this post was submitted on 25 Dec 2023
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Advent Of Code

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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 2023

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submitted 10 months ago* (last edited 10 months ago) by CameronDev to c/advent_of_code
 

Day 25: Snowverload

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[โ€“] cvttsd2si 4 points 10 months 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€“] UFODivebomb 2 points 10 months ago

Impressive!

[โ€“] [email protected] 2 points 10 months 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€“] [email protected] 1 points 10 months ago* (last edited 10 months 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<String>;
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::<usize>();
        }
    }
}
[โ€“] [email protected] 1 points 10 months ago

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

[โ€“] [email protected] 1 points 10 months ago

Python

import networkx as nx

from .solver import Solver


class Day25(Solver):

  def __init__(self):
    super().__init__(25)

  def presolve(self, input: str):
    self.graph = nx.Graph()
    for line in input.splitlines():
      from_, to_line = line.split(': ')
      for to in to_line.split(' '):
        self.graph.add_edge(from_, to)

  def solve_first_star(self) -> int | str:
    cut_value, partition = nx.algorithms.stoer_wagner(self.graph)
    return len(partition[0]) * len(partition[1])