Day 14 2/25/98 ``I'm just trying to confuse you.''
proof: use euler's formula with a constraint on the maximum number of faces f £ 2/3m
2/3n is a bit arbitrary.
Square grid's have a Ön seperator and tree's have a 1 seperator.
Not every graph has a good edge seperator. For an n-star, there are only n/3 edge seperators. If every node has degree at most 3, then you can edge seperate.
The geometric dual is constructed by drawing an edge crossing each edge in G . Then connect all these new edges together not crossing the old edges in the process.
The geometric dual of K4 is K4 .
Convert all edges into directed edges ``darts''. Number the darts. Cycle through the darts at a particular node, then at each node.
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Which means there are 4 connected nodes, the first has darts '1','2', and'3'. The cycling through the darts you go from 1 to 3 to 2 to 1 to 3 to 2 to ....
The reflection permutation moves you along a dart.
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f* = Rf = (174)(3058)(269)
Reasoning: 1® R8® f7® R5® f4® R2® f1 (continuing forever) Taking every other element, you get: (174) . Continue similarly for each cycle. 2® R4® f6® R0® f9® R3® f2 yields (269) .
Each cycle (xy...z) is associated with a face of the graph.
Start with the graph:
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which goes to:
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The permutation representation works for more than planar graphs. One interesting question is how many handles the surface which the graph can be embedded on has. In other words: how many holes does the graph have? Examples surfaces: sphere, torus, 2-torus, ...
Use euler's formula: f-m+n = 2 where f = #faces , m = #edges , n = #vertices
This can be calculated quickly from the cycle representation.