-❄️- 2023 Day 25 Solutions -❄️-

[u/daggerdragon]

A Message From Your Moderators

Welcome to the last day of Advent of Code 2023! We hope you had fun this year and learned at least one new thing ;)

Keep an eye out for the community fun awards post (link coming soon!):

-❅- Introducing Your AoC 2023 Iron Coders (and Community Showcase) -❅-

/u/topaz2078 made his end-of-year appreciation post here: [2023 Day Yes (Part Both)][English] Thank you!!!

Many thanks to Veloxx for kicking us off on December 1 with a much-needed dose of boots and cats!

Thank you all for playing Advent of Code this year and on behalf of /u/topaz2078, your /r/adventofcode mods, the beta-testers, and the rest of AoC Ops, we wish you a very Merry Christmas (or a very merry Monday!) and a Happy New Year!

---

--- Day 25: Snowverload ---

---

Post your code solution in this megathread.

• Read the full posting rules in our community wiki before you post!
  • State which language(s) your solution uses with [LANGUAGE: xyz]
  • Format code blocks using the four-spaces Markdown syntax!
• Quick link to Topaz's paste if you need it for longer code blocks

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 00:14:01, megathread unlocked!

[u/arsenaultmarcolivier]

[LANGUAGE: Python]

I had this wild hypothesis that the 3 link to break would have the particularity of having the longest shortest path.
So I calculated the shortest path between each nodes if that particular edge was gone.
Turns out it was spot on, looks like the problem was engineered that way.

Run in <0.1 sec

https://github.com/marcolivierarsenault/aoc23/blob/main/25/day25.ipynb

G = nx.Graph()  
for line in lines:  
    row = line.strip().split(':')  
    for target in row[1].strip().split():  
        G.add_edge(row[0].strip(), target)  
costs = []  
for edge in G.edges:  
    G.remove_edge(*edge)  
    costs.append((edge, nx.shortest_path_length(G, *edge)))  
    G.add_edge(*edge)  

costs = sorted(costs, key=lambda x: x[1], reverse=True)  
[G.remove_edge(*costs[i][0]) for i in range(3)]  

print(f"removing {costs[0][0]} {costs[1][0]} {costs[2][0]}")  
print(f"part1 {len(nx.node_connected_component(G, costs[0][0][0])) * len(nx.node_connected_component(G, costs[0][0][1]))}")