Skip to main content

Cycles on Lattices

So, first a small correction to the last post : a 'honeycomb' layout is a cycle on a hexagonal lattice, while the other layouts there are not on-lattice. I've renamed them 'flower' layouts, as I'm not sure if they have a name - they're not at all new, I just haven't looked! To be clear:

these are two different layouts of the same 12-cycle. As it happens, the honeycomb layout is also a [5, 5, 5]-flower layout with straight edges on the first edge of the outer cycle. However, not every honeycomb is a flower.

There might be many ways to do it, but one way to make honeycomb layouts is to find cycles on a hexagonal lattice. Assuming, of course, that you have such a lattice already - it's not terribly hard to make one, but connecting it up properly is a bit fiddly. Luckily, it turned out that making the dual of a triangular lattice is slightly easier. To see how this works, consider the three possible (regular) planar lattices:

The grey dots are actually the dual lattice of the black lattice. Click-for-bigger to see the lines connecting the dual points. Of course, a square lattice (on the left) is self-dual, while triangular and hexagonal lattices are duals of each other.

Fine, so given such a hex-lattice, how to get cycles of the right size? Well, the simplest way might be to find all cycles in a grid of some size, and just take ones of whatever size. This ... works - but not very well. Here are the 10-cycles from the hex-grid above:


clearly they are all the same! In fact, this is one of the smaller 'families' (orbits) of cycles. There are 78 cycles of size 12, and 2,082 of size 22. Here is a table:
Clearly, this is not the right approach.

Comments

Popular posts from this blog

Adamantane, Diamantane, Twistane

After cubane, the thought occurred to look at other regular hydrocarbons. If only there was some sort of classification of chemicals that I could use look up similar structures. Oh wate, there is . Anyway, adamantane is not as regular as cubane, but it is highly symmetrical, looking like three cyclohexanes fused together. The vertices fall into two different types when colored by signature: The carbons with three carbon neighbours (degree-3, in the simple graph) have signature (a) and the degree-2 carbons have signature (b). Atoms of one type are only connected to atoms of another - the graph is bipartite . Adamantane connects together to form diamondoids (or, rather, this class have adamantane as a repeating subunit). One such is diamantane , which is no longer bipartite when colored by signature: It has three classes of vertex in the simple graph (a and b), as the set with degree-3 has been split in two. The tree for signature (c) is not shown. The graph is still bipartite accordin

Király's Method for Generating All Graphs from a Degree Sequence

After posting about the Hakimi-Havel  theorem, I received a nice email suggesting various relevant papers. One of these was by Zoltán Király  called " Recognizing Graphic Degree Sequences and Generating All Realizations ". I have now implemented a sketch of the main idea of the paper, which seems to work reasonably well, so I thought I would describe it. See the paper for details, of course. One focus of Király's method is to generate graphs efficiently , by which I mean that it has polynomial delay. In turn, an algorithm with 'polynomial delay' takes a polynomial amount of time between outputs (and to produce the first output). So - roughly - it doesn't take 1s to produce the first graph, 10s for the second, 2s for the third, 300s for the fourth, and so on. Central to the method is the tree that is traversed during the search for graphs that satisfy the input degree sequence. It's a little tricky to draw, but looks something like this: At the top

General Graph Layout : Putting the Parts Together

An essential tool for graph generation is surely the ability to draw graphs. There are, of course, many methods for doing so along with many implementations of them. This post describes one more (or perhaps an existing method - I haven't checked). Firstly, lets divide a graph up into two parts; a) the blocks, also known as ' biconnected components ', and b) trees connecting those blocks. This is illustrated in the following set of examples on 6 vertices: Trees are circled in green, and blocks in red; the vertices in the overlap between two circles are articulation points. Since all trees are planar, a graph need only have planar blocks to be planar overall. The layout then just needs to do a tree layout  on the tree bits and some other layout on the embedding of the blocks. One slight wrinkle is shown by the last example in the image above. There are three parts - two blocks and a tree - just like the one to its left, but sharing a single articulation point. I had