Advent of Code 2023 - December 14th

Advent of Code 2023 (25 Part Series)

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 14th

The puzzle of day 14 is quite fun and very doable. I probably could shave off more lines of code but as I’m still running behind with the other days I’m not going to spend more time on this puzzle.

My pitfall for this puzzle: Used a lot of code, almost reaching the hard limit I set for myself at the start (100 lines.)

Solution here, do not click if you want to solve the puzzle first yourself

#!/usr/bin/env python3 
with open('input.txt') as infile:
    lines = infile.readlines()
    rows = []
    for line in lines:
        rows.append([c for c in line.strip()])
    platform = tuple(tuple(r) for r in rows)

def tilt_north(p):
    pl = [list(r) for r in p]
    while True:
        changed = False
        for idx, row in enumerate(pl):
            if idx <= len(pl) - 2:
                next_row = pl[idx + 1]
                for idx_x in range(len(row)):
                    if row[idx_x] == '.' and next_row[idx_x] == 'O':
                        row[idx_x] = 'O'
                        next_row[idx_x] = '.'
                        changed = True
        if not changed:
            break
    return tuple(tuple(r) for r in pl)

def tilt_south(p):
    pl = [list(r) for r in p]
    while True:
        changed = False
        for idx_y in range(len(pl) - 1, 0, -1):
            row = pl[idx_y]
            prev_row = pl[idx_y - 1]
            for idx_x in range(len(row)):
                if row[idx_x] == '.' and prev_row[idx_x] == 'O':
                    row[idx_x] = 'O'
                    prev_row[idx_x] = '.'
                    changed = True
        if not changed:
            break
    return tuple(tuple(r) for r in pl)

def tilt_east(p):
    pl = [list(r) for r in p]
    while True:
        changed = False
        for idx_y, row in enumerate(pl):
            for idx_x in range(len(row) - 1, 0, -1):
                if row[idx_x] == '.' and row[idx_x - 1] == 'O':
                    row[idx_x] = 'O'
                    row[idx_x - 1] = '.'
                    changed = True
        if not changed:
            break
    return tuple(tuple(r) for r in pl)

def tilt_west(p):
    pl = [list(r) for r in p]
    while True:
        changed = False
        for idx_y, row in enumerate(pl):
            for idx_x in range(len(row) - 1):
                if row[idx_x] == '.' and row[idx_x + 1] == 'O':
                    row[idx_x] = 'O'
                    row[idx_x + 1] = '.'
                    changed = True
        if not changed:
            break
    return tuple(tuple(r) for r in pl)

def calc_load(p):
    load = 0
    for idx, row in enumerate(p):
        load += (len(p) - idx) * len([c for c in row if c == 'O'])
    return load 

def cycle(p):
    return tilt_east(tilt_south(tilt_west(tilt_north(p))))

full_cycles = []
index = -1
cycle_index = -1
for i in range(1000000000):
    platform = cycle(platform)
    if platform in full_cycles:
        cycle_index = full_cycles.index(platform)
        index = i
        break
    full_cycles.append(platform)
cycles_left = (1000000000 - index) % (index - cycle_index)
for i in range(cycles_left - 1):
    platform = cycle(platform)
print(calc_load(platform))

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That’s it! See you again tomorrow!

Advent of Code 2023 (25 Part Series)

原文链接：Advent of Code 2023 – December 14th

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