So, a student asked me about a homework question that is a sub-problem of the structure generation problem. Basically, it was to

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

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

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

Happy Holidays, anyway...

*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...

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Here comes another holiday...