this post was submitted on 13 Dec 2024
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Advent Of Code

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An unofficial home for the advent of code community on programming.dev! Other challenges are also welcome!

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.

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Day 13: Claw Contraption

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[โ€“] Acters@lemmy.world 3 points 11 months ago* (last edited 11 months ago) (2 children)

Python

Execution time: ~<1 millisecond (800 microseconds on my machine)

Good old school linear algebra from middle school. we can solve this really really fast. With minimal changes from part 1!

FastCode[ paste ]

from time import perf_counter_ns
import string

def profiler(method):
    def wrapper_method(*args: any, **kwargs: any) -> any:
        start_time = perf_counter_ns()
        ret = method(*args, **kwargs)
        stop_time = perf_counter_ns() - start_time
        time_len = min(9, ((len(str(stop_time))-1)//3)*3)
        time_conversion = {9: 'seconds', 6: 'milliseconds', 3: 'microseconds', 0: 'nanoseconds'}
        print(f"Method {method.__name__} took : {stop_time / (10**time_len)} {time_conversion[time_len]}")
        return ret

    return wrapper_method

@profiler
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,r = divmod(Px - Bx * y, Ax)
            if r == 0:
                part1_total_cost += 3*x + y
        y,r = divmod((Ay * (Px+10000000000000) - Ax * (Py+10000000000000)), (Ay * Bx - Ax * By))
        if r == 0:
            x,r = divmod((Px+10000000000000) - Bx * y, Ax)
            if r == 0:
                part2_total_cost += 3*x + y

    return part1_total_cost,part2_total_cost

if __name__ == "__main__":
    with open('input', 'r') as f:
        input_data = f.read().strip().replace(',', '').split('\n\n')
    part_one, part_two = main(input_data)
    print(f"Part 1: {part_one}\nPart 2: {part_two}")

[โ€“] Sparrow_1029@programming.dev 2 points 11 months ago (1 children)

This is a really excellent, clean solution! Would you mind breaking down how the piece of linear algebra works (for a shmo like me who doesn't remember that stuff frum school heh ๐Ÿ˜…)

[โ€“] Acters@lemmy.world 4 points 11 months ago

https://lemmy.world/comment/13950499

take the two equations, solve for y, and make sure y is fully divisible.