In this series, I'll share my progress with the 2023 version of Advent of Code.
Check the first post for a short intro to this series.
You can also follow my progress on GitHub.
December 21st
The puzzle of day 21 is more of a math puzzle than it is a programming puzzle. The code is very straightforward but I needed a lot of hints for part two, which is basically solving a polynomial equation.
My pitfall for this puzzle: The second part would have taken me way too much time to figure it out on my own, therefore I needed to turn to Reddit for some hints.
Solution here, do not click if you want to solve the puzzle first yourself
#!/usr/bin/env python3
grid = []
with open('input.txt') as infile:
lines = infile.readlines()
for y, line in enumerate(lines):
grid.append([c for c in line.strip()])
if 'S' in line:
S = (y, line.index('S'))
free_positions = set()
for y, row in enumerate(grid):
for x, col in enumerate(row):
if col == '.' or col == 'S':
free_positions.add((y, x))
def show_grid(grid, positions):
for y, row in enumerate(grid):
line = ''
for x, col in enumerate(row):
if (y, x) in positions:
line += 'O'
else:
line += grid[y][x]
print(line)
def free(pos):
return (pos[0] % len(grid), pos[1] % len(grid[0])) in free_positions
def calc(n):
positions = [S]
for _ in range(n):
new_positions = set()
for p in positions:
south = (p[0] + 1, p[1])
if free(south) and not south in new_positions:
new_positions.add(south)
west = (p[0], p[1] - 1)
if free(west) and not west in new_positions:
new_positions.add(west)
north = (p[0] - 1, p[1])
if free(north) and not north in new_positions:
new_positions.add(north)
east = (p[0], p[1] + 1)
if free(east) and not east in new_positions:
new_positions.add(east)
positions = new_positions
return len(positions)
r1 = calc(65)
r2 = calc(65 + 131)
r3 = calc(65 + 2 * 131)
a = (r3 + r1 - (2 * r2)) / 2
b = ((4 * r2) - (3 * r1) - r3) / 2
c = r1
print((a * pow(202300, 2)) + (b * 202300) + c)
That's it! See you again tomorrow!
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