Skip to main content

PVR numbering scheme not solution to all woes : film at 11

On a whim, I decided to try generating all adjacency matrices with the property that they are PVR numbered. The short summary of that link is that a matrix can be expressed as a sequence of positive integers by considering each row of the matrix as a binary number.

The point of doing this (I thought) was that you can number a molecule in such a way that the adjacency matrix is PVR-numbered, and that this is canonical. So my cunning plan was to generate all sequences of n numbers that are partially ordered, choosing them from [1, 2n] to give all non-redundant (simple) graphs with n vertices.

Unfortunately, it seems like this can't work:


This image shows all adjacency matrices for n = 3 which are PVR-numbered. They are made by backtracking through all sequences of integers with a partial order, pruning the solutions using the symmetry of the matrix as a constraint.

Anyway, the point is that the first two graphs are clearly isomorphic! More simply, they both represent propane. Maybe this is well known, but it's a surprise to me...

Comments

The method used by the previous deterministic structure generator used a similar approach... If just looking at rows may not be enough... what happens if you would look at the ordering of the columns too?
Anonymous said…
What a great web log. I spend hours on the net reading blogs, about tons of various subjects. I have to first of all give praise to whoever created your theme and second of all to you for writing what i can only describe as an fabulous article. I honestly believe there is a skill to writing articles that only very few posses and honestly you got it. The combining of demonstrative and upper-class content is by all odds super rare with the astronomic amount of blogs on the cyberspace.

Popular posts from this blog

Listing Degree Restricted Trees

Although stack overflow is generally just an endless source of questions on the lines of "HALP plz give CODES!? ... NOT homeWORK!! - don't close :(" occasionally you get more interesting ones. For example this one that asks about degree-restricted trees. Also there's some stuff about vertex labelling, but I think I've slightly missed something there.

In any case, lets look at the simpler problem : listing non-isomorphic trees with max degree 3. It's a nice small example of a general approach that I've been thinking about. The idea is to:
Given N vertices, partition 2(N - 1) into N parts of at most 3 -> D = {d0, d1, ... }For each d_i in D, connect the degrees in all possible ways that make trees.Filter out duplicates within each set generated by some d_i. Hmm. Sure would be nice to have maths formatting on blogger....

Anyway, look at this example for partitioning 12 into 7 parts:

At the top are the partitions, in the middle the trees (colored by degree) …

Common Vertex Matrices of Graphs

There is an interesting set of papers out this year by Milan Randic et al (sorry about the accents - blogger seems to have a problem with accented 'c'...). I've looked at his work before here.

[1] Common vertex matrix: A novel characterization of molecular graphs by counting
[2] On the centrality of vertices of molecular graphs

and one still in publication to do with fullerenes. The central idea here (ho ho) is a graph descriptor a bit like path lengths called 'centrality'. Briefly, it is the count of neighbourhood intersections between pairs of vertices. Roughly this is illustrated here:


For the selected pair of vertices, the common vertices are those at the same distance from each - one at a distance of two and one at a distance of three. The matrix element for this pair will be the sum - 2 - and this is repeated for all pairs in the graph. Naturally, this is symmetric:


At the right of the matrix is the row sum (∑) which can be ordered to provide a graph invarian…

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…