this post was submitted on 13 Dec 2024
17 points (90.5% liked)

Advent Of Code

920 readers
25 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

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
 

Day 13: Claw Contraption

Megathread guidelines

  • Keep top level comments as only solutions, if you want to say something other than a solution put it in a new post. (replies to comments can be whatever)
  • You can send code in code blocks by using three backticks, the code, and then three backticks or use something such as https://topaz.github.io/paste/ if you prefer sending it through a URL

FAQ

you are viewing a single comment's thread
view the rest of the comments
[โ€“] Acters@lemmy.world 3 points 1 week ago (1 children)

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.

[โ€“] Sparrow_1029@programming.dev 1 points 1 week ago (1 children)

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!

[โ€“] Acters@lemmy.world 2 points 1 week ago (1 children)

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.

Ooh that is tricky of them. Good catch!