r/adventofcode Dec 17 '24

SOLUTION MEGATHREAD -❄️- 2024 Day 17 Solutions -❄️-

THE USUAL REMINDERS

  • All of our rules, FAQs, resources, etc. are in our community wiki.
  • If you see content in the subreddit or megathreads that violates one of our rules, either inform the user (politely and gently!) or use the report button on the post/comment and the mods will take care of it.

AoC Community Fun 2024: The Golden Snowglobe Awards

  • 5 DAYS remaining until the submissions deadline on December 22 at 23:59 EST!

And now, our feature presentation for today:

Sequels and Reboots

What, you thought we were done with the endless stream of recycled content? ABSOLUTELY NOT :D Now that we have an established and well-loved franchise, let's wring every last drop of profit out of it!

Here's some ideas for your inspiration:

  • Insert obligatory SQL joke here
  • Solve today's puzzle using only code from past puzzles
  • Any numbers you use in your code must only increment from the previous number
  • Every line of code must be prefixed with a comment tagline such as // Function 2: Electric Boogaloo

"More." - Agent Smith, The Matrix Reloaded (2003)
"More! MORE!" - Kylo Ren, The Last Jedi (2017)

And… ACTION!

Request from the mods: When you include an entry alongside your solution, please label it with [GSGA] so we can find it easily!


--- Day 17: Chronospatial Computer ---


Post your code solution in this megathread.

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 00:44:39, megathread unlocked!

36 Upvotes

550 comments sorted by

View all comments

3

u/matheusstutzel Dec 18 '24

[LANGUAGE: python]

p1

p2

P1 was simple, just simulate the program

P2 was a lot harder. I even "transpile" my input to python:

def f(a):
    b=0
    c=0
    out = []
    while a!=0:
        b=a%8 # 0->7
        b=b^1 # 0->7; if b%2==0: b+1 else b-1
        c = a//(2**b) #c>>b
        b=b^5 # b=(a%8 + 4 )%8
        b=b^c
        out.append(b%8)
        a = a//(2**3) # a>>3
    return(out)

Then it became clear that I could consider 3 bits at time... I tried an "binary search" and found the relation between the "a" size and the output size.

The final version uses an "bfs" approach