Misunderstanding Embeddings, and Whitney Flips

So I should correct something that I posted a while ago about embedding 3-connected graphs. The most obvious examples of this class of graph are polyhedra - tetrahedra, cubes, etc - and maybe it's obvious that there is only one way to embed these in the plane. So in that sense, 'enumerating' the embeddings for these graphs is quite easy ... there's just one to count.

Of course, this embedding can be drawn with any of its cycles as an outer face; this is what gave rise to the different looking drawings. I guess the way to think about it is that the embedding is on the surface of a sphere - where there is no 'privileged' face to call the outer face - and that the drawing on the plane just picks one of the faces to 'squash flat' as I put it back in (wow) 2011.

Anyway - on to Whitney flips! There are a class of graphs that can be embedded in the plane in different ways (that is, the combinatorial map is different for the same outer face). These are subject to a Whitney flip, named after the mathematician who laid the foundation for matroids among other things. As an example see the image above.

The only graphs that have this flip are ones with a 2-vertex cutset - the top and bottom vertices in the image. Of course, finding these is a whole other problem, which I may or may not get around to describing...

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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...

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