Advent of Code 2023 - December 23rd

robvanderleek

Rob van der Leek

Posted on January 1, 2024

Advent of Code 2023 - December 23rd

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 23rd

The puzzle of day 23 had me locked for a couple of days. My original plan was to do an optimized run through the maze looking for longer and longer paths, but that solution did not terminate for part two. In the end I implemented the graph approach that is mentioned a lot on Reddit.

My pitfall for this puzzle: The puzzle input appears (at least to me) to be suitable to use an optimized part-one solution for part two of the puzzle. But I can guarantee that will not work at all 😉

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

maze = []
with open('input.txt') as infile:
    lines = infile.readlines()
    for line in lines:
        maze.append([c for c in line.strip()])

start = (0, 1)
end = (len(maze) - 1, len(maze[0]) - 2)
longest = []

def in_maze(pos):
    return pos[0] >= 0 and pos[0] < len(maze) and \
        pos[1] >= 0 and pos[1] < len(maze[0])

def valid_new_pos(pos, visited):
    return pos not in visited and in_maze(pos) and maze[pos[0]][pos[1]] != '#'

def get_neighbours(pos, visited):
    result = []
    new_pos = (pos[0] + 1, pos[1])
    if valid_new_pos(new_pos, visited):
        result.append(new_pos)
    new_pos = (pos[0] - 1, pos[1])
    if valid_new_pos(new_pos, visited):
        result.append(new_pos)
    new_pos = (pos[0], pos[1] + 1)
    if valid_new_pos(new_pos, visited):
        result.append(new_pos)
    new_pos = (pos[0], pos[1] - 1)
    if valid_new_pos(new_pos, visited):
        result.append(new_pos)
    return result 

def build_graph(maze):
    visited_nodes = set()
    heads = [start] 
    edges = set() 
    while heads:
        head = heads.pop(0)
        visited_nodes.add(head)
        for p in get_neighbours(head, set()):
            visited = [head]
            new_head = p
            nbs = get_neighbours(new_head, visited)
            while len(nbs) == 1:
                visited.append(new_head)
                new_head = nbs[0]
                nbs = get_neighbours(new_head, visited)
            if len(nbs) > 1:
                if not new_head in visited_nodes:
                    edges.add((head, new_head, len(visited))) 
                    heads.append(new_head)
            if new_head == end:
                edges.add((head, new_head, len(visited)))
    return edges    

def move(head):
    global longest
    pos = head[0]
    path = head[1]
    if pos == end and len(path) > len(longest):
        longest = path 
    else:
        result = []
        new_pos = (pos[0] + 1, pos[1])
        if valid_new_pos(new_pos, head):
            result.append(((pos[0] + 1, pos[1]), path + [pos]))
        new_pos = (pos[0] - 1, pos[1])
        if valid_new_pos(new_pos, head):
            result.append(((pos[0] - 1, pos[1]), path + [pos]))
        new_pos = (pos[0], pos[1] + 1)
        if valid_new_pos(new_pos, head):
            result.append(((pos[0], pos[1] + 1), path + [pos]))
        new_pos = (pos[0], pos[1] - 1)
        if valid_new_pos(new_pos, head):
            result.append(((pos[0], pos[1] - 1), path + [pos]))
        return result

def explore(edges):
    longest = 0
    heads = [(start, 0, {start})]
    while heads:
        pos, l, visited = heads.pop()
        if pos == end:
            if l > longest:
                longest = l
        else:
            outgoing = [e for e in edges if e[0] == pos or e[1] == pos]
            for out in outgoing:
                if out[0] == pos and not out[1] in visited:
                    heads.append((out[1], l + out[2], visited | {out[1]}))
                if out[1] == pos and not out[0] in visited:
                    heads.append((out[0], l + out[2], visited | {out[0]}))
    return longest

print(explore(build_graph(maze)))
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That's it! See you again tomorrow!

💖 💪 🙅 🚩
robvanderleek
Rob van der Leek

Posted on January 1, 2024

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