this post was submitted on 07 Dec 2023
16 points (100.0% liked)
Advent Of Code
920 readers
116 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
- 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. [2024 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
Python
Oh boy. bitwise nonsense. Ok, can you explain it to me? I'm terrible at bitwise stuff.
Sure! This generates a number for every hand, so that a better hand gets a higher number. The resulting number will contain 11 hexadecimal digits:
The hand itself is 5 hexadecimal digits for every card, 0 for "2" to b for "ace".
This way the higher combination always has a higher number, and hands with the same combination are ordered by the order of the cards in the hand.
Wow, this is exactly what I did, but in C#. That's cool.
Cool!
That is a really cool solution. Thanks for the explanation! I took a much more... um... naive path lol.
I think you have the same solution, basically, just the details are a bit different. I like how you handled the joker, I didn't realise you could just multiply your best streak of cards to get the best possible combination.
I didn't multiply the streak, I just took the jokers and added them to the highest hand already in the list. Is that not what you did? It looked the same to me.
This is what I meant, but I phrased it poorly :)
In my solution I reimplement the logic of identifying the hand value, but with the presence of joker (instead of just reusing the same logic).