this post was submitted on 25 Dec 2023
11 points (86.7% liked)

Advent Of Code

1105 readers
13 users here now

An unofficial home for the advent of code community on programming.dev!

Advent of Code is an annual Advent calendar of small programming puzzles for a variety of skill sets and skill levels that can be solved in any programming language you like.

AoC 2024

Solution Threads

M T W T F S S
1
2 3 4 5 6 7 8
9 10 11 12 13 14 15
16 17 18 19 20 21 22
23 24 25

Rules/Guidelines

Relevant Communities

Relevant Links

Credits

Icon base by Lorc under CC BY 3.0 with modifications to add a gradient

console.log('Hello World')

founded 2 years ago
MODERATORS
 

Day 25: Snowverload

Megathread guidelines

  • Keep top level comments as only solutions, if you want to say something other than a solution put it in a new post. (replies to comments can be whatever)
  • You can send code in code blocks by using three backticks, the code, and then three backticks or use something such as https://topaz.github.io/paste/ if you prefer sending it through a URL

FAQ

all 6 comments
sorted by: hot top controversial new old
[โ€“] cvttsd2si@programming.dev 4 points 2 years ago (1 children)

Scala3

all done!

def parseLine(a: String): List[UnDiEdge[String]] = a match
    case s"$n: $r" => r.split(" ").map(_ ~ n).toList
    case _ => List()

def removeShortestPath(g: Graph[String, UnDiEdge[String]], ns: List[String]) =
    g.removedAll(g.get(ns(0)).shortestPathTo(g.get(ns(1))).map(_.edges.map(_.outer)).getOrElse(List()))

def removeTriple(g: Graph[String, UnDiEdge[String]], ns: List[String]) =
    List.fill(3)(ns).foldLeft(g)(removeShortestPath)

def division(g: Graph[String, UnDiEdge[String]]): Option[Long] =
    val t = g.componentTraverser()
    Option.when(t.size == 2)(t.map(_.nodes.size).product)

def task1(a: List[String]): Long = 
    val g = Graph() ++ a.flatMap(parseLine)
    g.nodes.toList.combinations(2).map(a => removeTriple(g, a.map(_.outer))).flatMap(division).take(1).toList.head
[โ€“] sjmulder@beehaw.org 2 points 2 years ago

Python (Not C...)

(Posting from an alt because SDF is having issues)

A graph problem ๐Ÿ˜ฌโ€‹. There are well-known algorithms but none trivial to implement in C.

I tried dividing the nodes into two groups and then moving the node with the worst in-group/out-group edge balance - a Wish version of the Kernighan-Lin algorithm. As expected, it got stuck in a local optimum for the real input and random prodding didn't help.

Programming is programming and libraries are fair game so I went to Python and used NetworkX which was my first time using the library but so easy to use! Maybe I'll go back and do it in C someday and implement the algorithm myself.

#!/usr/bin/env python3
import sys
import re
from networkx import Graph
from networkx.algorithms.community import greedy_modularity_communities

G = Graph()

for line in sys.stdin.readlines():
  words = re.findall("\\w+", line)
  for to in words[1:]:
    G.add_edge(words[0], to)

coms = greedy_modularity_communities(G, best_n=2)

print("25:", len(coms[0]) * len(coms[1]))

https://github.com/sjmulder/aoc/tree/master/2023/python/day25.py

[โ€“] Treeniks@lemmy.ml 1 points 2 years ago* (last edited 2 years ago) (1 children)

Rust

github codeberg gitlab

First tried a really slow brute force, but after waiting many minutes heard others talk of Karger's Algorithm, so I implemented that.

use rand::prelude::*;
use std::collections::HashSet;

type Graph = (V, E);

type V = HashSet;
type E = Vec<(String, String)>;

fn parse_input(input: &str) -> Graph {
    let mut vertices = HashSet::new();
    let mut edges = Vec::new();

    for line in input.trim().lines() {
        let mut it = line.split(':');

        let left = it.next().unwrap();
        vertices.insert(left.to_owned());

        for right in it.next().unwrap().trim().split_whitespace() {
            vertices.insert(right.to_owned());
            edges.push((left.to_owned(), right.to_owned()));
        }
    }

    (vertices, edges)
}

fn part1(input: &str) -> usize {
    let (vertices, edges) = parse_input(input);

    let mut rng = rand::thread_rng();

    // Karger's Algorithm
    loop {
        let mut vertices = vertices.clone();
        let mut edges = edges.clone();
        while vertices.len() > 2 {
            let i = rng.gen_range(0..edges.len());
            let (v1, v2) = edges[i].clone();

            // contract the edge
            edges.swap_remove(i);
            vertices.remove(&v1);
            vertices.remove(&v2);

            let new_v = format!("{}:{}", v1, v2);
            vertices.insert(new_v.clone());

            for (e1, e2) in edges.iter_mut() {
                if *e1 == v1 || *e1 == v2 {
                    *e1 = new_v.clone()
                }
                if *e2 == v1 || *e2 == v2 {
                    *e2 = new_v.clone()
                }
            }

            // remove loops
            let mut j = 0;
            while j < edges.len() {
                let (e1, e2) = &edges[j];
                if e1 == e2 {
                    edges.swap_remove(j);
                } else {
                    j += 1;
                }
            }
        }

        if edges.len() == 3 {
            break vertices
                .iter()
                .map(|s| s.split(':').count())
                .product::();
        }
    }
}
[โ€“] aoidenpa@lemmy.world 1 points 2 years ago

Same. I am happy that it was a straightforward algorithm.