2022-01-14 11:21:56 +01:00

172 lines
5.9 KiB
Python

# --- Day 3: Toboggan Trajectory ---
# With the toboggan login problems resolved, you set off toward the airport.
# While travel by toboggan might be easy, it's certainly not safe: there's very
# minimal steering and the area is covered in trees. You'll need to see which
# angles will take you near the fewest trees.
# Due to the local geology, trees in this area only grow on exact integer
# coordinates in a grid. You make a map (your puzzle input) of the open squares
# (.) and trees (#) you can see. For example:
# ..##.......
# #...#...#..
# .#....#..#.
# ..#.#...#.#
# .#...##..#.
# ..#.##.....
# .#.#.#....#
# .#........#
# #.##...#...
# #...##....#
# .#..#...#.#
# These aren't the only trees, though; due to something you read about once
# involving arboreal genetics and biome stability, the same pattern repeats to
# the right many times:
# ..##.........##.........##.........##.........##.........##....... --->
# #...#...#..#...#...#..#...#...#..#...#...#..#...#...#..#...#...#..
# .#....#..#..#....#..#..#....#..#..#....#..#..#....#..#..#....#..#.
# ..#.#...#.#..#.#...#.#..#.#...#.#..#.#...#.#..#.#...#.#..#.#...#.#
# .#...##..#..#...##..#..#...##..#..#...##..#..#...##..#..#...##..#.
# ..#.##.......#.##.......#.##.......#.##.......#.##.......#.##..... --->
# .#.#.#....#.#.#.#....#.#.#.#....#.#.#.#....#.#.#.#....#.#.#.#....#
# .#........#.#........#.#........#.#........#.#........#.#........#
# #.##...#...#.##...#...#.##...#...#.##...#...#.##...#...#.##...#...
# #...##....##...##....##...##....##...##....##...##....##...##....#
# .#..#...#.#.#..#...#.#.#..#...#.#.#..#...#.#.#..#...#.#.#..#...#.# --->
# You start on the open square (.) in the top-left corner and need to reach the
# bottom (below the bottom-most row on your map).
# The toboggan can only follow a few specific slopes (you opted for a cheaper
# model that prefers rational numbers); start by counting all the trees you
# would encounter for the slope right 3, down 1:
# From your starting position at the top-left, check the position that is right
# 3 and down 1. Then, check the position that is right 3 and down 1 from there,
# and so on until you go past the bottom of the map.
# The locations you'd check in the above example are marked here with O where
# there was an open square and X where there was a tree:
# ..##.........##.........##.........##.........##.........##....... --->
# #..O#...#..#...#...#..#...#...#..#...#...#..#...#...#..#...#...#..
# .#....X..#..#....#..#..#....#..#..#....#..#..#....#..#..#....#..#.
# ..#.#...#O#..#.#...#.#..#.#...#.#..#.#...#.#..#.#...#.#..#.#...#.#
# .#...##..#..X...##..#..#...##..#..#...##..#..#...##..#..#...##..#.
# ..#.##.......#.X#.......#.##.......#.##.......#.##.......#.##..... --->
# .#.#.#....#.#.#.#.O..#.#.#.#....#.#.#.#....#.#.#.#....#.#.#.#....#
# .#........#.#........X.#........#.#........#.#........#.#........#
# #.##...#...#.##...#...#.X#...#...#.##...#...#.##...#...#.##...#...
# #...##....##...##....##...#X....##...##....##...##....##...##....#
# .#..#...#.#.#..#...#.#.#..#...X.#.#..#...#.#.#..#...#.#.#..#...#.# --->
# In this example, traversing the map using this slope would cause you to
# encounter 7 trees.
# Starting at the top-left corner of your map and following a slope of right 3
# and down 1, how many trees would you encounter?
with open("files/P3.txt", "r") as f:
forest = [line.split() for line in f.read().strip().split("\n")]
def part_1() -> None:
trees_found, pos = 0, 0
for idx, line in enumerate(forest):
extended_line = ",".join(line * 32).replace(",", "")
if extended_line[pos] == "#":
trees_found += 1
pos += 3
print(f"We found {trees_found} trees in our way")
# --- Part Two ---
# Time to check the rest of the slopes - you need to minimize the probability
# of a sudden arboreal stop, after all.
# Determine the number of trees you would encounter if, for each of the
# following slopes, you start at the top-left corner and traverse the map all
# the way to the bottom:
# Right 1, down 1.
# Right 3, down 1. (This is the slope you already checked.)
# Right 5, down 1.
# Right 7, down 1.
# Right 1, down 2.
# In the above example, these slopes would find 2, 7, 3, 4, and 2 tree(s)
# respectively; multiplied together, these produce the answer 336.
# What do you get if you multiply together the number of trees encountered on
# each of the listed slopes?
def part_2() -> None:
pos = 0
trees_first = 0
# first slope: right 1, down 1
for idx, line in enumerate(forest):
extended_line = ",".join(line * 11).replace(",", "")
if extended_line[pos] == "#":
trees_first += 1
pos += 1
# reset pos
pos = 0
trees_second = 0
# second slope: right 3, down 1
for idx, line in enumerate(forest):
extended_line = ",".join(line * 32).replace(",", "")
if extended_line[pos] == "#":
trees_second += 1
pos += 3
# reset pos
pos = 0
trees_third = 0
# third slope: right 5, down 1
for idx, line in enumerate(forest):
extended_line = ",".join(line * 55).replace(",", "")
if extended_line[pos] == "#":
trees_third += 1
pos += 5
# reset pos
pos = 0
trees_fourth = 0
# fourth slope: right 7, down 1
for idx, line in enumerate(forest):
extended_line = ",".join(line * 77).replace(",", "")
if extended_line[pos] == "#":
trees_fourth += 1
pos += 7
# reset pos
pos = 0
trees_fifth = 0
# fifth slope: right 1, down 2
for idx, line in enumerate(forest):
if idx % 2 == 0:
# break
extended_line = ",".join(line * 6).replace(",", "")
if extended_line[pos] == "#":
trees_fifth += 1
pos += 1
total_trees = (
trees_first * trees_second * trees_third * trees_fourth * trees_fifth
)
print(f"We found {total_trees} trees in our way")
if __name__ == "__main__":
part_1()
part_2()