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.


chembioinfo 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 :)