this post was submitted on 21 Dec 2023
13 points (93.3% liked)
Advent Of Code
761 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
you are viewing a single comment's thread
view the rest of the comments
view the rest of the comments
Nim
My part 2 solution assumes the input has an unimpeded shortest path from the center of each garden section to its corner, and to the center of its neighbor. The possible destinations will form a diamond pattern, with "radius" equal to the number of steps. I broke down the possible section permutations:
Sections that are completely within the interior of the diamond
Sections containing the points of the diamond
Depending on the number of steps, there may be sections adjacent to the point sections, that have two corners outside of the diamond
Edge sections. These will form a zig-zag pattern to cover the diamond boundary.
I determined how many of each of these should be present based on the number of steps, used my code from part 1 to get a destination count for each type, and then added them all up.