there are at least two papers that were written to explain the algorithm!

In any case, the lexicographically-sorted canonical string form of the labelled tree does not work for signatures in general. Signatures can represent cycles using tree nodes that are numbered - as with smiles notation where you have C1CCCC1 for a 5-membered ring.

### 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…