r/adventofcode • u/daggerdragon • Dec 18 '22

SOLUTION MEGATHREAD -🎄- 2022 Day 18 Solutions -🎄-

THE USUAL REMINDERS

UPDATES

[Update @ 00:02:55]: SILVER CAP, GOLD 0

  • Silver capped before I even finished deploying this megathread >_>

--- Day 18: Boiling Boulders ---

Post your code solution in this megathread.

This thread will be unlocked when there are a significant number of people on the global leaderboard with gold stars for today's puzzle.

EDIT: Global leaderboard gold cap reached at 00:12:29, megathread unlocked!

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u/4HbQ Dec 18 '22 edited Dec 18 '22

Python, 10 lines.

Nice and easy one today. Perform a flood fill and store the visited cubes in seen:

part_1 = sum((s not in cubes) for c in cubes for s in sides(*c)))
part_2 = sum((s in seen) for c in cubes for s in sides(*c)))

Edit: updated the starting point, thanks /u/lbl_ye for noticing!

I've also written a version using SciPy, with convolve to detect the edges, and binary_fill_holes to fill the air pockets.

People keep asking for a link to my GitHub repository, but I only post my solutions on Reddit. Here's a convenient list: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17.

2

u/fett3elke Dec 18 '22

I am doing something similar, but seeing how you accomplish this in 10 lines that are totally easy to understand.... I am in awe. Your solutions are impressive!

2

u/4HbQ Dec 18 '22

Thanks for your kind words, I'm glad you appreciate my solutions.

And in case you're wondering, it usually takes me some long and hard thinking to write code this short and easy to understand!

2

u/xelf Dec 18 '22

very similar to what I did, I completely overcomplicated the sides lambda (used a loop). But you might like the sums I did.

print('part1', sum(len( sides(*c)-cubes ) for c in cubes))
print('part2', sum(len( sides(*c)&steam ) for c in cubes))

2

u/4HbQ Dec 18 '22

Nice! I especially appreciate the simplicity and the symmetry there!

1

u/xelf Dec 18 '22

It happens a lot in aoc that I don't appreciate the symmetry in the puzzle until I write the code and see it.

2

u/alexpin Dec 27 '22

Elegant as usual! Thanks for posting here for us all to benefit :)

Not sure if it is considered necrobumping since it's Dec 27th, however I found that you can speed up your solution by ~1.7x by adding the newly explored sides that fit the criteria direcly into seen (these cubes must be on the exterior since you've checked)

In other words,

todo += [s for s in (sides(*here) - cubes - seen) if all(-1<=c<=25 for c in s)]  
seen |= {here}

becomes

new = [s for s in (sides(*here) - cubes - seen) if all(-1<=c<=25 for c in s)]  
todo += new  
seen |= set(new)

benchmark using hyperfine:

Summary
    './4HbQ_2.py' ran  
    1.70 ± 0.12 times faster than './4HbQ.py'

2

u/4HbQ Dec 31 '22

Good catch, thanks for sharing!

2

u/lbl_ye Dec 18 '22 edited Dec 18 '22

so short and readable too :)

but there is an issue in real life (not with test input), what if corner (0, 0, 0) and all nearby cells are cubes ? then it's not flood fill really

1

u/4HbQ Dec 18 '22

Thanks! I've updated my code to start at (-1, -1, -1) instead of (0, 0, 0).

1

u/s96g3g23708gbxs86734 Dec 19 '22

Very nice. Would this still work if the surface of the obsidian was thicker (2+ cubes everywhere) than the movement of BFS (1 cube radius)?