this post was submitted on 06 Dec 2024
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Advent Of Code

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Day 6: Guard Gallivant

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[–] Pyro 4 points 2 weeks ago* (last edited 2 weeks ago) (1 children)

Python

Part 1: Simulate the guard's walk, keeping track of visited positions
Part 2: Semi brute-force. Try to place an obstacle at every valid position in the guard's original path and see if it leads to a loop.

import os
from collections import defaultdict

# paths
here = os.path.dirname(os.path.abspath(__file__))
filepath = os.path.join(here, 'input.txt')

# read input
with open(filepath, mode='r', encoding='utf8') as f:
    data = f.read()
rows = data.splitlines()

# bounds
m = len(rows)
n = len(rows[0])

# directions following 90 degree clockwise turns
#   up, right, down, left
DIRECTIONS = [(-1, 0), (0, 1), (1, 0), (0, -1)]

# find position of guard
guard_i, guard_j = -1, -1
for i in range(m):
    for j in range(n):
        if rows[i][j] == '^':
            guard_i, guard_j = i, j
            break
    if guard_i != -1:
        break


def part1(guard_i, guard_j):
    # keep track of visited positions
    visited = set()
    visited.add((guard_i, guard_j))

    dir_idx = 0     # current direction index

    # loop while guard is in map
    while True:
        delta_i, delta_j = DIRECTIONS[dir_idx]
        next_gi, next_gj = guard_i + delta_i, guard_j + delta_j   # next pos
        
        # if out of bounds, we are done
        if not (0 <= next_gi < m) or not (0 <= next_gj < n):
            break
        # change direction when obstacle encountered
        if rows[next_gi][next_gj] == "#":
            dir_idx = (dir_idx + 1) % 4
            continue
        # update position and visited
        guard_i, guard_j = next_gi, next_gj
        visited.add((guard_i, guard_j))

    print(f"{len(visited)=}")


def part2(guard_i, guard_j):
    # keep track of visited positions
    visited = set()
    visited.add((guard_i, guard_j))

    dir_idx = 0 # current direction index
    loops = 0   # loops encountered

    # walk through the path
    while True:
        delta_i, delta_j = DIRECTIONS[dir_idx]
        next_gi, next_gj = guard_i + delta_i, guard_j + delta_j # next pos
        
        # if out of bounds, we are done
        if not (0 <= next_gi < m) or not (0 <= next_gj < n):
            break
        # change direction when obstacle encountered
        if rows[next_gi][next_gj] == "#":
            dir_idx = (dir_idx + 1) % 4
            continue
        # if a position is not already in path,
        # put a obstacle there and see if guard will loop
        if (next_gi, next_gj) not in visited and willLoop(guard_i, guard_j, dir_idx):
            loops += 1
        # update position and visited
        guard_i, guard_j = next_gi, next_gj
        visited.add((guard_i, guard_j))
    
    print(f"{loops=}")

# used in part 2
# returns whether placing an obstacle on next pos causes a loop or not
def willLoop(guard_i, guard_j, dir_idx) -> bool:
    # obstacle pos
    obs_i, obs_j = guard_i + DIRECTIONS[dir_idx][0], guard_j + DIRECTIONS[dir_idx][1]

    # keep track of visited pos and the direction of travel
    visited: defaultdict[tuple[int, int], list[int]] = defaultdict(list)
    visited[(guard_i, guard_j)].append(dir_idx)
    
    # walk until guard exits map or loops
    while True:
        delta_i, delta_j = DIRECTIONS[dir_idx]
        next_gi, next_gj = guard_i + delta_i, guard_j + delta_j # next pos
        
        # if out of bounds, no loop
        if not (0 <= next_gi < m) or not (0 <= next_gj < n):
            return False
        # change direction when obstacle encountered
        if rows[next_gi][next_gj] == "#" or (next_gi == obs_i and next_gj == obs_j):
            dir_idx = (dir_idx + 1) % 4
            continue
        # we are looping if we encounter a visited pos in a visited direction
        if (next_gi, next_gj) in visited and dir_idx in visited[(next_gi, next_gj)]:
            return True
        
        # update position and visited
        guard_i, guard_j = next_gi, next_gj        
        visited[(guard_i, guard_j)].append(dir_idx)


part1(guard_i, guard_j)
part2(guard_i, guard_j)

[–] CameronDev 3 points 2 weeks ago (4 children)

How long did brute force take? Mine was 9s on an m1 with rust.

[–] Deebster 3 points 2 weeks ago* (last edited 2 weeks ago)

My rust code ran in 6s on my phone (Samsung A35 running under Termux). When I'm back at a computer it'd be interesting to compare times properly.

[–] Pyro 2 points 2 weeks ago* (last edited 2 weeks ago) (1 children)

About 15-20 seconds, not too bad.

[–] CameronDev 1 points 2 weeks ago* (last edited 2 weeks ago)

I got mine down to 3s, but it wasn't a very smart loop detection. All I did was count steps and stop after 10000. The 9 second run was 100000 steps, which is obviously a bit excessive.

Does save iterating over the list of past visits, so probably a good shortcut.

[–] [email protected] 1 points 2 weeks ago* (last edited 2 weeks ago)

I did a similar approach (place obstacles on guards path). Takes about ~~80s~~ 10-15s in 11th Gen Intel(R) Core(TM) i7-11800H. Motivated by the code above, I also restricted the search to start right before the obstacle rather than the whole path which took it down from 80s to 10-15s

[–] [email protected] 1 points 2 weeks ago (1 children)

Mine was 9s

That's about how long it takes for my python solution to complete.

[–] CameronDev 2 points 2 weeks ago (2 children)

How did you detect loops? I just ran for 100000 steps to see if I escaped, got my time down to 3s by doing only 10000 steps.

[–] [email protected] 2 points 2 weeks ago* (last edited 2 weeks ago)

I added each visited position/direction to a set, and when a 'state' is reached again you have entered a loop:

v = set()
while t[g.r][g.c] != 'X':
    state = (g.r, g.c, g.d)
    if state in v:
        acc += 1
        break
    v.add(state)
    g.move(t)

You can view my full solution here.

[–] [email protected] 2 points 2 weeks ago (1 children)

Not who you asked but: I save coordinates and direction into a vector each time the guard faces a #. Also every time the guard faces a #, I check if the position exists in the vector, if true, it’s an infinite loop. 78ms rust aolution.

[–] CameronDev 1 points 2 weeks ago

That's probably quite optimal, compared with checking every state in the path, or running off a fixed number of steps