Festive Chemical Structure Generation : Necklaces and Trees!

So, a student asked me about a homework question that is a sub-problem of the structure generation problem. Basically, it was to count the number of chemical structures with exactly one cycle given the elemental formula. Of course, the best solution here is probably to use the Polyá Enumeration Theorem since all that was asked for was a count (enumeration) of the structures.

Naturally, I have a different way to do this - especially since I don't really understand the mathematics of PET enough to implement it. So:

The image shows a rough overview of how I might list all of the structures with a single cycle. It takes a number of necklaces (one shown), and a number of trees, and glues the one to the other in all possible ways. The word 'necklace' here is specifically the combinatorial object; so  the cyclic sequence CNCNO is the same as CNOCN since you can rotate one to get the other.

One tricky decision here is whether to add multiple bonds to the necklace before or after adding the trees. It seems like this would make a difference to how fast the algorithm was - if you add the bonds afterwards, you might reject many of the possible attachments. Hard to say.

The other aspect to consider is the connection of the parts. If we consider necklaces (or 'cycles') and trees as types of block, then the problem is connecting together blocks into a tree. This is essentially the reverse of the approach detailed in this post - using a block decomposition tree to guide the assembly of the blocks:

Although, now I come to look at it, it seems like the attachment points on the necklace would drive the underlying block-tree. So perhaps this is only relevant for graphs that contain multiple cycles - which starts to become a much more difficult problem!

Happy Holidays, anyway...

Anonymous said…
Very easy to understand with your words and very vivid. When I read, I also can't help to imagine the picture in my head. If all chemsitry teacher can teach like you, there will be no students dislike chemistry anymore. BOC Sciences
Here comes another holiday...

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…