Skip to main content

Schmidt's Bridge Finding Algorithm

While out looking for an algorithm to check for bridges in graphs, I found this paper by Jens Schmidt. It's very simple, and claims to be as fast as Tarjan's classic algorithm.

The basic idea is to make something he calls a 'chain decomposition' of the graph, where a 'chain' is either a cycle or a path. This is quite similar to the 'parts' of a graph I described here, except more formal than my slightly ad-hoc definitions.

Bridges, then are just those edges that are not in a chain. For example, take the graph from the paper:



Click for bigger, as always. First make a spanning tree from the graph, and store the back-edges (curved edges in the picture). Then go through the back-edges in order of the depth-first index, and make chains from them and the tree edges. A chain that ends up back at the start is a cycle, otherwise it is an edge. The chains A-E are shown in the middle, along with a bridge F.

The author also claims that the chain decomposition can be used in further algorithms, so it seems quite useful. My implementation is currently here, although it could probably be improved.

Comments

Popular posts from this blog

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…

Generating Trees

Tree generation is a well known (and solved!) problem in computer science. On the other hand, it's pretty important for various problems - in my case, making tree-like fusanes. I'll describe here the slightly tortuous route I took to make trees.

Firstly, there is a famous theorem due to Cayley that the number of (labelled) trees on n vertices is nn - 2 which can be proved by using Prüfer sequences. That's all very well, you might well say - but what does all this mean?

Well, it's not all that important, since there is a fundamental problem with this approach : the difference between a labelled tree and an unlabelled tree. There are many more labeled trees than unlabeled :


There is only one unlabeled tree on 3 vertices, but 3 labeled ones
this is easy to check using the two OEIS sequences for this : A000272 (labeled) and A000055 (unlabeled). For n ranging from 3 to 8 we have [3, 16, 125, 1296, 16807, 262144] labeled trees and [1, 2, 3, 6, 11, 23] unlabeled ones. Only 23 …