14
🦶️ - 2023 DAY 21 SOLUTIONS - 🦶️
(programming.dev)
you are viewing a single comment's thread
view the rest of the comments
view the rest of the comments
this post was submitted on 21 Dec 2023
14 points (93.8% liked)
Advent Of Code
736 readers
1 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 2023
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
- Follow the programming.dev instance rules
- Keep all content related to advent of code in some way
- If what youre posting relates to a day, put in brackets the year and then day number in front of the post title (e.g. [2023 Day 10])
- When an event is running, keep solutions in the solution megathread to avoid the community getting spammed with posts
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 1 year ago
MODERATORS
Rust
https://github.com/Treeniks/advent-of-code/blob/master/2023/day21/rust/src/main.rs
I reused my Grid struct from day 17 for part 1, just to realize that I'll need to expand the grid for part 2 so I awkwardly hacked it to be a
Vec>instead of a linearVec.I solved task 2 by reading through the reddit thread and trying to puzzle together what I was supposed to do. Took me a while to figure it out, even with literally looking at other people's solutions. I wrote a lengthy comment about it for anyone that's still struggling, but I honestly still don't really understand why it works. I think I wouldn't have solved it if I didn't end up looking at other solutions. Not a fan of the "analyze the input and notice patterns in them" puzzles.
Agreed, i get annoyed when I can't actually solve the problem. I would be ok if the inputs are trivial special cases, as long as feasible (but harder) generalized solutions still existed.
If you wonder why the function is a quadratic, I suggest drawing stuff on a piece of paper. Essentially, if there were no obstacles, the furthest reachable cells would form a large diamond, which is tiled by some copies of the diamond in the input and some copies of the corners. As these have constant size, and the large diamond will grow quadratically with steps, you need a quadratic number of copies (by drawing, you can see that if
steps = k * width + width/2, then there arefloor((2k + 1)^2/2)copies of the center diamond, andceil((2k + 1)^2/2)copies of each corner around).What complicates this somewhat is that you don't just have to be able to reach a square in the number of steps, but that the parity has to match: By a chessboard argument, you can see any given square only every second step, as each step you move from a black tile to a white one or vice versa. And the parities flip each time you cross a boundary, as the input width is odd. So actually you have to either just guess the coefficients of a quadratic, as you and @hades@lemm.ee did, or do some more working out by hand, which will give you the explicit form, which I did and can't really recommend.