advent-of-code/2024-python/output/__init__.py
Anders Englöf Ytterström e6307795a4 Solve 2024:14 p1-2 "Restroom Redoubt"
After working overtime to solve day 12, this brief
task was a much welcome change.

For pt 1, the code uses integer division and
modulus to update all positions. The tricky part
was to determine which quadrant got the middle
column and row (since dimensions was odd numbers,
and either left or right quandrants is +1 longer),
which was straightforward to verify by the test
cases.

For pt 2, the remains of the code used to visually
identify the tree is left under the `find_easter_egg`
toggle.

Basically, the code prints the grid with
all robots marked as a `#`, and free space as `.`.`

This wording of the puzzle is important:

> very rarely, _most of the robots_ should arrange
> themselves into a picture of a Christmas tree.

"most of the robots" means the tree will be quite
visible, which also means it is safe to asume a
significant cluster of "#" would appear over time.

By keeping track of the counter (seconds the robots
has traveled), it was evident that the cluster of `#`
occoured th first time at i=64 and every 103th
time forward.

In other words, `i % 103 == 64` will most likely
give a tree. The print statement is therefore
limited to those i's, and at `i == easter_egg_appearance`
the tree is visible.

So, to be extra clear: 64, 103 and pt2 answer is
unique to my puzzle input. If this code is used
with any other puzzle input, these numbers will
most likely vary.

For the fun, the code also contains the `display_easter_egg`
flag to actually print the tree. This is provided
to indicate how big the actual tree is: 33*36 chars.

Also, the `sints` utility function was added to
extract all signed ints from a string.
2025-01-05 00:06:18 +01:00

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import functools
import re
# Directions/Adjacents for 2D matrices, in the order UP, RIGHT, DOWN, LEFT
D = [
(-1, 0),
(0, 1),
(1, 0),
(0, -1),
]
Di = [
(-1, -1),
(-1, 1),
(1, -1),
(1, 1),
]
# Directions for 2D matrices, as a dict with keys U, R, D, L
DD = {
"U": (-1, 0),
"R": (0, 1),
"D": (1, 0),
"L": (0, -1),
}
# Adjacent relative positions including diagonals for 2D matrices, in the order NW, N, NE, W, E, SW, S, SE
ADJ = [
(-1, -1),
(-1, 0),
(1, -1),
(0, -1),
(0, 1),
(1, 1),
(1, 0),
(1, -1),
]
def ints(s):
"""Extract all integers from a string"""
return [int(n) for n in re.findall(r"\d+", s)]
def sints(s):
"""Extract all signed integers from a string"""
return [int(n) for n in re.findall(r"-?\d+", s)]
def mhd(a, b):
"""Calculates the Manhattan distance between 2 positions in the format (y, x) or (x, y)"""
ar, ac = a
br, bc = b
return abs(ar - br) + abs(ac - bc)
def matrix(d):
"""Transform a string into an iterable matrix. Returns the matrix, row count and col count"""
m = [tuple(r) for r in d.split()]
return m, len(m), len(m[0])
def mdbg(m):
"""Print-debug a matrix"""
for r in m:
print("".join(r))
def vdbg(seen, h, w):
"""Print-debug visited positions of a matrix"""
for r in range(h):
print("".join(["#" if (r, c) in seen else "." for c in range(w)]))
def cw(y, x):
"""Flip a (y, x) direction counterwise: U->R, R->D, D->L, L->U.
>>> cw(-1, 0)
(0, 1)
>>> cw(0, 1)
(1, 0)
>>> cw(1, 0)
(0, -1)
>>> cw(0, -1)
(-1, 0)
"""
return (x, y) if y == 0 else (x, -y)
def ccw(y, x):
"""Flip a (y, x) direction counterwise: U->L, L->D, D->R, R->U.
>>> ccw(-1, 0)
(0, -1)
>>> ccw(0, -1)
(1, 0)
>>> ccw(1, 0)
(0, 1)
>>> ccw(0, 1)
(-1, 0)
"""
return (x, y) if x == 0 else (-x, y)
def bfs(S, E=None):
"""BFS algorithm, equal weighted nodes"""
seen = set()
q = [(S, 0)]
g = {} # graph, required to be provided at some point
while q:
m, w = q.pop(0)
if m in seen:
continue
seen.add(m)
# investigate here
for s in g[m]:
q.append((s, w + 1))
# return insights
def dijkstras(grid, start, target):
"""
1. Create an array that holds the distance of each vertex from the starting
vertex. Initially, set this distance to infinity for all vertices except
the starting vertex which should be set to 0.
2. Create a priority queue (heap) and insert the starting vertex with its
distance of 0.
3. While there are still vertices left in the priority queue, select the vertex
with the smallest recorded distance from the starting vertex and visit its
neighboring vertices.
4. For each neighboring vertex, check if it is visited already or not. If it
isnt visited yet, calculate its tentative distance by adding its weight
to the smallest distance found so far for its parent/previous node
(starting vertex in case of first-level vertices).
5. If this tentative distance is smaller than previously recorded value
(if any), update it in our distances array.
6. Finally, add this visited vertex with its updated distance to our priority
queue and repeat step-3 until we have reached our destination or exhausted
all nodes.
"""
import heapq
target = max(grid)
seen = set()
queue = [(start, 0)]
while queue:
cost, pos, direction, steps = heapq.heappop(queue)
y, x = pos
dy, dx = direction
if pos == target:
return cost
if ((pos, "and stuff")) in seen:
continue
seen.add((pos, "and stuff"))
neighbors = []
for n in neighbors:
heapq.heappush(queue, ("stuffs"))
return -1