Anders Englöf Ytterström
1e807b5daf
Initial version of the code tried to solve pt 1 using BFS, which took way too long even for the test data. After some fiddling with algebra with pen and paper, I realized this 2 formulaes (using first example): 94 * b1 + 22 * b2 == 840 34 * b1 + 67 * b2 == 540 *b1 = button A presses *b2 = button B presses ... could be rewritten to this single expression: (94 + 34) * b1 + (22 + 67) * b2 = 840 * 540 I failed to remember the algebra for solving x than y though, that I had to learn from the subreddit. In the code, this is the ratio part. Also, this solution using fractions is SICK. https://www.reddit.com/r/adventofcode/comments/1hd4wda/comment/m1tz3nf/?utm_source=share&utm_medium=web3x&utm_name=web3xcss&utm_term=1&utm_content=share_button
36 lines
935 B
Python
36 lines
935 B
Python
from output import ints
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def solve(data):
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offset = 10_000_000_000_000
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machines = data.split("\n\n")
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p1 = ratio(machines)
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p2 = ratio(machines, offset)
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return p1, p2
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def ratio(machines, offset=0):
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cost = 0
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for mid, config in enumerate(machines):
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a_x, a_y, b_x, b_y, goal_x, goal_y = ints(config)
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goal_x, goal_y = goal_x + offset, goal_y + offset
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ratio = (goal_x * a_y - goal_y * a_x) / (goal_y * b_x - goal_x * b_y)
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a_presses = round(goal_x / (b_x * ratio + a_x))
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b_presses = round(a_presses * ratio)
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a_presses, b_presses = int(a_presses), int(b_presses)
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if (a_x + a_y) * a_presses + (b_x + b_y) * b_presses == goal_x + goal_y:
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cost += 3 * a_presses + b_presses
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return cost
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if __name__ == "__main__":
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with open("./input/13.txt", "r") as f:
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inp = f.read().strip()
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p1, p2 = solve(inp)
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print(p1)
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print(p2)
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