Mercurial > repos > other > adventofcode2023
view day14.txt @ 21:46fb65f2cb94
Add Day 14 part 1 implementation
If you look at columns and see "how far can this go" then the
weight is just the sum of the partial triangle numbers of
highest_weight to highest_weight minus number of rocks that
can roll up to there.
Part 2 seems to involve actually moving the rocks, though!
author | IBBoard <dev@ibboard.co.uk> |
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date | Sat, 16 Dec 2023 10:29:36 +0000 |
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--- Day 14: Parabolic Reflector Dish --- Given movable objects (O) and stationary objects (#) with free space (.), "tilt" the map to the top so that the movable objects slide to the top of the map and then determine the total "load" - the sum of the number of rows from the movable object to the bottom of the map. Given: O....#.... O.OO#....# .....##... OO.#O....O .O.....O#. O.#..O.#.# ..O..#O..O .......O.. #....###.. #OO..#.... Sliding makes it: OOOO.#.O.. OO..#....# OO..O##..O O..#.OO... ........#. ..#....#.# ..O..#.O.O ..O....... #....###.. #....#.... The load calculation is: OOOO.#.O.. 10 OO..#....# 9 OO..O##..O 8 O..#.OO... 7 ........#. 6 ..#....#.# 5 ..O..#.O.O 4 ..O....... 3 #....###.. 2 #....#.... 1 Which gives 136. Given a huge map, what is the load? --- Part 2 --- Now we want to cycle them by shifting them all in a cycle north, west, south and east. After 1 cycle: .....#.... ....#...O# ...OO##... .OO#...... .....OOO#. .O#...O#.# ....O#.... ......OOOO #...O###.. #..OO#.... After 2 cycles: .....#.... ....#...O# .....##... ..O#...... .....OOO#. .O#...O#.# ....O#...O .......OOO #..OO###.. #.OOO#...O After 3 cycles: .....#.... ....#...O# .....##... ..O#...... .....OOO#. .O#...O#.# ....O#...O .......OOO #...O###.O #.OOO#...O What is the top load after 1,000,000,000 cycles?