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63 lines
1.8 KiB
Python
63 lines
1.8 KiB
Python
"""
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Project Euler Problem 115: https://projecteuler.net/problem=115
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NOTE: This is a more difficult version of Problem 114
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(https://projecteuler.net/problem=114).
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A row measuring n units in length has red blocks
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with a minimum length of m units placed on it, such that any two red blocks
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(which are allowed to be different lengths) are separated by at least one black square.
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Let the fill-count function, F(m, n),
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represent the number of ways that a row can be filled.
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For example, F(3, 29) = 673135 and F(3, 30) = 1089155.
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That is, for m = 3, it can be seen that n = 30 is the smallest value
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for which the fill-count function first exceeds one million.
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In the same way, for m = 10, it can be verified that
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F(10, 56) = 880711 and F(10, 57) = 1148904, so n = 57 is the least value
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for which the fill-count function first exceeds one million.
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For m = 50, find the least value of n
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for which the fill-count function first exceeds one million.
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"""
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from itertools import count
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def solution(min_block_length: int = 50) -> int:
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"""
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Returns for given minimum block length the least value of n
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for which the fill-count function first exceeds one million
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>>> solution(3)
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30
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>>> solution(10)
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57
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"""
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fill_count_functions = [1] * min_block_length
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for n in count(min_block_length):
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fill_count_functions.append(1)
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for block_length in range(min_block_length, n + 1):
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for block_start in range(n - block_length):
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fill_count_functions[n] += fill_count_functions[
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n - block_start - block_length - 1
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]
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fill_count_functions[n] += 1
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if fill_count_functions[n] > 1_000_000:
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break
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return n
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if __name__ == "__main__":
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print(f"{solution() = }")
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