@ericwastl I'm constantly blown away by how you're able to create novel puzzles nearly a decade into Advent of Code.

As with every year, thank you for working so hard on this project and sharing your puzzles with the hundreds of thousands around the world who enjoy and learn from them!

@bp Apparently, this is also equivalent to using Pick's theorem, where `i` is the internal area from whatever method you choose, `b` is our perimeter, and algebraically rearranging to solve for b + i. https://en.m.wikipedia.org/wiki/Pick%27s_theorem#Formula

Really liked this problem as well; it was super satisfying to derive this solution (and I'm quite surprised it hadn't appeared in any past years' Advent of Code problems).

@bp The intuition from your earlier posts helps explain this — take a simple rectangle. If you add perimeter/2, you're account for all the area on the "edges" you're missing, plus 1/2 for all 4 of the "outer corners". Each outer corner *actually* contributes 3/4 units, so you need 1/4 * 4 = 1 more unit (the +1).

For more complex polygons, each "inner corners" (1/4 unit area) necessitates an extra "outer corner" (3/4 unit area) so they are net equal to 2 edge pieces.