0

Here is a paper extending Pick's Theorem to cover the case where the lattice is hexagonal.

Paper: http://www.jstor.org/stable/2323889?seq=1#page_scan_tab_contents

The things that remains a puzzle to me is what is meant by the 'boundary characteristic'. The authors introduce this parameter as the "number of edge extended locally into the exterior of P minus those extended locally into the interior of P", where P is the polygon in question. Can anybody help explain this more thoroughly, ideally with a simple example?

(If you are looking for a solution for triangular lattices, here is a good thread I found: Pick's Theorem on a triangular (or hex) grid)

EDIT: Here is the relevant part of the paper I am talking about: enter image description here

Community
  • 1
Plato
  • 2,372

1 Answers1

1

We can just use the example $P = JKLMNQ$ given in the paper. The first term $c(J, P)$ is $2$. Why? Look at the three edges in the lattice coming out from $J$. Two of them are completely outside $P$, while the last one $JK$ is neither inside nor outside. Thus, $c(J, P) = 2 - 0 = 2$.

The next term $c(K, P)$ is $-1$. In this case, $KJ$ and $KL$ are neither inside nor outside, and the other edge is inside $P$. So $c(K, P) = 0 - 1 = -1$. Note that the last edge is not completely inside, but we are only looking at its behavior very close to $K$, so it still counts as inside.

You continue doing this for all boundary lattice points of $P$, and adding them up gives you $c(P)$.

arghbleargh
  • 501