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Day 25: Code Chronicle

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FAQ

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[-] lwhjp@lemmy.sdf.org 5 points 2 weeks ago

Haskell

A total inability to write code correctly today slowed me down a bit, but I got there in the end. Merry Christmas, everyone <3

import Data.Either
import Data.List
import Data.List.Split

readInput = partitionEithers . map readEntry . splitOn [""] . lines
  where
    readEntry ls =
      (if head (head ls) == '#' then Left else Right)
        . map (length . head . group)
        $ transpose ls

main = do
  (locks, keys) <- readInput <$> readFile "input25"
  print . length $ filter (and . uncurry (zipWith (<=))) ((,) <$> locks <*> keys)
[-] sjmulder@lemmy.sdf.org 5 points 2 weeks ago

C

Merry Christmas everyone!

Code

#include "common.h"

int
main(int argc, char **argv)
{
	static char buf[7];
	static short h[500][5];	/* heights */
	static short iskey[500];
	int p1=0, nh=0, i,j,k;

	if (argc > 1)
		DISCARD(freopen(argv[1], "r", stdin));
	
	for (nh=0; !feof(stdin) && !ferror(stdin); nh++) {
		assert(nh < (int)LEN(h));

		for (i=0; i<7; i++) {
			fgets(buf, 7, stdin);
			if (i==0)
				iskey[nh] = buf[0] == '#';
			for (j=0; j<5; j++)
				h[nh][j] += buf[j] == '#';
		}

		/* skip empty line */
		fgets(buf, 7, stdin);
	}

	for (i=0; i<nh; i++)
	for (j=0; j<nh; j++)
		if (iskey[i] && !iskey[j]) {
			for (k=0; k<5 && h[i][k] + h[j][k] <= 7; k++) ;
			p1 += k == 5;
		}

	printf("25: %d\n", p1);
	return 0;
}

https://codeberg.org/sjmulder/aoc/src/branch/master/2024/c/day25.c

Made the 1 second challenge with most of it to spare! 😎

$ time bmake bench                                                                                                      
day01  0:00.00  1912 Kb  0+88 faults                                                                                            
day02  0:00.00  1992 Kb  0+91 faults 
day03  0:00.00  1920 Kb  0+93 faults
day04  0:00.00  1912 Kb  0+90 faults 
day05  0:00.00  2156 Kb  0+91 faults
day06  0:00.03  1972 Kb  0+100 faults
day07  0:00.06  1892 Kb  0+89 faults
day08  0:00.00  1772 Kb  0+87 faults 
day09  0:00.02  2024 Kb  0+137 faults
day10  0:00.00  1876 Kb  0+87 faults 
day11  0:00.00  6924 Kb  0+3412 faults
day12  0:00.00  1952 Kb  0+103 faults
day13  0:00.00  1908 Kb  0+88 faults
day14  0:00.05  1944 Kb  0+92 faults                                                                                            
day15  0:00.00  2040 Kb  0+89 faults
day16  0:00.03  2020 Kb  0+250 faults
day17  0:00.00  1896 Kb  0+88 faults
day18  0:00.00  1952 Kb  0+107 faults
day19  0:00.01  1904 Kb  0+91 faults
day20  0:00.01  2672 Kb  0+325 faults
day21  0:00.00  1804 Kb  0+86 faults
day22  0:00.03  2528 Kb  0+371 faults
day23  0:00.02  2064 Kb  0+152 faults
day24  0:00.00  1844 Kb  0+89 faults
day25  0:00.00  1788 Kb  0+89 faults  
                                                                
real    0m0,359s
[-] mykl@lemmy.world 4 points 2 weeks ago* (last edited 2 weeks ago)

Dart

Quick and dirty, and slightly tipsy, code.

Happy Christmas everyone!

Thanks to Eric and the team at Advent of Code, to @Ategon@programming.dev and @CameronDev@programming.dev for giving us somewhere to share and discuss our solutions, and to everyone here for the friendly and supportive community.

See you all next year!

import 'package:collection/collection.dart';
import 'package:more/more.dart';

part1(List<String> lines) {
  var (w, h) = (lines.first.length, lines.indexOf(''));
  var (falsey: keys, truthy: locks) = (lines..insert(0, ''))
      .splitBefore((l) => l.isEmpty)
      .map((g) => [
            for (var x in 0.to(w)) [for (var y in 1.to(h + 1)) g[y][x]]
          ])
      .partition((g) => g[0][0] == '#');
  return keys
      .map((l) => locks.count((k) =>
          0.to(w).every((r) => (l[r] + k[r]).count((e) => e == '#') < 8)))
      .sum;
}
[-] VegOwOtenks@lemmy.world 3 points 2 weeks ago* (last edited 2 weeks ago)

Haskell

Have a nice christmas if you're still celebrating today, otherwise hope you had a nice evening yesterday.

import Control.Arrow
import Control.Monad (join)
import Data.Bifunctor (bimap)
import qualified Data.List as List

heights = List.transpose
        >>> List.map (pred . List.length . List.takeWhile (== '#'))

parse = lines
        >>> init
        >>> List.groupBy (curry (snd >>> (/= "")))
        >>> List.map (List.filter (/= ""))
        >>> List.partition ((== "#####") . head)
        >>> second (List.map List.reverse)
        >>> join bimap (List.map heights)

cartesianProduct xs ys = [(x, y) | x <- xs, y <- ys]

part1 = uncurry cartesianProduct
        >>> List.map (uncurry (List.zipWith (+)))
        >>> List.filter (List.all (<6))
        >>> List.length
part2 = const 0

main = getContents
        >>= print
        . (part1 &&& part2)
        . parse
[-] LeixB@lemmy.world 3 points 2 weeks ago

Haskell

Merry Christmas!

{-# LANGUAGE OverloadedStrings #-}

module Main where

import Data.Either
import Data.Text hiding (all, head, zipWith)
import Data.Text qualified as T
import Data.Text.IO as TIO

type Pins = [Int]

toKeyLock :: [Text] -> Either Pins Pins
toKeyLock v = (if T.head (head v) == '#' then Left else Right) $ fmap (pred . count "#") v

solve keys locks = sum [1 | k <- keys, l <- locks, fit k l]
  where
    fit a b = all (<= 5) $ zipWith (+) a b

main = TIO.getContents >>= print . uncurry solve . partitionEithers . fmap (toKeyLock . transpose . T.lines) . splitOn "\n\n"
[-] GiantTree@feddit.org 3 points 2 weeks ago

Kotlin

A fun and small challenge. First read all locks, transpose their profile and count the #s (-1 for the full row). Then do the same for the keys.

Lastly find all keys for all locks that do not sum to more than 5 with their teeth:

Code


val lockRegex = Regex("""#{5}(\r?\n[.#]{5}){6}""")
val keyRegex = Regex("""([.#]{5}\r?\n){6}#{5}""")

fun parseLocksAndKeys(inputFile: String): Pair<List<IntArray>, List<IntArray>> {
    val input = readResource(inputFile)
    val locks = lockRegex
        .findAll(input)
        .map {
            it
                .value
                .lines()
                .map { line -> line.toList() }
                .transpose()
                .map { line -> line.count { c -> c == '#' } - 1 }
                .toIntArray()
        }
        .toList()

    val keys = keyRegex
        .findAll(input)
        .map {
            it
                .value
                .lines()
                .map { line -> line.toList() }
                .transpose()
                .map { line -> line.count { c -> c == '#' } - 1 }
                .toIntArray()
        }
        .toList()

    return locks to keys
}

fun part1(inputFile: String): String {
    val (locks, keys) = parseLocksAndKeys(inputFile)

    val matches = locks.map { lock ->
        keys.filter { key ->
            for (i in lock.indices) {
                // Make sure the length of the key and lock do not exceed 5
                if (lock[i] + key[i] > 5) {
                    return@filter false
                }
            }
            true
        }
    }
        .flatten()
        .count()

    return matches.toString()
}

Also on GitHub

[-] mykl@lemmy.world 3 points 2 weeks ago

Uiua

A Christmas Day treat: a one-liner for you all to decipher!

"#####\n.####\n.####\n.####\n.#.#.\n.#...\n.....\n\n#####\n##.##\n.#.##\n...##\n...#.\n...#.\n.....\n\n.....\n#....\n#....\n#...#\n#.#.#\n#.###\n#####\n\n.....\n.....\n#.#..\n###..\n###.#\n###.#\n#####\n\n.....\n.....\n.....\n#....\n#.#..\n#.#.#\n#####"
/+β™­βŠž(/Γ—<8+)βˆ©Β°β–‘Β°βŠŸ βŠ•(░≑≑/+βŒ•@#)β‰ @#≑(⊒⊒). ⊜(β‰βŠœβˆ˜βŠΈβ‰ @\n)¬±⦷"\n\n". 
[-] Gobbel2000@programming.dev 2 points 2 weeks ago

Rust

Nice ending for this year. Lock and key arrays are just added together and all elements must be <= 5. Merry Christmas!

Solution

fn flatten_block(block: Vec<Vec<bool>>) -> [u8; 5] {
    let mut flat = [0; 5];
    for row in &block[1..=5] {
        for x in 0..5 {
            if row[x] {
                flat[x] += 1;
            }
        }
    }
    flat
}

fn parse(input: &str) -> (Vec<[u8; 5]>, Vec<[u8; 5]>) {
    let mut locks = Vec::new();
    let mut keys = Vec::new();
    for block_s in input.split("\n\n") {
        let block: Vec<Vec<bool>> = block_s
            .lines()
            .map(|l| l.bytes().map(|b| b == b'#').collect::<Vec<bool>>())
            .collect();
        assert_eq!(block.len(), 7);
        // Lock
        if block[0].iter().all(|e| *e) {
            locks.push(flatten_block(block));
        } else {
            keys.push(flatten_block(block));
        }
    }
    (locks, keys)
}

fn part1(input: String) {
    let (locks, keys) = parse(&input);
    let mut count = 0u32;
    for l in locks {
        for k in &keys {
            if l.iter().zip(k).map(|(li, ki)| li + ki).all(|sum| sum <= 5) {
                count += 1;
            }
        }
    }
    println!("{count}");
}

fn part2(_input: String) {
    println!("⭐");
}

util::aoc_main!();

Also on github

[-] Zikeji@programming.dev 2 points 2 weeks ago

Javascript

Spent 10 minutes debugging my solution until I reread and found out they wanted the number of keys that fit, not the ones that overlapped. Reading comprehension is not it tonight.

const [locks, keys] = require('fs').readFileSync(0, 'utf-8').split(/\r?\n\r?\n/g).filter(v => v.length > 0).map(s => s.split(/\r?\n/g).filter(v => v.length > 0)).reduce((acc, s) => {
    const lock = s[0].split('').every(v => v === '#');
    const schema = s.slice(1, -1);
    let rotated = [];
    for (let i = 0; i < schema[0].length; i += 1) {
        for (let j = 0; j < schema.length; j += 1) {
            if (!rotated[i]) rotated[i] = [];
            rotated[i].push(schema[j][i]);
        }
    }
    if (!lock) {
        rotated = rotated.map(v => v.reverse());
    }
    const pinHeights = [];
    for (const row of rotated) {
        const height = row.indexOf('.');
        pinHeights.push(height !== -1 ? height : 5);
    }
    if (lock) {
        acc[0].push(pinHeights);
    } else {
        acc[1].push(pinHeights);
    }
    return acc;
}, [[],[]]);

let fits = 0;
for (const lock of locks) {
    for (const key of keys) {
        let overlapped = false;
        for (let i = 0; i < lock.length; i += 1) {
            if ((lock[i] + key[i]) > 5) {
                overlapped = true;
            }
        }
        if (!overlapped) {
            fits += 1;
        }
    }
}

console.log('Part One', fits);
[-] Acters@lemmy.world 2 points 1 week ago* (last edited 1 week ago)

Python3

ah well this year ends with a simple ~12.5 ms solve and not too much of a brain teaser. Well at least I got around to solving all of the challenges.

Code

from os.path import dirname,realpath,join

from collections.abc import Callable
def profiler(method) -> Callable[..., any]:
    from time import perf_counter_ns
    def wrapper_method(*args: any, **kwargs: any) -> any:
        start_time = perf_counter_ns()
        ret = method(*args, **kwargs)
        stop_time = perf_counter_ns() - start_time
        time_len = min(9, ((len(str(stop_time))-1)//3)*3)
        time_conversion = {9: 'seconds', 6: 'milliseconds', 3: 'microseconds', 0: 'nanoseconds'}
        print(f"Method {method.__name__} took : {stop_time / (10**time_len)} {time_conversion[time_len]}")
        return ret
    return wrapper_method

@profiler
def solver(locks_and_keys_str: str) -> int:
    locks_list = []
    keys_list  = []

    for schematic in locks_and_keys_str.split('\n\n'):
        if schematic[0] == '#':
            locks_list.append(tuple([column.index('.') for column in zip(*schematic.split())]))
        else:
            keys_list.append(tuple([column.index('#') for column in zip(*schematic.split())]))
    
    count = 0
    for lock_configuration in locks_list:
        for key_configuration in keys_list:
            for i,l in enumerate(lock_configuration):
                if l>key_configuration[i]:
                    # break on the first configuration that is invalid
                    break
            else:
                # skipped when loop is broken
                count += 1
    
    return count

if __name__ == "__main__":
    BASE_DIR = dirname(realpath(__file__))
    with open(join(BASE_DIR, r'input'), 'r') as f:
        input_data = f.read().replace('\r', '').strip()
    result = solver(input_data)
    print("Day 25 final solve:", result)

[-] CameronDev@programming.dev 1 points 1 week ago

Congrats on reaching the finish line!

The bit that caught me out was that the key + lock should equal 5 in reality, instead of being up to 5 in the challenge.

[-] Acters@lemmy.world 2 points 1 week ago

Thanks! I quickly wrote it but didn't think to count things. I just took the index of where the edge was located at and ran with it.

So I don't understand what you mean by equal 5. Could you elaborate? Cause I must have read the challenge text differently.

[-] CameronDev@programming.dev 1 points 6 days ago

For a real world lock, the key height + pin height must equal the height of the barrel exactly. If it is taller or shorter, the lock will bind and not open.

https://upload.wikimedia.org/wikipedia/commons/thumb/e/e8/Pin_tumbler_no_key.svg/400px-Pin_tumbler_no_key.svg.png

For the challenge, as long as its not overlapping (too tall), its a valid key/lock pair.

[-] Acters@lemmy.world 2 points 6 days ago

Oh! I didn't think it that way, lol, I was thinking this quickly through. I didn't think of relating to physical locks because it clearly said it was virtual. But I guess, there could theoretically be a physical tumbler lock with 0-5 spacers, it would just be a tall lock. You know like how some master keys have it so that there are spacers for the master key or the client key to open the lock.

this post was submitted on 25 Dec 2024
15 points (100.0% liked)

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