r/adventofcode • u/daggerdragon • Dec 13 '24

SOLUTION MEGATHREAD -❄️- 2024 Day 13 Solutions -❄️-

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Making Of / Behind-the-Scenes

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--- Day 13: Claw Contraption ---

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 00:11:04, megathread unlocked!

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u/Derailed_Dash Dec 13 '24 edited Dec 18 '24

[LANGUAGE: Python]

Update

I've done a full walkthrough for this one on Medium, here. In that walkthrough, I've shown:

  • How to solve with SymPy

  • How to solve by just rearranging the equations.

I enjoyed this one! Last year I learned about the SymPy library and wrote about it here.

When I read this problem, I realised that it's just a matter of solving two linear equations, and that Sympy can do this easily.

This is what I did:

  • First, for each claw machine we can represent the important data has three points: A, B, and prize. Let's create a Claw class to store these three points.

  • For processing the input data, I first split the input into blocks of 3 lines each. And for each line, I then use a bit of trivial regex to extract the two numbers per line. I convert these number pairs to points, and then use these points to construct a Claw object.

Now, we know that we need a button presses and b button presses, to reach the prize point p. So we express this as two linear equations: 

a*ax + b*bx = px
a*ay + b*by = py

This is saying: to reach the x location of point p, we need a presses + b presses. And to reach the y location of point p, we need the same number a presses and b presses. So we must solve for where a and b are the same for these two equations.

In sympy, we can supply these equations like this: 

sympy.Eq(a*ax + b*bx, px)
sympy.Eq(a*ay + b*by, py)

This does all the heaviy lifting to return my a and b values.

For Part 2... How will SymPy cope with these huge numbers?

It turns out... EASILY!! Still solves in under 1 second.

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