advent-of-code/2024-python/output/__init__.py
Anders Englöf Ytterström cb622409f9 Solve 2024:3 p1-2 "Mull It Over"
Felt paranoid on this one, was expecting something
much worse for pt2.

The code that solved the puzzle was _extremely_
naive:

- It asumes puzzle input only contains mul() with
1-3 chars. The versioned code have `{1,3}` as safe
guard.
- p2 asumed initial chunk does not begin with
"n't". Would have been easy to handle though.
- p2 asumed junk strings as "don", "do", "don()t"
would not exist.

In other words, I was really lucky since I did not
look that closely on puzzle input beforehand.

Might have cut 60-90 seconds further if I had just
ran pt2 immidately instead of staring dumbly on the
test data (that changed for pt2).

Also, I got reminded that \d in regular expressions
is equal to `0-9`: it does not include commas and
punctations. The code that solved the puzzle was
paranoid and instead used `0-9`.

Managed to score ~1500th somehow despite this.
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),
]
# 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 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