DEV Community

Cover image for Advent of Code 2023 - December 16th
Rob van der Leek
Rob van der Leek

Posted on

Advent of Code 2023 - December 16th

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 16th

The puzzle of day 16 was a lot of instructions for a fairly trivial coding problem.

My pitfall for this puzzle: The code is very inelegant, lots of if statements, and unfortunately no time to refactor this one.

Solution here, do not click if you want to solve the puzzle first yourself
#!/usr/bin/env python3

with open('input-small.txt') as infile:
    lines = infile.readlines()

grid = []
for line in lines:
    grid.append([c for c in line.strip()])

def trace(beam, grid, visited):
    x = beam[1]
    y = beam[0]
    direction = beam[2]
    while True:
        if y < 0 or y >= len(grid) or x < 0 or x >= len(grid[0]):
            return []
        tile = grid[y][x]
        if (y, x, direction) in visited:
            return []
        else: 
            visited.append((y, x, direction))
        if tile == '|':
            if direction == 'left' or direction == 'right':
                return [(y - 1, x, 'up'), (y + 1, x, 'down')] 
        elif tile == '-':
            if direction == 'up' or direction == 'down':
                return [(y, x - 1, 'left'), (y, x + 1, 'right')]
        elif tile == '\\':
            if direction == 'up':
                return [(y, x - 1, 'left')]
            elif direction == 'down':
                return [(y, x + 1, 'right')]
            if direction == 'right':
                return [(y + 1, x, 'down')]
            elif direction == 'left':
                return [(y - 1, x, 'up')]
        elif tile == '/':
            if direction == 'up':
                return [(y, x + 1, 'right')]
            elif direction == 'down':
                return [(y, x - 1, 'left')]
            if direction == 'right':
                return [(y - 1, x, 'up')]
            elif direction == 'left':
                return [(y + 1, x, 'down')]
        if direction == 'right':
            x += 1
        elif direction == 'left':
            x -= 1
        elif direction == 'up':
            y -= 1
        elif direction == 'down':
            y += 1

def calc_energized(start, grid):
    beams = [start]
    visited = []
    while beams:
        beams.extend(trace(beams.pop(), grid, visited))
    visited = set([(v[0], v[1]) for v in visited])
    result = len(visited)
    print(f'{start} = {result}')
    return result

start_coords = []
start_coords.extend([(0, x, 'down') for x in range(len(grid[0]))])
start_coords.extend([(y, len(grid[0]) - 1, 'left') for y in range(len(grid))])
start_coords.extend([(len(grid) - 1, x, 'up') for x in range(len(grid[0]))])
start_coords.extend([(y, 0, 'right') for y in range(len(grid))])

print(max([calc_energized(s, grid) for s in start_coords]))
Enter fullscreen mode Exit fullscreen mode

That's it! See you again tomorrow!

Image of Timescale

🚀 pgai Vectorizer: SQLAlchemy and LiteLLM Make Vector Search Simple

We built pgai Vectorizer to simplify embedding management for AI applications—without needing a separate database or complex infrastructure. Since launch, developers have created over 3,000 vectorizers on Timescale Cloud, with many more self-hosted.

Read more

Top comments (0)

Sentry image

See why 4M developers consider Sentry, “not bad.”

Fixing code doesn’t have to be the worst part of your day. Learn how Sentry can help.

Learn more