This challenge gives me a map of asteroids. I’ll need to play with different ways to find which ones are directly in the path of others, first to see which asteroids can see the most others, and then to destroy them one by one with a laser.

## Challenge

The puzzle can be found here. I’m given a map, where . represents space, and # represents an asteroid. For example:

.#..#
.....
#####
....#
...##


An asteroid can see another one unless there is a third one directly in the line of sight (which is to say, the center of the asteroid falls on the line between the center of the first two).

In the first challenge, I’ll find the asteroid with direct line of sight to the most other asteroids. In part two, I’ll swing a rotating laser around, which blows up the first asteroid it can see. I need to find the 200th asteroid blown up.

## Solution

### Part 1

I started by reading the input file, and creating a list of asteroids:

with open(sys.argv[1], "r") as f:

asteroids = []
for y, line in enumerate(lines):
for x, xval in enumerate(line):
if xval == "#":
asteroids.append((x,y))


Next, I needed a function to see if one asteroid can see another. I’ll use the idea that for x, y, and z, y is between x and z only if dist(x,y) + dist(y,z) == dist(x, z). Since floating point math gets tough, I’ll give it a bit of space (0.0001) seemed to work nicely. For a pair of asteroids, I’ll loop over the others and see if any of them are between the two:

def dist(a1, a2):
return math.sqrt(pow(a1[0] - a2[0], 2) + pow(a1[1] - a2[1], 2))

def can_see(a1, a2):
if a1 == a2:
return False
for ast in asteroids:
if a1 == ast or a2 == ast:
continue
if abs((dist(a1, ast) + dist(ast, a2)) - dist(a1, a2)) < 0.0001:
return False
return True


I kept a dictiony of lists that allows me to note for each asteroid what others it can see. I’ll just loop over the list twice, checking each pair:

vis = defaultdict(list)
for i in range(len(asteroids)):
for j in range(i + 1, len(asteroids)):
if can_see(asteroids[i], asteroids[j]):
vis[asteroids[i]].append(asteroids[j])
vis[asteroids[j]].append(asteroids[i])


Now I can just use max with a custom key based on length of the list:

base = max(vis, key=lambda x: len(vis[x]))
print(f"Part 1: {len(vis[base])} (at asteroid located at {base})")


The samples ran pretty quickly, but the puzzle input took almost 30 seconds:

$time ./day10.py 10-puzzle_input.txt Part 1: 227 (at asteroid located at (11, 13)) real 0m29.170s user 0m29.158s sys 0m0.012s  ### Part 2 Starting from the base I found, now I need to think in terms of angles. I’ll write a loop to go over each asteroid, and calculate the angle between it and the base using some basic trig. I’ll add 90 to get it so that 0 is up. I’ll use a dictionary I called angels to store a list of asteroids at each angle. I’ll not only store the location, but the distance to base to make sorting easier: angles = defaultdict(list) for ast in asteroids: if ast == base: continue angle = (math.degrees(math.atan2(ast[1] - base[1], ast[0] - base[0])) + 90) % 360 angles[angle].append((dist(ast, base), ast)) for angle in angles: angles[angle] = sorted(angles[angle], reverse=True)  In a second loop, I’ll make sure each list of asteroids at a given angel is sorted so that the closest is at the end. Now I’ll loop over the angles using pop to remove the last item from the list (the closest to the base). i = 0 for angle in sorted(angles): boom = angles[angle].pop() i += 1 if i == 200: print(f"Part 2: {boom[1]}") sys.exit()  It turns out that I hit the 200th asteroid in my first pass. Had I not, I would have needed to wrap this all in a while True, and handled skipping over empty lists. This second part runs instantly (the time is all from part 1): $ time ./day10.py 10-puzzle_input.txt
Part 1: 227 (at asteroid located at (11, 13))
Part 2: (6, 4)

real    0m31.261s
user    0m31.223s
sys     0m0.000s


## Final Code

#!/usr/bin/env python3

import math
import sys
from collections import defaultdict

def dist(a1, a2):
return math.sqrt(pow(a1[0] - a2[0], 2) + pow(a1[1] - a2[1], 2))

def can_see(a1, a2):
if a1 == a2:
return False
for ast in asteroids:
if a1 == ast or a2 == ast:
continue
if abs((dist(a1, ast) + dist(ast, a2)) - dist(a1, a2)) < 0.0001:
return False
return True

with open(sys.argv[1], "r") as f:

asteroids = []
for y, line in enumerate(lines):
for x, xval in enumerate(line):
if xval == "#":
asteroids.append((x, y))

vis = defaultdict(list)
for i in range(len(asteroids)):
for j in range(i + 1, len(asteroids)):
if can_see(asteroids[i], asteroids[j]):
vis[asteroids[i]].append(asteroids[j])
vis[asteroids[j]].append(asteroids[i])

base = max(vis, key=lambda x: len(vis[x]))
print(f"Part 1: {len(vis[base])} (at asteroid located at {base})")

angles = defaultdict(list)
for ast in asteroids:
if ast == base:
continue
angle = (math.degrees(math.atan2(ast[1] - base[1], ast[0] - base[0])) + 90) % 360
angles[angle].append((dist(ast, base), ast))

for angle in angles:
angles[angle] = sorted(angles[angle], reverse=True)

i = 0
for angle in sorted(angles):
boom = angles[angle].pop()
i += 1
if i == 200:
print(f"Part 2: {boom[1]}")
sys.exit()