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The official awful.systems Advent of Code 2023 thread
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this post was submitted on 01 Dec 2023
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The beauty is you don't need to keep track of the corners at all: ultimately the area contributed by the perimeter is ( 1/2 * perimeter ) + 1. The short justification is that is if was just ( 1/2 * perimeter ), for every inside corners you overcount by 1/4 and for every outside corner you undercount. And there is exactly 4 more outside corners that inside ones, always. You can justify that by having an arrow follow the eddges, utlmately the arrow must make 1 full turn, each outside corner adds 1/4 turn. each inside corner removes 1/4 turn.I knew there was a better way! Thanks!
perhaps
A more elegant proof might show that starting with a rectangle, you have 4 corners contributing 1/4. You can push out parts of the edges of the rectangle to generate more corners, but they will always be in pairs of opposite types.