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For a given triangulation (combinatorial Type I. or Type II.) of a $2$-sphere, what is the number of unique polygonal covers with $n$ polygons where ($n$ goes from $2$ to $N$)?

Under polygonal cover, I mean, you take a particular triangulations of the sphere with $N$ triangles, then select a few connected triangles and "mark them" as a polygon, then move to the next ones. Obviously, one possibility is having a triangle, plus all the rest, so technically two triangles. The second trivial cover is the other side, where all triangles are marked separately, so we have $n = N.$ But there are exponentially many options (I guess), and I would be interested if they can be enumerated with some close form or not.

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  • $\begingroup$ Surely it depends on the triangulation, so what would your variables for the closed form be? $\endgroup$
    – Peter Taylor
    Commented Sep 20, 2023 at 16:26
  • $\begingroup$ @PeterTaylor , I'm the one who's asking! Apart from the joke, I would assume, that for every triangulation of N triangles, one could find all the poligonial covers. Or all the possible forests on the triangulation. Is it wrong to assume ? $\endgroup$
    – Kregnach
    Commented Sep 20, 2023 at 19:52
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    $\begingroup$ Not forests: elements in the poset formed by edge contractions. And yes, for small graphs one can calculate. E.g. of the simple cubic graphs on 8 vertices two are non-planar and the others have distributions [(2, 39), (3, 146), (4, 207), (5, 146), (6, 58), (7, 12), (8, 1)], [(2, 54), (3, 220), (4, 283), (5, 176), (6, 62), (7, 12), (8, 1)], [(2, 63), (3, 268), (4, 345), (5, 202), (6, 66), (7, 12), (8, 1)]. $\endgroup$
    – Peter Taylor
    Commented Sep 21, 2023 at 8:23

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