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I am reading an article about counting hexagonal p-minos.

In page 3 (of the pdf) we define a boundoing hexagon for a p-mino, In page 6 the article claims that delta - the number of cells inside and on the boundry of the bounding hexagon equal to $s1+s3+s5$ choose $2$ -...

Does anyone have any idea for why this is true ? I have no clue to even why it looks like something in the form of a choose b -...

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up vote 1 down vote accepted

The formula comes from combining two ideas:

  1. In a triangular lattice, an equilateral triangle with $k$ lattice points in the first row contains a total of $$k+(k-1)+(k-2)+\dots+1=\binom{k+1}{2}$$ lattice points: $k$ in the first row, $k-1$ in the second row, and so on.

  2. The hexagon can be thought of as taking one larger triangle and cutting a small triangle off of each of the corners (figure $8$ in their paper). The larger triangle has $(s_1+1)+(s_5+1)+(s_3+1)-2=s_1+s_3+s_5-1$ points in the first row ($(s_1+1)$ points from the triangle of side $s_1$, etc., then subtract $2$ since we counted two vertices twice). The triangles we subtract off have $s_1$, $s_3$, and $s_5$ points in their first row.

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Thank you vey much, so "the number of cells" means the number of lattice points ? maybe you also understand why K<=C ?(and what is the meaning of K..I looked up the word 'skew' and didn't find a meaning that suits) – Belgi Apr 6 '12 at 14:22
$K$ represents the difference between opposite sides -- how far the hexagon is from being a symmetrical one where opposite sides are equal. Geometrically, you can think of the $K \leq \min\{A,B,C\}$ here as representing that the sides of a hexagon can't change too quickly if the angles aren't too small. Every time we move (for example) to the right by one unit, the height of the hexagon in the corresponding vertical slice can change by at most $1$. So the difference in the leftmost and rightmost vertical slices ($K$) can be at most as large as the horizontal width of the hexagon ($B$). – Kevin Costello Apr 6 '12 at 17:31
On cells vs. points: To each cell the article associates a set of coordinates $(x,y,z)$. It's natural to think of these as points in a triangular lattice (the actual lattice underlying this is the dual lattice formed by associating to each hexagonal cell its center and connecting the centers of adjacent hexagons). Counting the number of hexagons intersecting a region then corresponds to counting the number of lattice points in a region. – Kevin Costello Apr 6 '12 at 18:48

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