# How to enumerate unique lattice polygons for a given area using Pick's Theorem?

Pick's Theorem

Suppose that a polygon has integer coordinates for all of its vertices. Let $$i$$ be the number of integer points interior to the polygon, and let $$b$$ be the number of integer points on its boundary (including both vertices and points along the sides). Then the area $$A$$ of this polygon is:

$$A = i + \frac{b}{2} - 1$$

Example

Let integer area $$A = 5$$, these are the possible pairs of boundary $$b$$ and interior $$i$$ points that satisfy Pick's Notice that some pairs $$(b,i)$$ have multiple unique shapes For $$b=6,i=3$$ above, found $$7$$ unique shapes so far. Are there others that I missed? I'm just visually checking and haven't written any code to determine the exact number, yet.

The Pick's integer solutions are

$$\forall A>0, \quad 1 < n \leq (A+1) : b = 2n, \, i = (A + 1) - n, \quad n,A \in \mathbb{Z}$$

Brute-Force Approach

1. Find all possible pairs of boundary $$b$$ and interior $$i$$ points that satisfy Pick's Theorem.
2. For each $$(b, i)$$ pair, attempt to generate all unique simple lattice polygons.
3. Count the unique shapes for the given area $$A$$.

Question

Given an integer area $$A$$, I'm curious about how many unique shapes can be formed on a $$2D$$ lattice?

• Even counting rotated/reflected figures as distinct, a formula would be of interest. [I would think also difficult to find] Sep 18 at 20:16
• Let $a(n)$ be the number of lattice polygons of area $n$ (modulo some symmetries, of course). How many of the $a(n)$ can you compute? If you can get through $a(5)$ or $a(6)$, you can probably check OEIS for more information Sep 18 at 21:26
• Depends on the definition of a unique shape. Up to isometries? Then even the number of triangles of area $1/2$ is infinite. Up to affine transformations? Then the answer is different. Sep 19 at 12:52

There are infinitely many unique shapes for any area $$A$$. Let $$A=1$$ and we can make triangles with base of 2 and height of 1 and move the top vertex over a lattice point. This can be done for any size base of $$2A$$ and height 1 for a triangle of area $$A$$. Add in an interior point and this game can continue for any $$i$$ you choose also. And for more interior points, you will move the top green dot along, keeping the rest fixed. Shown are $$A = 2,3,$$ and $$4$$, with $$i = 1,1,2$$ respectively. Complementing N. Owad's answer -- the number of polygons of a given area becomes finite if you consider them up to unimodular equivalence, i.e. up to lattice-preserving affine transformations. The number of convex polygons up to unimodular equivalence with area $$1/2,1,3/2,2,5/2,3\ldots$$ is $$1, 2, 3, 7, 6, 13,\ldots$$ -- see https://oeis.org/A187015 and refs. therein, especially Balletti.