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 24th
The puzzle of day 24 was too hard for me 😞 This was more of a math puzzle than it was a programming puzzle. I used this excellent write-up for the first part of the puzzle but could not get the second part to work. In the end, I used this script for the second part.
My pitfall for this puzzle: Unfortunately, I do not have the experience (and time) to solve this kind of puzzle.
Solution here, do not click if you want to solve the puzzle first yourself
#!/usr/bin/env python3
import random
hail = []
with open('input.txt') as infile:
lines = infile.readlines()
for line in lines:
parts = line.strip().split(' @ ')
pos = tuple([int(n) for n in parts[0].split(', ')])
vel = tuple([int(n) for n in parts[1].split(', ')])
hail.append((pos, vel))
pairs = []
for i in range(len(hail) - 1):
for j in range(i + 1, len(hail)):
pairs.append((hail[i], hail[j]))
def collide(h1, h2):
(x1, y1, z1) = h1[0]
(vx1, vy1, vz1) = h1[1]
(x2, y2, z2) = h2[0]
(vx2, vy2, vz2) = h2[1]
if vx1 == 0 or vx2 == 0:
return None
s1 = vy1 / vx1
s2 = vy2 / vx2
if s1 == s2:
return None
c1 = y1 - s1 * x1
c2 = y2 - s2 * x2
x = (c2 - c1) / (s1 - s2)
t1 = (x - x1) / vx1
t2 = (x - x2) / vx2
if t1 < 0 or t2 < 0:
return None
y = s1 * (x - x1) + y1
return(x, y, int(t1))
def update(hail, dvx, dvy):
return (hail[0], (hail[1][0] + dvx, hail[1][1] + dvy, hail[1][2]))
def predict(hail, t, dvz):
return hail[0][2] + t * (hail[1][2] + dvz)
def solve():
while True:
sel = random.choices(hail, k=4)
for dvx in range(-500, 500):
for dvy in range(-500, 500):
hail0 = update(sel[0], dvx, dvy)
c1 = collide(hail0, update(sel[1], dvx, dvy))
c2 = collide(hail0, update(sel[2], dvx, dvy))
c3 = collide(hail0, update(sel[3], dvx, dvy))
if c1 and c2 and c3 and c1[0] == c2[0] and c2[0] == c3[0] and \
c1[1] == c2[1] and c2[1] == c3[1]:
for dvz in range(-500, 500):
z1 = predict(sel[1], c1[2], dvz)
z2 = predict(sel[2], c2[2], dvz)
z3 = predict(sel[3], c3[2], dvz)
if z1 == z2 and z2 == z3:
return c1[0] + c1[1] + z1
total = 0
minrange = 200000000000000
maxrange = 400000000000000
for p in pairs:
pos = collide(p[0], p[1])
if pos:
if pos[0] >= minrange and pos[0] <= maxrange and \
pos[1] >= minrange and pos[1] <= maxrange:
total += 1
print(total)
print(solve())
That's it! See you again tomorrow!
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