118 lines
3.3 KiB
Python
118 lines
3.3 KiB
Python
from lib import get_data
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def part_1_with_numpy(data):
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import numpy as np
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def make_base_matrix(pattern, n):
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xss = []
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for round in range(n):
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xs = [pattern[((i + 1) // (round + 1)) % len(pattern)] for i in range(n)]
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xss.append(xs)
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return np.array(xss)
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pattern = [0, 1, 0, -1]
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input = int(data.strip())
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v = np.array(list(map(int, str(input))))
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m = make_base_matrix(pattern, len(v))
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func = np.vectorize(lambda x: abs(x) % 10)
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for _ in range(100):
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v = func(np.dot(m, v))
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print("".join(map(str, v[:8].tolist())))
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def phase(digits_in):
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pattern = [0, 1, 0, -1]
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digits_out = []
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for round in range(len(digits_in)):
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i, out = 0, 0
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while i < len(digits_in):
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pattern_i = ((i + 1) // (round + 1)) % len(pattern)
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out += pattern[pattern_i] * digits_in[i]
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i += 1
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out = abs(out) % 10
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digits_out.append(out)
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return digits_out
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def phase_with_offset(digits_in, pattern, offset):
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digits_out = digits_in.copy()
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for round in range(offset, len(digits_in)):
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i, out = 0, 0
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print(round)
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pattern_value = pattern[((i + 1) // (round + 1)) % len(pattern)]
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if pattern_value == 0:
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i += round
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while i < len(digits_in):
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pattern_i = ((i + 1) // (round + 1)) % len(pattern)
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pattern_value = pattern[pattern_i]
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if pattern_value != 0:
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out += pattern_value * sum(
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digits_in[i : min(i + 1 + round, len(digits_in))]
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)
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# for j in range(i, min(i + 1 + round, len(digits_in))):
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# out += (pattern_value * digits_in[j])
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i += round + 1
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out = abs(out) % 10
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digits_out[round] = out
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return digits_out
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def part_1(data):
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pattern = [0, 1, 0, -1]
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input = list(map(int, (data.strip())))
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for _ in range(100):
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input = phase_with_offset(input, pattern, 0)
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print("".join(map(str, input[:8])))
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input = list(map(int, (data.strip()))) * 10_000
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offset = int("".join(map(str, input[:7]))) - 200
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for i in range(100):
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print(i)
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input = phase_with_offset(input, pattern, offset)
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print(input)
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return
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# What do I know?
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#
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# 1. There is a solution. Other people have solved it.
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# 2. The solution is not crazy. It will be rather obvious.
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# 3. 6_500_000 * 6_500_000 is definitely too much to brute force.
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# 4. Can we go from O(N^2) to O(N) somehow? Yes, that's what we have to do.
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# The whole point of FFT is to get from O(N^2) to O(N*log(N)). Now,
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# how exactly do we do that?
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#
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# Ways to improve performance:
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#
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# 1. Speed up `phase` significantly. Yes, but how?
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# 2. Only compute a subset of the lists? - No!
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# 3. Discover some kind of pattern? - No!
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#
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# Assumptions:
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#
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# 1. I need every digit of the previous round. - False!
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# 2. I cannot just operate on a subset. - False!
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#
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# Non-approaches:
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#
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# 1. Fancy recursive algorithm that selectively picks fields.
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# 2. Pattern detection or subset consideration.
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def main():
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data = get_data(__file__)
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# part_1_with_numpy(data)
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part_1(data)
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if __name__ == "__main__":
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main()
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