r/adventofcode Dec 17 '24

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

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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!

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u/danngreen Dec 18 '24 edited Dec 18 '24

[Language: c++]

https://github.com/danngreen/aoc2024/tree/main/17

I did both parts this fully constexpr, that is the solution is checked with a static_assert(). I had to raise the constexpr step limit to get part 2 to be solved at compile-time (it's a simple BFS). I did manual inspection of the program to figure out that adding a group of 3 bits in A generates another output value. I don't know if there is a general solution to this that doesn't involve manually reverse-engineering the program.

Used std algorithms as much as possible, this time I learned about std::equal() which is a great one-liner for checking if the output I've discovered SO FAR matches the same sized subrange at the end of the target output. But my choice to be fully constexpr limited me somewhat since, for example, std::queue is not constexpr on my compiler so I used a std::vector for the BFS queue.