Anders Englöf Ytterström
d7d5312786
Turns out the initial trial of using Manhattan distance was required to solve part 2, since the BFS implementation is impossible to scale up from 2 robots to 25. Recursion and memoization makes the execution time acceptable. Line 53 makes all the difference here. The order of `<|v|^|>` matters when constructing a sequence. Many hours was spent trying to find the correct priority. For the example input, especially 456A and 379A were volatile.
67 lines
1.2 KiB
Python
67 lines
1.2 KiB
Python
from functools import cache
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DKP = {
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"^": (0, 1),
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"A": (0, 2),
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"<": (1, 0),
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"v": (1, 1),
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">": (1, 2),
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}
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NKP = {
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"7": (0, 0),
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"8": (0, 1),
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"9": (0, 2),
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"4": (1, 0),
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"5": (1, 1),
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"6": (1, 2),
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"1": (2, 0),
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"2": (2, 1),
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"3": (2, 2),
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"0": (3, 1),
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"A": (3, 2),
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}
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def solve(data):
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codes = data.split()
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p1 = 0
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p2 = 0
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for code in codes:
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num = int(code[:-1])
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p1 += num * unfold(code, 1 + 2, num=True)
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p2 += num * unfold(code, 1 + 25, num=True)
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return p1, p2
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@cache
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def unfold(sequence, iterations, num=False):
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if iterations == 0:
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return len(sequence)
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s = "A"
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total = 0
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pad, invalid = (NKP, (3, 0)) if num else (DKP, (0, 0))
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for d in sequence:
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total += unfold(seq(pad[s], pad[d], invalid), iterations - 1)
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s = d
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return total
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@cache
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def seq(S, E, invalid):
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y1, x1 = S
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y2, x2 = E
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seq = "<" * (x1 - x2) + "v" * (y2 - y1) + "^" * (y1 - y2) + ">" * (x2 - x1)
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if (y2, x1) == invalid or (y1, x2) == invalid:
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seq = seq[::-1]
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return seq + "A"
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if __name__ == "__main__":
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with open("./input/21.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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