Final post on Combinatorial Maps of Cuneane

EDIT : Updated with new image.
Well, unless I can make a more horrifying diagram than this:

..but I think that's unlikely. An explanation - if such a thing is possible - is that the red/purple arrows correspond to a kind of inside-out operation. So the triangular map (M3) at the bottom can be converted to the square (M4) by choosing the purple darts (2, 14, 16, 5) as the boundary, and putting everything else on the inside. The cycle (1, 4, 9, 20, 22) is red because it is the boundary of the pentagonal map (M5).

So, it seems like ϕ(M4) is α(ϕ(M5)) - so the cycle (0,8,23,21,5) in ϕ(M4) is (1,9,22,20,4) in ϕ(M5). That is, because (α[0] = 1, α[8] = 9, α[23] = 22 ...). However ϕ(M3) = ϕ(M5), which is confusing. Oh well

So I have updated the image so that all three 'maps' have the same cycles. Which means they have the same ϕ and therefore the same σ. I guess this means I misunderstood what a CM actually is : you can get 'different' embeddings with the same σ. Here is a summary diagram:

Which is just the same diagram, without all the numbers. Colored arrows in the faces show cycles of darts chosen as the bounding faces according to the same colored arrows joining maps. Note that the 4-cycle in CM3 chosen as the bounds for CM4 has to be flipped. So does the 5-cycle in CM3 chosen as the bounds of CM5. Perhaps there is still a transformation wrong here somewhere.

ANOTHER EDIT : After flipping CM5, and labelling the cycles (A-F/a-f : clockwise is uppercase, anticlockwise is lowercase) it is better.

The transitions are quite simple, really. Take a clockwise face, and make it anticlockwise (eg: B->b, red arrow from CM3 to CM5) and make the anticlockwise outer face clockwise (d -> D). These changes are on the red/purple arrows between maps. Under each map is a summary of the cycles, which really just shows that the outer face is anticlockwise and the others are clockwise.

Anonymous said…
This provokes a lot of ideas and question in ones mind.
It also reminds me of flow optimisation problem in computer science.
Anonymous said…
I love this game :)

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 …