Day 24, Part 1
It involves a floor tiled with hexagonal tiles. The tiles have a white side and a black side, and can flip from one to the other. The puzzle involves starting from a center tile, and following directions (east, northeast, west, etc.) to get to another tile, which must be flipped. The answer to the puzzle is how many tiles are flipped after following all the directions.
So! I’ve never had to work with a hexagonal grid before, but so many games have one, it must be a solved problem. I google “represent hex grid in array” and land on a Stack Overflow question, which leads me to this brilliant page, “Hexagonal Grids” by Amit Patel. This is nothing short of a miracle, telling me everything I need to know in order to do things with hexagonal grids.
After reading that page and thinking about it for a while, I decide that I will use cube coordinates (x, y, z) and that I don’t even need to store a grid. I just need to store the destination coordinates that each instruction from the input takes me to. A tile is white at the end, if its coordinates are reached an even number of times (including 0), and a tile is black if its coordinates are reached an odd number of times.
I could store the destination coordinates in a hashmap from coordinate to count, but I wonder if there is a multiset similar to the multimap I used a few days ago. There is. With that, I can write the code for Part 1:
Day 24, Part 2
In a surprise twist, the second part of the puzzle is yet another Conway’s Game of Life, this time on the hex grid! So no more storing the coordinates of the flipped tiles in a multiset. I will need to have some sort of payday loans in IN array to store the grid, and calculate the number of occupied neighbour cells, as we have done on several previous puzzles.
The Hexagonal Grids page comes through once again. Isn’t this great, that I knew nothing about hexagonal grids before encountering this puzzle, and there’s just a page on the internet that explains all of it well enough that I can use it to solve this puzzle! I will use axial coordinates (meaning, just discard one of the cube coordinates) and store the grid in a rectangular ndarray . The only question is how big the array has to be.
I decide, as in Day 17, that a good upper bound is probably the size of the starting pattern, plus the number of iterations of the game, extended in each direction. So, for the example pattern in the puzzle description, the highest number in any of the three coordinates is 3 (and ?3), and the number of iterations is 100, so we’d want a grid of 103?103.
I store the map as a 2-dimensional ndarray , with coordinates (q, r) equal to (x, y) in the cube coordinate scheme (I just drop the z coordinate.) I store the offset of the center tile in (qref, rref).
I make a constructor that takes the multiset from Part 1 as input, and an iterate() method that calculates one iteration of the map and updates the class. The calc_neighbours() and iterate() code is practically copied from Day 11 except that we only shift the map in six directions instead of eight. (Which six directions those are, I get from the hex grids page.)
The Day 20 puzzle was a bit disappointing, since I spent so much more time on it than any of the other puzzles, and the solution wasn’t even particularly good. I’m not sure what made it so much more difficult, but I suspect that I just didn’t find the right data structure to read the data into.