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There are two issues here, both related to the layout of the grid on top of the faces of an icosahedron (see this blog post for more details):
Because of these issues, there are a number of cells at each resolution that may have more or less than 6 vertexes. At even resolutions, there are hexagons with 6 vertexes and pentagons with 5; at odd resolutions, there are hexagons with 6, 7, or 8 vertexes and pentagons with 10. See the "Literal Edge Cases" section here for examples of the different types of cell distortion. |
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Hi,
I'm trying to compute the distances between a
lat,lon
point and its sequence of assigned hexagons in different resolutions.To this end I have been using the following code:
what I've noticed however was that h3.cell_to_bondary doesn't return 6 hexagon points as expected but instead if I print:
I see that the first two levels are 5 and 7.
Could you please explain why and what does this mean?
Thanks,
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