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Two things I am thinking. I have to analyze them better, but for sure you can help me before I'll find the answer myself. The first thing is that maps are usually represented using the graphs without considering multiple edges, which is good for coloring but I am not sure abount counting. For example, considering multiple edges the dual graph wouldn't have triangular faces. The second thing is about reversibility: do all triangulations have duals 3-regular planar graphs? !maps and triangulations
I made a correction to the question (as pointed out by jc). I was interested in "regular maps" without forcing 3-vertex-connectivity. Sorry for the mistake. The abstract of the paper reports: "In the paper, we enumerate three classes of cubic planar maps with no loops or multiple edges: 1-connected; 2-connected; 3-connected and triangle-free." The results I found manually give these results for 3, 4 and 5 faces: 1 maps, 3 maps, 20 maps.
Yes, you are correct. My mistake. I want to consider only "regular maps" without forcing 3-vertex-connectivity. Only vertexes having 3 edges. For example in the next picture, starting from a map of four faces (3 + the ocean) and adding one face, I would like to count maps excluding those generated from symmetries. 4coloring.files.wordpress.com/2011/04/counting-maps-example.png I'll remove 3-connected from the question.
@all: Thanks for the info. Yes, it is the problem I'm facing. But to get the "number of colorings" the only method I found is to compute the chromatic polynomial, which is known only for few graphs and it is hard to find for more complex cases. Do you know of papers that directly approach the computation of the "number of colorings without exchanges of colors"? I've implemented a brute force algorithm to color a given map with four colors. I'll try to extend it to find all possible colorings manually ... excluding exchanges. youtube.com/user/mariostefanutti#p/u/2/YmYGFxtj2es