Day 13: Claw Contraption

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[-] Acters@lemmy.world 3 points 5 days ago* (last edited 5 days ago)

we are solving for y first. If there is a y then x is found easily.

(Ax)*x + (Bx)*y = Px and (Ay)*x + (By)*y = Py

Because of Ax or Ay and Bx or By, lets pretend that they are not (A*x)*x and (A*y)*y for both. they are just names. could be rewritten as: (Aleft)*x + (Bleft)*y = Pleft and (Aright)*x + (Bright)*y = Pright

but I will keep them short. solving for y turns into this:

y = (Ay*Px - Ax*Py) / (Ay*Bx - Ax*By)

if mod of 1 is equal to 0 then there is a solution. We can be confident that x is also a solution, too. Could there be an edge case? I didn't proof it, but it works flawlessly for my solution.

Thankfully, divmod does both division and mod of 1 at the same time.

[-] Sparrow_1029@programming.dev 2 points 5 days ago* (last edited 5 days ago)

Thank you so much for your explanation! I kind of found some clues looking up perp dot products & other vector math things, but this breaks it down very nicely.

I implemented your solution in rust, and part 2 failed by +447,761,194,259 (this was using signed 64-bit integers, i64). When I changed it to use signed 64 bit floating-point f64 and checked that the solution for x produces a whole number it worked.

[-] Acters@lemmy.world 3 points 5 days ago

Did you run my python code as is? I would hope it works for everyone. though, I am unsure what the edge cases are, then.

[-] Sparrow_1029@programming.dev 1 points 4 days ago

I did run your code as-is in an ipython REPL to check. These were the results:

REPL session

# With unmodified `main` function & `import string` not shown
In [4]: with open("inputs/day13.txt", "r") as f:
   ...:     input_data = f.read().strip().replace(',', '').split('\n\n')
   ...:

In [5]: part_one, part_two = main(input_data)

In [6]: part_one
Out[6]: 39748

In [7]: part_two
Out[7]: 74926346266863

# Then I modified the function to check if x is fractional
In [8]: def main(input_data):
   ...:     part1_total_cost = 0
   ...:     part2_total_cost = 0
   ...:     for machine in input_data:
   ...:         Ax,Ay,Bx,By,Px,Py = [ int(l[2:]) for l in machine.split() if l[-1] in string.digits ]
   ...:         y,r = divmod((Ay * Px - Ax * Py), (Ay * Bx - Ax * By))
   ...:         if r == 0:
   ...:             x = (Px - Bx * y) / Ax
   ...:             if x % 1 == 0:
   ...:                 part1_total_cost += 3*x + y
   ...:         y,r = divmod((Ay * (Px+10000000000000) - Ax * (Py+10000000000000)), (Ay * Bx - Ax * By))
   ...:         if r == 0:
   ...:             x = ((Px+10000000000000) - Bx * y) / Ax
   ...:             if x % 1 == 0:
   ...:                 part2_total_cost += 3*x + y
   ...:
   ...:     return part1_total_cost,part2_total_cost
   ...:

In [9]: part_one, part_two = main(input_data)

In [10]: part_one
Out[10]: 39748.0

In [11]: part_two
Out[11]: 74478585072604.0  # Correct answer for pt 2 of my input

If you're curious to check against my puzzle input, it's here

Thank you again for the back & forth, and for sharing your solution!

[-] Acters@lemmy.world 2 points 4 days ago

there is exactly ONE "machine" that causes your result to be incorrect. ONLY for part 2.

Button A: X+67, Y+67
Button B: X+16, Y+73
Prize: X=4877, Y=7214

I see now what your corner case causes. so when my script solves for y first. it will be exact. BUT when you solve for x, it will be not divisible... makes sense now. I didn't expect this. This only occurs because of part 2! so dastardly. well, that was interesting. I guess I am forced to add that extra check... rip those microsecond gains.

[-] Sparrow_1029@programming.dev 1 points 3 days ago

Ooh that is tricky of them. Good catch!

this post was submitted on 13 Dec 2024

18 points (90.9% liked)

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